Which has the greater density, 5 g of mercury or 4000 g of water?

A. 5 g of mercury
B. 4000 g of water

A. 5 g of mercury

Object 1 has an irregular shape. Its density is 4000 kg/m^3 .

Object 2 has the same shape and dimensions as object 1, but it is twice as massive. What is the density of object 2?

8000 kg/m^3

Object 1 has an irregular shape. Its density is 4000 kg/m^3 .

Object 3 has the same mass and the same shape as object 1, but its size in all three dimensions is twice that of object 1. What is the density of object 3?

500 kg/m^3

To explore the bottom of a 10-m-deep lake, your friend Tom proposes to get a long garden hose, put one end on land and the other in his mouth for breathing underwater, and descend into the depths. Susan, who overhears the conversation, reacts with horror and warns Tom that he will not be able to inhale when he is at the lake bottom.

Why is Susan so worried?

A. The pressure at a depth of 10 m is 3.5×10^5Pa. This is almost 3.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

B. The pressure at a depth of 10 m is 2.5×10^5Pa. This is almost 2.5 times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

C. The pressure at a depth of 10 m is 3×10^5Pa. This is almost three times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

D. The pressure at a depth of 10 m is 2×10^5Pa. This is almost two times greater than atmospheric pressure. But he air pressure in the hose will be only slightly higher than atmospheric pressure, because the density of air is so low. So Tom will have great difficulty breathing in the low-pressure air when the large pressure of the water is pressing in on his chest.

D. The pressure at a depth of 10 m is 2×10^5Pa.

Rank in order, from largest to smallest, the pressures at A, B, and C in (Figure 1).

A=B=C

Refer to (Figure 1). Rank in order, from largest to smallest, the pressures at D, E, and F.

D>F>E

A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pressure and one that reads gauge pressure. The tank is then brought to the top of a mountain.

For the gauge that reads absolute pressure in the tank, does the pressure reading decrease, increase, or remain the same?
A. The pressure reading decreases.
B. The pressure reading increases.
C. The pressure reading remains the same.
D. More information is needed to answer.

C. The pressure reading remains the same.

A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pressure and one that reads gauge pressure. The tank is then brought to the top of a mountain.

For the gauge that reads gauge pressure, does the pressure reading decrease, increase, or remain the same?

A. The pressure reading decreases.
B. The pressure reading increases.
C. The pressure reading remains the same.
D. More information is needed to answer.

B. The pressure reading increases.

A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pressure and one that reads gauge pressure. The tank is then brought to the top of a mountain.

Does the difference between the two readings decrease, increase, or remain the same?

A. The difference decreases.
B. The difference increases.
C. The difference remains the same.
D. More information is needed to answer.

A. The difference decreases.

Imagine a square column of the atmosphere, 1 mm on a side, that extends all the way to the top of the atmosphere.

How much does the column of air weigh in newtons?

0.1 N

Imagine a square column of the atmosphere, 10 m on a side, that extends all the way to the top of the atmosphere.

How much does this column of air weigh in newtons?

10,000,000 N

In (Figure 1), is pA larger, smaller, or equal to pB? Choose the correct explanation.
A. Two points on a horizontal line, connected by a single liquid in hydrostatic equilibrium, are at the same pressure. pA=pB.
B. Two points on any line, connected by a single liquid in hydrostatic equilibrium, are at the same pressure. pA=pB.
C. Point A has to support the weight of the ship, while point B does not. pA>pB.
D. The ship is lighter that the water since it doesn't go down. Point A has to support the weight of the ship, while point B has to support the water. pA<pB.

A. Two points on a horizontal line, connected by a single liquid in hydrostatic equilibrium, are at the same pressure. pA=pB.

A beaker of water rests on a scale. A metal ball is then lowered into the beaker using a string tied to the ball. The ball doesn't touch the sides or bottom of the beaker, and no water spills from the beaker. Does the scale reading decrease, increase, or stay the same?
A. The scale reading decreases. The fluid exerts an upward force on the ball. Thus, by Newton's third law, the ball exerts a downward force on the fluid. This downward force decreases the scale reading.
B. The scale reading increases. The fluid exerts an upward force on the ball. Thus, by Newton's third law, the ball exerts a downward force on the fluid. This downward force raises the scale reading.
C. The scale reading stays the same. The fluid exerts an upward force on the ball which is equal to the ball's weight. There is no force to change the scale reading.
D. The scale reading stays the same. The tension force of the string always balances the ball, so the ball doesn't affect the scale reading.

B. The scale reading increases

Objects A, B, and C in (Figure 1) have the same volume. Rank in order, from largest to smallest, the sizes of the buoyant forces FA, FB, and FC on A, B, and C.

A=B=C

A heavy lead block and a light aluminum block of equal size sit at rest at the bottom of a pool of water. Is the buoyant force on the lead block greater than, less than, or equal to the buoyant force on the aluminum block? Choose the best explanation.
A. For objects that are completely submerged the buoyant force is proportional to the mass of the object. The buoyant force on the lead block is greater than the buoyant force on the aluminum block.
B. For objects that are completely submerged the buoyant force is inversely proportional to the density of the object. Since the two blocks are the same size (volume), then the buoyant force on the lead block is less than the buoyant force on the aluminum block.
C. For objects that are completely submerged the buoyant force is proportional to the volume of the object. Since the two blocks are the same size (volume), then the buoyant force is the same on both blocks.
D. For objects that are completely submerged the buoyant force is proportional to the density of the object. Since the two blocks are the same size (volume), then the buoyant force on the lead block is greater than the buoyant force on the aluminum block.

C. For objects that are completely submerged the buoyant force is proportional to the volume of the object. Since the two blocks are the same size (volume), then the buoyant force is the same on both blocks.

When you place an egg in water, it sinks. If you add salt to the water, after some time the egg floats. Choose the correct explanation.
A. Adding salt to the water decreases its density. When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats.
B. Adding salt to the water increases its volume. When the volume of the water matches that of the egg, the egg becomes neutrally buoyant and floats.
C. Adding salt to the water decreases its volume. When the volume of the water matches that of the egg, the egg becomes neutrally buoyant and floats.
D. Adding salt to the water increases its density. When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats.

D. Adding salt to the water increases its density. When the density of the water matches that of the egg, the egg becomes neutrally buoyant and floats.

The water of the Dead Sea is extremely salty, which gives it a very high density of 1240 kg/m^3. Explain why a person floats much higher in the Dead Sea than in ordinary water as shown in (Figure 1).
Adding salt to water _______________ its density.
How high a person floats in the water depends on the ratio of the density of the person to the density of the water. The density of the person _______________.
With denser water, however, the person would ____________________ in the water.

increases, doesn't change, float higher

Fish can adjust their buoyancy with an organ called the swim bladder. The swim bladder is a flexible gas-filled sac; the fish can increase or decrease the amount of gas in the swim bladder so that it stays neutrally buoyant-neither sinking nor floating. Suppose the fish is neutrally buoyant at some depth and then goes deeper.

What would happen to the volume of air in the swim bladder if the fish didn't change the amount of air in it?
A. The volume of air would decrease because of the decreasing pressure.
B. The volume of air would decrease because of the increasing pressure.
C. The volume of air would increase because of the decreasing pressure.
D. The volume of air would increase because of the increasing pressure.

B. The volume of air would decrease because of the increasing pressure.

Fish can adjust their buoyancy with an organ called the swim bladder. The swim bladder is a flexible gas-filled sac; the fish can increase or decrease the amount of gas in the swim bladder so that it stays neutrally buoyant-neither sinking nor floating. Suppose the fish is neutrally buoyant at some depth and then goes deeper.

Will the fish need to add or remove gas from the swim bladder to maintain its neutral buoyancy?
A. The fish will need to add gas, otherwise the fish will go deeper with decreasing speed.
B. The fish will need to remove gas, otherwise the fish will go deeper with decreasing speed.
C. The fish will need to remove gas, otherwise the fish will go deeper with increasing speed.
D. The fish will need to add gas, otherwise the fish will go deeper with increasing speed.

D. The fish will need to add gas, otherwise the fish will go deeper with increasing speed.

Is it possible for a fluid in a tube to flow in the direction from low pressure to high pressure?
A. No. According to Pascal's principle, the pressure is the same at all points of the fluid.
B. Yes. If a horizontal pipe has a constant diameter, the fluid will flow at constant speed in it, for example, from left to right. According to Bernoulli's principle, the pressure in the right part of the fluid will be higher than in the left part.
C. Yes. If a horizontal pipe has an increase in its diameter, the fluid will flow slower in this wider part. According to Bernoulli's principle, the pressure in this slow-moving part of the fluid will be higher than in the upstream, fast-moving part.
D. No. According to Bernoulli's principle, the fluid always flows from high pressure to low pressure.

C. Yes. If a horizontal pipe has an increase in its diameter, the fluid will flow slower in this wider part. According to Bernoulli's principle, the pressure in this slow-moving part of the fluid will be higher than in the upstream, fast-moving part.

Wind blows over a house as shown in (Figure 1). A window on the ground floor is open.

Is there an air flow through the house? If so, does the air flow in the window and out the chimney, or in the chimney and out the window? Choose the correct explanations.
A. The pressure is increased at the chimney due to the movement of the wind above. Thus, the air will flow in the chimney and out the window.
B. The pressure is reduced at the chimney due to the movement of the wind above. Thus, the air will flow in the window and out the chimney.
C. The pressure is increased at the chimney due to the movement of the wind above. Thus, the air will flow in the window and out the chimney.
D. The pressure is reduced at the chimney due to the movement of the wind above. Thus, the air will flow in the chimney and out the window.
E. The pressures at the chimney and the window are the same. Thus, the air will not flow.B

B. The pressure is reduced at the chimney due to the movement of the wind above. Thus, the air will flow in the window and out the chimney.

(Figure 1) shows a 100 g block of copper (ρ = 8900 kg/m^3 ) and a 100 g block of aluminum (ρ = 2700 kg/m^3 ) connected by a massless string that runs over two massless, frictionless pulleys. The two blocks exactly balance, since they have the same mass. Now suppose that the whole system is submerged in water. What will happen?
A. The copper block will fall, the aluminum block will rise.
B. The aluminum block will fall, the copper block will rise.
C. Nothing will change.
D. Both blocks will rise.A

A. The copper block will fall, the aluminum block will rise.

Masses A and B rest on very light pistons that enclose a fluid, as shown in (Figure 1). There is no friction between the pistons and the cylinders they fit inside. Which mass is greater?
A. Mass A is greater.
B. Mass B is greater.
C. Mass A and mass B are the same.
D. The answer depends on the density of the fluid.

B. Mass B is greater.

If you dive under-water, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm2 (7×10^−5m^2), and it can sustain a force of about 7 N without rupturing.
If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum?
A. 0.3 m
B. 1 m
C. 3 m
D. 10 m

D. 10 m

An 9.0 lb bowling ball has a diameter of 8.5 inches.
When lowered into water, this ball will
A. have neutral buoyancy.
B. float.
C. sink.

B. float

A large beaker of water is filled to its rim with water. A block of wood is then carefully lowered into the beaker until the block is floating. In this process, some water is pushed over the edge and collects in a tray. The weight of the water in the tray is
A. equal to the weight of the block.
B. greater than the weight of the block.
C. less than the weight of the block.

A. equal to the weight of the block.

An object floats in water, with 70 % of its volume submerged.

What is its approximate density?

A. 300 kg/ m^3

B. 700 kg/ m^3

C. 1000 kg/m^3

D. 1300 kg/m^3

B. 700 kg/ m^3

An object floats in water, with 80 % of its volume submerged.
What is its approximate density?
A. 200 kg/m^3
B. 800 kg/m^3
C. 1000 kg/m^3
D. 1200 kg/m^3

B. 800 kg/m^3

Suppose a syringe is being used to squirt water as shown in (Figure 1). The water is ejected from the nozzle at 10 m/s .
At what speed is the plunger of the syringe being depressed?
A. 0.01 m/s
B. 0.1 m/s
C. 1.0 m/s
D. 10 m/s

B. 0.1 m/s

Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm.
Ignoring viscosity, what is the pressure difference between the wide and narrow sections of the pipe?
A. 850 Pa
B. 3400 Pa
C. 9300 Pa
D. 12700 Pa
E. 13500 Pa

D. 12700 Pa

A 15-m-long garden hose has an inner diameter of 2.5 cm. One end is connected to a spigot; 20∘C waters flow from the other end at a rate of 1.2 L/s.
What is the gauge pressure at the spigot end of the hose?
A. 1900 Pa
B. 2700 Pa
C. 4200 Pa
D. 5800 Pa
E. 7300 Pa

A. 1900 Pa

Which has more mass, a mole of CO gas or a mole of O2 gas?
A. CO
B. O2

B. O2

A tire is inflated to a gauge pressure of 35 psi . The absolute pressure in the tire is
A. Less than 35 psi .
B. Equal to 35 psi .
C. Greater than 35 psi .

C. Greater than 35 psi

The number of atoms in a container is increased by a factor of 3 while the temperature is held constant.
What happens to the pressure?
A. The pressure decreases by a factor of 6.
B. The pressure decreases by a factor of 3.
C. The pressure stays the same.
D. The pressure increases by a factor of 3.
E. The pressure increases by a factor of 6.

D. The pressure increases by a factor of 3.

The number of atoms in a container is increased by a factor of 5 while the temperature is held constant.

What happens to the pressure?

A. The pressure decreases by a factor of 10.

B. The pressure decreases by a factor of 5.

C. The pressure stays the same.

D. The pressure increases by a factor of 5.

E. The pressure increases by a factor of 10.

D. The pressure increases by a factor of 5.

A gas is compressed by an isothermal process that decreases its volume by a factor of 2. In this process, the pressure
A. increases by a factor of less than 2.
B. increases by a factor of more than 2.
C. does not change.
D. increases by a factor of 2.

D. increases by a factor of 2

The thermal energy of a container of helium gas is halved.
What happens to the temperature, in kelvin?
A. It decreases to one-fourth its initial value.
B. It decreases to one-half its initial value.
C. It stays the same.
D. It increases to twice its initial value.

B. It decreases to one-half its initial value.

A gas is compressed by an adiabatic process that decreases its volume by a factor of 2.
In this process, the pressure

A. increases by a factor of more than 2.
B. does not change.
C. increases by a factor of 2.
D. increases by a factor of less than 2.

A. increases by a factor of more than 2.

Suppose you do a calorimetry experiment to measure the specific heat of a penny. You take a number of pennies, measure their mass, heat them to a known temperature, and then drop them into a container of water at a known temperature. You then deduce the specific heat of a penny by measuring the temperature change of the water. Unfortunately, you didn't realize that you dropped one penny on the floor while transferring them to the water. How will this affect your calculation of the penny's heat?
A. This will cause you to underestimate the specific heat.
B. This will cause you to overestimate the specific heat.
C. This will not affect your calculation of specific heat.A

A. This will cause you to underestimate the specific heat.

A cup of water is heated with a heating coil that delivers 100 W of heat. In one minute, the temperature of the water rises by 23 ∘C. What is the mass of the water?
A. 62 g
B. 120 g
C. 620 g
D. 1.2 kg

A. 62 g

A cup of water is heated with a heating coil that delivers 100 W of heat. In one minute, the temperature of the water rises by 20 ∘C. What is the mass of the water?

A. 72 g

B. 140 g

C. 720 g

D. 1.4 kg

A. 72 g

Three identical beakers each hold 1000 g of water at 20 ∘C. 100 g of liquid water at 0∘C is added to the first beaker, 100 g of ice at 0 ∘C is added to the second beaker, and the third beaker gets 100 g of aluminum at 0∘C. The contents of which container end up at the lowest final temperature?
A. The first beaker.
B. The second beaker.
C. The third beaker.
D. All end up at the same temperature.

B. The Second beaker

100 g of ice at 0∘C and 100 g of steam at 100∘C interact thermally in a well-insulated container.
What is the final state of the system?
A. An ice-water mixture at 0∘C
B. Water at a temperature between 0∘C and 50∘C.
C. Water at 50∘C.
D. Water at a temperature between 50∘C and 100∘C.
E. A water-steam mixture at 100∘C.

E. A water-steam mixture at 100∘C.

Suppose the 650 W of radiation emitted in a microwave oven is absorbed by 320 g of water in a very lightweight cup.
Approximately how long will it take to heat the water from 20 ∘C to 50 ∘C?
A. 31 s
B. 62 s
C. 93 s
D. 120 sB

B. 62 s

Suppose the 600 W of radiation emitted in a microwave oven is absorbed by 330 g of water in a very lightweight cup. Approximately how long will it take to heat the water from 20 ∘C to 50 ∘C?

A. 35 s

B. 69 s

C. 100 s

D. 140 s

B. 69 s

40000 J of heat is added to 1.00 kg of ice at -10 ∘C. How much ice melts?
A. 8.50×10^−3 kg
B. 1.42×10^−2 kg
C. 5.74×10^−2 kg
D. 0.120 kg

C. 5.74×10^−2 kg

Steam at 100∘C causes worse burns than liquid water at 100∘C. This is because
A. The steam is hotter than the water.
B. Heat is transferred to the skin as steam condenses.
C. Steam has a higher specific heat than water.
D. Evaporation of liquid water on the skin causes cooling.

B. Heat is transferred to the skin as steam condenses.

Seasonal temperature changes in the ocean only affect the top layer of water, to a depth of 500 m or so. This "mixed" layer is thermally isolated from the cold, deep water below. The average temperature of this top layer of the world's oceans, which has area 3.6×108 km2 is approximately 17∘C.
In addition to seasonal temperature changes, the oceans have experienced an overall warming trend over the last century that is expected to continue as the earth's climate changes. A warmer ocean means a larger volume of water; the oceans will rise. Suppose the average temperature of the top layer of the world's oceans were to increase from a temperature Ti to a temperature Tf. The area of the oceans will not change, as this is fixed by the size of the ocean basin, so any thermal expansion of the water will cause the water level to rise, as shown in the figure. The original volume is the product of the original depth and the surface area, Vi = Adi. The change in volume is given by ΔV = A Δd. (Figure 1)

If the top 500 m of ocean water increased in temperature from 17 ∘C to 18 ∘C, what would be the resulting rise in ocean height?
A. 0.22 m
B. 0.11 m
C. 0.88 m
D. 0.44 m

B. 0.11 m

Seasonal temperature changes in the ocean only affect the top layer of water, to a depth of 500 m or so. This "mixed" layer is thermally isolated from the cold, deep water below. The average temperature of this top layer of the world's oceans, which has area 3.6×10^8 km2 is approximately 17∘C.
In addition to seasonal temperature changes, the oceans have experienced an overall warming trend over the last century that is expected to continue as the earth's climate changes. A warmer ocean means a larger volume of water; the oceans will rise. Suppose the average temperature of the top layer of the world's oceans were to increase from a temperature Ti to a temperature Tf. The area of the oceans will not change, as this is fixed by the size of the ocean basin, so any thermal expansion of the water will cause the water level to rise, as shown in the figure. The original volume is the product of the original depth and the surface area, Vi = Adi. The change in volume is given by ΔV = A Δd. (Figure 1)

Approximately how much energy would be required to raise the temperature of the top layer of the oceans by 1∘C? (1 m3 of water has a mass of 1000 kg.)
A. 1×10^24 J
B. 1×10^15 J
C. 1×10^21 J
D. 1×10^18 J

A. 1×10^24 J

Seasonal temperature changes in the ocean only affect the top layer of water, to a depth of 500 m or so. This "mixed" layer is thermally isolated from the cold, deep water below. The average temperature of this top layer of the world's oceans, which has area 3.6×108 km2 is approximately 17∘C.
In addition to seasonal temperature changes, the oceans have experienced an overall warming trend over the last century that is expected to continue as the earth's climate changes. A warmer ocean means a larger volume of water; the oceans will rise. Suppose the average temperature of the top layer of the world's oceans were to increase from a temperature Ti to a temperature Tf. The area of the oceans will not change, as this is fixed by the size of the ocean basin, so any thermal expansion of the water will cause the water level to rise, as shown in the figure. The original volume is the product of the original depth and the surface area, Vi = Adi. The change in volume is given by ΔV = A Δd. (Figure 1)

Water's coefficient of expansion varies with temperature (Figure 2). For water at 2∘C, an increase in temperature of 1∘C would cause the volume to
A. decrease.
B. stay the same.
C. increase.

A. decrease.

Seasonal temperature changes in the ocean only affect the top layer of water, to a depth of 500 m or so. This "mixed" layer is thermally isolated from the cold, deep water below. The average temperature of this top layer of the world's oceans, which has area 3.6×108 km2 is approximately 17∘C.
In addition to seasonal temperature changes, the oceans have experienced an overall warming trend over the last century that is expected to continue as the earth's climate changes. A warmer ocean means a larger volume of water; the oceans will rise. Suppose the average temperature of the top layer of the world's oceans were to increase from a temperature Ti to a temperature Tf. The area of the oceans will not change, as this is fixed by the size of the ocean basin, so any thermal expansion of the water will cause the water level to rise, as shown in the figure. The original volume is the product of the original depth and the surface area, Vi = Adi. The change in volume is given by ΔV = A Δd. (Figure 1)

The ocean is mostly heated from the top, by light from the sun. The warmer surface water doesn't mix much with the colder deep ocean water. This lack of mixing can be ascribed to a lack of
A. radiation.
B. conduction.
C. evaporation.
D. convection.

D. convection

The positive charge in (Figure 1) is +Q. What is the negative charge if the electric field at the dot is zero?

-4Q

You are given two metal spheres on portable insulating stands, a glass rod, and a piece of silk. How to give the spheres exactly equal but opposite charges?
A. Put the two metal spheres in contact with each other. Then rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Touch one of the spheres with the glass rod then separate the spheres. Touch another sphere with the piece of silk.
B. Make sure spheres are not in contact with each other. Rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Bring the rod near, but not touching, one of the spheres. Do the same with piece of silk and another sphere. Remove the galss rod and the piece of silk.
C. Put the two metal spheres in contact with each other. Then rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Bring the rod near, but not touching, one of the spheres. Separate the spheres and remove the rod.
D. Make sure spheres are not in contact with each other. Rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Touch one of the spheres with the glass rod then touch another one with the piece of silk.

C. Put the two metal spheres in contact with each other. Then rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Bring the rod near, but not touching, one of the spheres. Separate the spheres and remove the rod.

A metal rod A and a metal sphere B, on insulating stands, touch each other as shown in (Figure 1).
They are originally neutral. A positively charged rod is brought near (but not touching) the far end of A. While the charged rod is still close, A and B are separated. The charged rod is then withdrawn. Is the sphere then positively charged, negatively charged, or neutral?
A. neutral
B. positively charged
C. Negatively charged

B. positively charged

A 10 nC charge sits at a point in space where the magnitude of the electric field is 1900 N/C . What will the magnitude of the field be if the 10 nC charge is replaced by a 20 nC charge? Assume the system is big enough to consider the charges as small test charges.

1900 N/C

A 10 nC charge sits at a point in space where the magnitude of the electric field is 1400 N/C . What will the magnitude of the field be if the 10 nC charge is replaced by a 20 nC charge? Assume the system is big enough to consider the charges as small test charges.

1400 N/C

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. Suppose the distance between the plates is doubled, with the charged particle remaining in the center. Does the force on this particle increase, decrease, or stay the same? Explain.
If we assume the size of the parallel plates is large compared to the separation, then doubling the plate separation will ____________________________ the field inside the capacitor.
That means this transformation will _______________________________ the force.

not change; not change

Two charged particles are separated by 10 cm. Suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?

4

Two charged particles are separated by 10 cm. Suppose the charge on each particle is quadrupled. By what factor does the electric force between the particles change?

16

A small positive charge q experiences a force of magnitude F1 when placed at point 1 in (Figure 1). What is the magnitude of the force on charge q at point 3?

3 F

A small positive charge q experiences a force of magnitude F1 when placed at point 1 in (Figure 1). What is the magnitude of the force on a charge 3q at point 1?

3 F

A small positive charge q experiences a force of magnitude F1 when placed at point 1 in (Figure 1). What is the magnitude of the force on a charge 2q at point 2?

4 F

A small positive charge q experiences a force of magnitude F1 when placed at point 1 in (Figure 1). What is the magnitude of the force on a charge −2q at point 2?

4 F

Is there a point between a 10 nC charge and a 20 nC charge at which the electric field is zero?

If so, which charge is this point closer to?
A. 10 nC
B. 20 nC

yes; A. 10 nC

Is there a point between a 10 nC charge and a -20 nC charge at which the electric field is zero? (Note the change of sign of the second charge.)

no

Two lightweight, electrically neutral conducting balls hang from threads. Note that parts A through D are independent; these are not actions taken in sequence. Choose the diagram in (Figure 1) that shows how the balls hang after both are touched by a negatively charged rod.
A. A
B. B
C. C
D. D
E. E

B

Two lightweight, electrically neutral conducting balls hang from threads. Note that parts A through D are independent; these are not actions taken in sequence. Choose the diagram that shows how the balls hang after ball 1 is touched by a negatively charged rod, and ball 2 is touched by a positively charged rod.
A. A
B. B
C. C
D. D
E. E

C

Two lightweight, electrically neutral conducting balls hang from threads. Note that parts A through D are independent; these are not actions taken in sequence.
Choose the diagram that shows how the balls hang after both are touched by a negatively charged rod, but ball 2 picks up more charge than ball 1.
A. A
B. B
C. C
D. D
E. E

B

Two lightweight, electrically neutral conducting balls hang from threads. Note that parts A through D are independent; these are not actions taken in sequence.
Choose the diagram that shows how the balls hang after only ball 1 is touched by a negatively charged rod.
A. A
B. B
C. C
D. D
E. E

C

All the charges in (Figure 1) have the same magnitude.
In which case does the electric field at the dot have the largest magnitude?
A. A
B. B
C. C
D. D

A

All the charges in (Figure 1) have the same magnitude.
In which case does the electric field at the dot have the largest magnitude?
A. A
B. B
C. C
D. D

B

All the charges in (Figure 1) have the same magnitude. In which case does the electric field at the dot have the largest magnitude?

A. A
B. B
C. C
D. D

D

A glass bead charged to + 3.1 nC exerts an 9.0×10^−4N repulsive electric force on a plastic bead 2.6 cm away. What is the charge on the plastic bead?

A. + 2.2 nC

B. + 7.7 nC

C. + 22 nC

D. + 770 nC

C. + 22 nC

A glass bead charged to + 3.1 nC exerts an 8.0×10−4N repulsive electric force on a plastic bead 2.9 cm away.

What is the charge on the plastic bead?

A. + 2.4 nC

B. + 8.7 nC

C. + 24 nC

D. + 870 nC

C. +24 nC

A + 7.6 nC point charge and a - 2.9 nC point charge are 3.5 cm apart. What is the electric field strength at the midpoint between the two charges?

A. 1.5×10⁵ N/C

B. 2100 N/C

C. 6200 N/C

D. 3.1×10⁵ N/C

D. 3.1×10⁵ N/C

A + 7.4 nC point charge and a - 2.7 nC point charge are 3.5 cm apart. What is the electric field strength at the midpoint between the two charges?

A. 1.5×10⁵ N/C

B. 2000 N/C

C. 5900 N/C

D. 3.0×10⁵ N/C

D. 3.0×10⁵ N/C

Three point charges are arranged as shown in (Figure 1).

Which arrow best represents the direction of the electric field vector at the position of the dot?

A

B

C

D

E

A

A positive charge is brought near to a dipole, as shown in (Figure 1). If the dipole is free to rotate, it

A. begins to rotate in a clockwise direction.

B. remains stationary.

C. begins to rotate in a counterclockwise direction.

A. begins to rotate in a clockwise direction.

Flow cytometry, illustrated in (Figure 1), is a technique used to sort cells by type. The cells are placed in a saline solution, which is a conductor, then forced from a nozzle. The stream soon breaks up into small droplets, each containing one cell. A metal collar surrounds the stream right at the point where the droplets separate from the stream. Charging the collar polarizes the conducting liquid, causing the droplets to become charged as they break off from the stream. A laser beam probes the solution just upstream from the charging collar, looking for the presence of certain types of cells. All droplets containing one particular type of cell are given the same charge by the charging collar. Droplets with other desired types of cells receive a different charge, and droplets with no desired cell receive no charge. The charged droplets then pass between two parallel charged electrodes where they receive a horizontal force that directs them into different collection tubes, depending on their charge.

If the charging collar has a positive charge, what will be the net charge on a droplet separating from the stream?

A. Positive

B. Negative

C. Neutral

D. The charge will depend on the type of cell

B. Negative

Flow cytometry, illustrated in (Figure 1), is a technique used to sort cells by type. The cells are placed in a saline solution, which is a conductor, then forced from a nozzle. The stream soon breaks up into small droplets, each containing one cell. A metal collar surrounds the stream right at the point where the droplets separate from the stream. Charging the collar polarizes the conducting liquid, causing the droplets to become charged as they break off from the stream. A laser beam probes the solution just upstream from the charging collar, looking for the presence of certain types of cells. All droplets containing one particular type of cell are given the same charge by the charging collar. Droplets with other desired types of cells receive a different charge, and droplets with no desired cell receive no charge. The charged droplets then pass between two parallel charged electrodes where they receive a horizontal force that directs them into different collection tubes, depending on their charge.

Which of the following describes the charges on the droplets that end up in the five tubes, moving from left to right?

A. +2q,+q,0,−q,−2q

B. +q,+2q,0,−2q,−q

C. −q,−2q,0,+2q,+q

D. −2q,−q,0,+q,+2q

D. −2q,−q,0,+q,+2q

Flow cytometry, illustrated in (Figure 1) , is a technique used to sort cells by type. The cells are placed in a conducting saline solution which is then forced from a nozzle. The stream breaks up into small droplets, each containing one cell. A metal collar surrounds the stream right at the point where the droplets separate from the stream. Charging the collar polarizes the

conducting liquid, causing the droplets to be-come charged as they break off from the stream. A laser beam probes the solution just upstream from the charging collar, looking for the presence of certain types of cells. All droplets containing one particular type of cell are given the same charge by the charging collar. Droplets with other desired types of cells receive a different charge, and droplets with no desired cell receive no charge. The charged droplets then pass between two parallel charged electrodes where they receive a horizontal force that directs them into different collection tubes, depending on their charge.

Because the droplets are conductors, a droplet's positive and negative charges will separate while the droplet is in the region between the deflection plates. Suppose a neutral droplet passes between the plates. The droplet's dipole moment will point

A. down.

B. left.

C. up.

D. right.

D. right.

Flow cytometry, illustrated in (Figure 1) , is a technique used to sort cells by type. The cells are placed in a conducting saline solution which is then forced from a nozzle. The stream breaks up into small droplets, each containing one cell. A metal collar surrounds the stream right at the point where the droplets separate from the stream. Charging the collar polarizes the

conducting liquid, causing the droplets to be-come charged as they break off from the stream. A laser beam probes the solution just upstream from the charging collar, looking for the presence of certain types of cells. All droplets containing one particular type of cell are given the same charge by the charging collar. Droplets with other desired types of cells receive a different charge, and droplets with no desired cell receive no charge. The charged droplets then pass between two parallel charged electrodes where they receive a horizontal force that directs them into different collection tubes, depending on their charge.

Another way to sort the droplets would be to give each droplet the same charge, then vary the electric field between the deflection plates. For the apparatus as sketched, this technique will not work because

A. the droplets would all repel each other, and this force would dominate the deflecting force.

B. several droplets are between the plates at one time, and they would all feel the same force.

C. a droplet with a net charge would always experience a net force between the plates.

D. the cells in the solution have net charges that would affect the droplet charge.

B. several droplets are between the plates at one time, and they would all feel the same force.

As shown in (Figure 1), two protons are launched with the same speed from point 1 inside a parallel-plate capacitor. One proton moves along the path from 1 to 2, the other from 1 to 3. Points 2 and 3 are the same distance from the positive plate.

Is ΔU1→2, the change in potential energy along the path 1→2, larger, smaller, or equal to ΔU1→3?

A. ΔU1→2>ΔU1→3

B. ΔU1→2<ΔU1→3

C. ΔU1→2=ΔU1→3

D. It's impossible to determine.

C. ΔU1→2=ΔU1→3

As shown in (Figure 1), two protons are launched with the same speed from point 1 inside a parallel-plate capacitor. One proton moves along the path from 1 to 2, the other from 1 to 3. Points 2 and 3 are the same distance from the positive plate.

Is the proton's speed v2 at point 2 larger than, smaller than, or equal to the proton's speed v3at point 3?

A. v2>v3

B. v2<v3

C.v2=v3

D. It's impossible to determine.

C. v2=v3

Each part of (Figure 1) shows one or more point charges. The charges have equal magnitudes.

For case (a), if a positive charge is moved from position i to position f, does the electric potential energy increase, decrease, or stay the same?

A. Electric potential energy increases.

B. Electric potential energy decreases.

C. Electric potential energy stays the same.

D. It's impossible to determine.

C. Electric potential energy stays the same.

Each part of (Figure 1) shows one or more point charges. The charges have equal magnitudes.

For case (b), if a positive charge is moved from position i to position f, does the electric potential energy increase, decrease, or stay the same?

A. Electric potential energy increases.

B. Electric potential energy decreases.

C. Electric potential energy stays the same.

D. It's impossible to determine.

B. Electric potential energy decreases.

Each part of (Figure 1) shows one or more point charges. The charges have equal magnitudes.

For case (c), if a positive charge is moved from position i to position f, does the electric potential energy increase, decrease, or stay the same?

A. Electric potential energy increases.

B. Electric potential energy decreases.

C. Electric potential energy stays the same.

D. It's impossible to determine.

A. Electric potential energy increases.

Each part of (Figure 1) shows one or more point charges. The charges have equal magnitudes.

For case (d), if a positive charge is moved from position i to position f, does the electric potential energy increase, decrease, or stay the same?

A. Electric potential energy increases.

B. Electric potential energy decreases.

C. Electric potential energy stays the same.

D. It's impossible to determine.

B. Electric potential energy decreases.

Each part of the figure shows three points in the vicinity of two point charges. The charges have equal magnitudes. (Figure 1)

(a) Rank in order, from largest to smallest, the potentials V1, V2, and V3.

V2>V1=V3

Each part of the figure shows three points in the vicinity of two point charges. The charges have equal magnitudes. (Figure 1)

(b) Rank in order, from largest to smallest, the potentials V1, V2, and V3.

V1>V2>V3

Each part of the figure shows three points in the vicinity of two point charges. The charges have equal magnitudes. (Figure 1)

(c) Rank in order, from largest to smallest, the potentials V1, V2, and V3.

V2>V1=V3

Each part of the figure shows three points in the vicinity of two point charges. The charges have equal magnitudes. (Figure 1)

(d) Rank in order, from largest to smallest, the potentials V1, V2, and V3.

V1=V2=V3

Rank in order, from largest to smallest, the electric field strengths E1, E2, E3, and E4 at the four labeled points in (Figure 1).

E3>E1>E2=E4

Rank in order, from highest to lowest, the electric potentials at points a, b, and c.

a=c>b

A proton is launched from point 1 in (Figure 1) with an initial velocity of 3.9×10^5 m/s.

By how much has its kinetic energy changed, in eV, by the time it passes through point 2? Assume negative answer if kinetic energy decreases and positive if kinetic energy increases.

-200 eV

A 5.0 nC positive point charge is located at point A in the figure.(Figure 1)

What is the electric potential at point B?

A. 45 V

B. 45 sin30∘V

C. 45 cos30∘V

D. 45 tan30∘V

A. 45 V

V

A 100 V battery is connected across the plates of a parallel-plate capacitor.

If a sheet of Teflon is slid between the plates, without disconnecting the battery, the electric field between the plates

A. increases

B. decreases

C. remains the same

C. remains the same

The electric potential is 300 V at x = 0 cm, is -100 V at x = 5 cm, and varies linearly with x.

If a positive charge is released from rest at x = 2.5 cm, and is subject only to electric forces, what will the charge do?

A. Move to the right.

B. Move to the left.

C. Stay at x = 2.5 cm

D. Not enough information to tell.

A. Move to the right

The figure shows equipotential lines in a region of space. The equipotential lines are spaced by the same difference in potential, and several of the potentials are given.(Figure 1)

What is the potential at point c?

A. -400 V

B. -350 V

C. -100 V

D. 350 V

E. 400 V

A. - 400 V

The figure shows equipotential lines in a region of space. The equipotential lines are spaced by the same difference in potential, and several of the potentials are given.(Figure 1)

At which point, a, b, or c, is the magnitude of the electric field the greatest?

A. a

B. b

C. c

C

The figure shows equipotential lines in a region of space. The equipotential lines are spaced by the same difference in potential, and several of the potentials are given.(Figure 1)

What is the approximate magnitude of the electric field at point c?

A. 100 V/m

B. 300 V/m

C. 800 V/m

D. 1500 V/m

E. 3000 V/m

C. 800 V/m

The figure shows equipotential lines in a region of space. The equipotential lines are spaced by the same difference in potential, and several of the potentials are given.(Figure 1)

The direction of the electric field at point b is closest to which direction?

A. Right

B. Up

C. Left

D. Down

D. Down

A bug zapper consists of two metal plates connected to a high-voltage power supply. The voltage between the plates is set to give an electric field slightly less than 1×10⁶ V/m . When a bug flies between the two plates, it increases the field enough to initiate a spark that incinerates the bug.

If a bug zapper has a 7000 V power supply, what is the approximate separation between the plates?

A. 7×10⁻² cm

B. 0.7 cm

C. 7 cm

D. 70 cm

B. 0.7 cm

An atom of helium and one of argon are singly ionized - one electron is removed from each. The two ions are then accelerated from rest by the electric field between two plates with a potential difference of 150 V.

What happens after the two ions accelerate from one plate to the other?

A. The helium ion has more kinetic energy.

B. The argon ion has more kinetic energy.

C. Both ions have the same kinetic energy.

D. There is not enough information to say which ion has more kinetic energy.

C. Both ions have the same kinetic energy.

The dipole moment of the heart is shown at a particular instant in (Figure 1).

Which of the following potential differences will have the largest positive value?

A. V1−V2

B. V1−V3

C. V2−V1

D. V3−V1

A. V1−V2

The wires in (Figure 1) are all made of the same material; the length and radius of each wire is noted.

Rank in order, from largest to smallest, the resistances R1 to R5 of these wires.

R4>R1=R5>R3>R2

The two circuits in (Figure 1) use identical batteries and wires of equal diameters.

Rank in order, from largest to smallest, the currents I1, I2, I3, and I4 at points 1 to 4.

I1=I2>I3=I4

The two circuits in (Figure 1) use identical batteries and wires of equal diameters.

Rank in order, from largest to smallest, the currents I1 to I7 at points 1 to 7.

I4=I7>I1=I2=I3=I5=I6

A copper wire is stretched so that its length increases and its diameter decreases.

What is the result?

A. The wire's resistance decreases, but its resistivity stays the same.

B. The wire's resistivity decreases, but its resistance stays the same.

C. The wire's resistance increases, but its resistivity stays the same.

D. The wire's resistivity increases, but its resistance stays the same.

C. The wire's resistance increases, but its resistivity stays the same.

The potential difference across a length of wire is increased.

Which of the following does not increase as well?

A. The electric field in the wire.

B. The power dissipated in the wire.

C. The resistance of the wire.

D. The current in the wire.

C. The resistance of the wire.

(Figure 1) shows a side view of a wire of varying circular cross section. Rank in order the currents flowing in the three sections.

A. I1>I2>I3

B. I3>I2>I1

C. I1=I2=I3

D. I1>I3>I2

C. I1=I2=I3

A person gains weight by adding fat - and therefore adding girth - to his body and his limbs, with the amount of muscle remaining constant.

How will this affect the electrical resistance of his limbs?

A. the resistance will stay the same

B. the resistance will increase

C. the resistance will decrease

C. the resistance will decrease

You've probably observed that the most common time for an incandescent lightbulb to fail is the moment when it is turned on. Let's look at the properties of the bulb's filament to see why this happens.
The current in the tungsten filament of a lightbulb heats the filament until it glows. The filament is so hot that some of the atoms on its surface fly off and end up sticking on a cooler part of the bulb. Thus the filament gets progressively thinner as the bulb ages. There will certainly be one spot on the filament that is a bit thinner than elsewhere. This thin segment will have a higher resistance than the surrounding filament. More power will be dissipated at this spot, so it won't only be a thin spot, it also will be a hot spot.
Now, let's look at the resistance of the filament. The graph in (Figure 1) shows data for the current in a lightbulb as a function of the potential difference across it. The graph is not linear, so the filament is not an ohmic material with a constant resistance. However, we can define the resistance at any particular potential difference ΔV to be R=ΔV/I. This ratio, and hence the resistance, increases with ΔV and thus with temperature.
When the bulb is turned on, the filament is cold and its resistance is much lower than during normal, high-temperature operation. The low resistance causes a surge of higher-than-normal current lasting a fraction of a second until the filament heats up. Because power dissipation is I2R, the power dissipated during this first fraction of a second is much larger than the bulb's rated power. This current surge concentrates the power dissipation at the high-resistance thin spot, perhaps melting it and breaking the filament.

For the given bulb, what is the approximate resistance of the bulb at a potential difference of 3.0 V?

A. 18 Ω

B. 27 Ω

C. 36 Ω

D. 9.0 Ω

B. 27 ohms

You've probably observed that the most common time for an incandescent lightbulb to fail is the moment when it is turned on. Let's look at the properties of the bulb's filament to see why this happens.
The current in the tungsten filament of a lightbulb heats the filament until it glows. The filament is so hot that some of the atoms on its surface fly off and end up sticking on a cooler part of the bulb. Thus the filament gets progressively thinner as the bulb ages. There will certainly be one spot on the filament that is a bit thinner than elsewhere. This thin segment will have a higher resistance than the surrounding filament. More power will be dissipated at this spot, so it won't only be a thin spot, it also will be a hot spot.
Now, let's look at the resistance of the filament. The graph in (Figure 1) shows data for the current in a lightbulb as a function of the potential difference across it. The graph is not linear, so the filament is not an ohmic material with a constant resistance. However, we can define the resistance at any particular potential difference ΔV to be R=ΔV/I. This ratio, and hence the resistance, increases with ΔV and thus with temperature.
When the bulb is turned on, the filament is cold and its resistance is much lower than during normal, high-temperature operation. The low resistance causes a surge of higher-than-normal current lasting a fraction of a second until the filament heats up. Because power dissipation is I2R, the power dissipated during this first fraction of a second is much larger than the bulb's rated power. This current surge concentrates the power dissipation at the high-resistance thin spot, perhaps melting it and breaking the filament.

For the given bulb, what is the approximate resistance of the bulb at a potential difference of 4.0 V ?

A. 31 Ω

B. 15 Ω

C. 43 Ω

D. 9.0 Ω

A. 31 ohms

You've probably observed that the most common time for an incandescent lightbulb to fail is the moment when it is turned on. Let's look at the properties of the bulb's filament to see why this happens.
The current in the tungsten filament of a lightbulb heats the filament until it glows. The filament is so hot that some of the atoms on its surface fly off and end up sticking on a cooler part of the bulb. Thus the filament gets progressively thinner as the bulb ages. There will certainly be one spot on the filament that is a bit thinner than elsewhere. This thin segment will have a higher resistance than the surrounding filament. More power will be dissipated at this spot, so it won't only be a thin spot, it also will be a hot spot.
Now, let's look at the resistance of the filament. The graph in (Figure 1) shows data for the current in a lightbulb as a function of the potential difference across it. The graph is not linear, so the filament is not an ohmic material with a constant resistance. However, we can define the resistance at any particular potential difference ΔV to be R=ΔV/I. This ratio, and hence the resistance, increases with ΔV and thus with temperature.
When the bulb is turned on, the filament is cold and its resistance is much lower than during normal, high-temperature operation. The low resistance causes a surge of higher-than-normal current lasting a fraction of a second until the filament heats up. Because power dissipation is I2R, the power dissipated during this first fraction of a second is much larger than the bulb's rated power. This current surge concentrates the power dissipation at the high-resistance thin spot, perhaps melting it and breaking the filament.

As the bulb ages, the resistance of the filament

A. stays the same

B. decreases

C. increases

C. increases.

Which of the curves best represents the expected variation in current as a function of time in the short time interval immediately after the bulb is turned on?

B

There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which of the following strategies would increase the lifetime of a bulb without making it dimmer? Check all that apply.

A. Reducing the average current through the bulb.

B. Increasing the average current through the bulb.

C. Limiting the maximum current through the bulb.

D. Limiting the minimum current through the bulb.

C. Limiting the maximum current through the bulb.

The switch in (Figure 1) is closed. Does the current through the battery increase, decrease, or stay the same?

A. The current increases.

B. The current stays the same.

C. The current decreases.

A. The current increases.

Does the current through R1 increase, decrease, or stay the same? Select the correct answer and explanation.

A. When the switch is closed resistor R1 "sees" the same potential difference, so the current through R1 stays the same.

B. When the switch is closed resistor R1 "sees" less potential difference, so the current through R1decreases.

C. When the switch is closed resistor R1 "sees" greater potential difference, so the current through R1 stays increases.

D. There is not enough information to answer the question.

A. When the switch is closed resistor R1 "sees" the same potential difference, so the current through R1 stays the same.

A voltmeter is (incorrectly) inserted into a circuit as shown in the figure. (Figure 1). What is the current in the circuit?

A. nearly zero

B. 0.9 A

C. 9 A

D. extremely large

A. nearly zero

A voltmeter is (incorrectly) inserted into a circuit as shown in the figure. (Figure 1). What does the voltmeter read?

9 V

An ammeter is (incorrectly) inserted into a circuit as shown in the figure. (Figure 1). What is the current through the 5.0 resistor?

A. nearly zero

B. 0.6 A

C. 0.9 A

D. 1.8 A

A. nearly zero

What is the current in the circuit of the figure? (Figure 1)

A. 1.0 A

B. 1.7 A

C. 2.5 A

D. 4.2 A

A. 1.0 A

Which resistor in the figure dissipates the most power? (Figure 1)

A. The 4.0 Ω resistor

B. The 6.0 Ω resistor

C. Both dissipate the same power.

B. The 6.0 ohm resistor

A metal wire of resistance R is cut into two pieces of equal length. The two pieces are connected together side by side. What is the resistance of the two connected wires?

A. R/4

B. R/2

C. R

D. 2R

E. 4R

A. R/4

What is the value of resistor R in the figure?

A. 4.0 Ω

B. 12 Ω

C. 36 Ω

D. 72 Ω

E. 96 Ω

B. 12 ohms

Two capacitors are connected in series. They are then reconnected to be in parallel.

What is the capacitance of the parallel combination?

A. Less than that of the series combination.

B. More than that of the series combination.

C. The same as that of the series combination.

D. Could be more or less than that of the series combination depending on the values of the capacitances.

B. More than that of the series combination.

A cell’s membrane thickness doubles but the cell stays the same size.

How do the resistance and the capacitance of the cell membrane change?

A. The resistance and the capacitance decrease.

B. The resistance decreases, the capacitance increases.

C. The resistance and the capacitance increase.

D. The resistance increases, the capacitance decreases.

D. The resistance increases, the capacitance decreases

A defibrillator is designed to pass a large current through a patient's torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient's torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient's torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.
A typical defibrillator has a 31 μF capacitor charged to 5000 V . The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient's torso is 100 Ω .

Which pair of graphs in the figure best represents the capacitor voltage and the current through the torso as a function of time after the switch is closed? (Figure 1)

A. A

B. B

C. C

D. D

A

A defibrillator is designed to pass a large current through a patient's torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient's torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient's torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.
A typical defibrillator has a 31 μF capacitor charged to 5000 V . The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient's torso is 100 Ω .

For the values noted in the passage above, what is the time constant for the discharge of the capacitor?

A. 3.1 us

B. 160 us

C. 3.1 ms

D. 160 ms

C. 3.1 ms

A defibrillator is designed to pass a large current through a patient's torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient's torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient's torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.
A typical defibrillator has a 32 μF capacitor charged to 5000 V . The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient's torso is 120 Ω .

For the values noted in the passage above, what is the time constant for the discharge of the capacitor?

A. 3.8 μs

B. 190 μs

C. 3.8 ms

D. 190 ms

C. 3.8 ms

A defibrillator is designed to pass a large current through a patient's torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient's torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient's torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.
A typical defibrillator has a 31 μF capacitor charged to 5000 V . The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient's torso is 100 Ω .

If a patient receives a series of jolts, the resistance of the torso may increase. How does such a change affect the initial current and the time constant of subsequent jolts?

A. The initial current and the time constant both increase.

B. The initial current decreases, the time constant increases.

C. The initial current increases, the time constant decreases.

D. The initial current and the time constant both decrease.

B. The initial current decreases, the time constant increases.

A defibrillator is designed to pass a large current through a patient's torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient's torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient's torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.
A typical defibrillator has a 31 μF capacitor charged to 5000 V . The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient's torso is 100 Ω .

In some cases, the defibrillator may be charged to a lower voltage. How will this affect the time constant of the discharge?

A. The time constant will increase.

B. The time constant will not change.

C. The time constant will decrease.

B. The time constant will not change.

In (Figure 1), suppose the magnet on the right is fixed in place and the magnet on the left is free to pivot about its center.

Will the magnet on the left start to rotate? If so, will it initially rotate clockwise or counterclockwise?

A. It will rotate in the counterclockwise direction.

B. It will rotate in the clockwise direction.

C. It will not rotate.

B. It will rotate in the clockwise direction.

If you were standing directly at the earth's north magnetic pole, in what direction would a compass point if it were free to swivel in any direction?

A. It would point west.

B. It would point down.

C. It would point up.

D. It would point east.

B. It would point down.

If you took a sample of magnetotactic bacteria from the northern hemisphere to the southern hemisphere, would you expect them to survive?

A. In the northern hemisphere, the vertical component of the earth's field points down, and in the southern hemisphere, it points up. So the bacteria are not likely to survive.

B. Both in the northern and in the southern hemispheres, the earth's field points down. So the bacteria are likely to survive.

C. Both in the northern and in the southern hemispheres, the earth's field points up. So the bacteria are likely to survive.

D. In the northern hemisphere, the vertical component of the earth's field points up, and in the southern hemisphere, it points down. So the bacteria are not likely to survive.

A. In the northern hemisphere, the vertical component of the earth's field points down, and in the southern hemisphere, it points up. So the bacteria are not likely to survive.

An electron is moving in a circular orbit in a uniform magnetic field. Is the kinetic energy of the electron changing? Select the correct answer and explanation.

A. Yes because the force is always perpendicular to the velocity.

B. Yes because the force is always parallel to the velocity.

C. No because the force is always parallel to the velocity.

D. No because the force is always perpendicular to the velocity.

D. No because the force is always perpendicular to the velocity.

The (Figure 1) shows a solenoid as seen in cross section. Compasses are placed at points 1, 2 and 3.

In which direction will each compass point when there is a large current in the direction shown?

A. The magnetic field points to the left for compass 1 and to the right for compasses 2 and 3 in accordance with the right-hand rule for fields.

B. The magnetic field points to the right for compasses 1 and 2 and to the left for compass 3 in accordance with the right-hand rule for fields.

C. The magnetic field points to the left for compasses 1 and 2 and to the right for compass 3 in accordance with the right-hand rule for fields.

D. The magnetic field points to the right for compass 1 and to the left for compasses 2 and 3 in accordance with the right-hand rule for fields.

E. The magnetic field points to the right for all three compasses in accordance with the right-hand rule for fields.T

F. he magnetic field points to the left for all three compasses in accordance with the right-hand rule for fields.

E. The magnetic field points to the right for all three compasses in accordance with the right-hand rule for fields.

One long solenoid is placed inside another solenoid. Both solenoids have the same length and the same number of turns of wire, but the outer solenoid has twice the diameter of the inner solenoid . Each solenoid carries the same current, but the two currents are in opposite directions as shown in (Figure 1).

What is the magnetic field at the center of the inner solenoid?

0 T

Describe the force on the charged particles after they enter the magnetic fields shown in (Figure 1).

A. The force is zero.

B. The force points toward the top of the page.

C. The force points toward the bottom of the page.

D. The force points out of the page.

E. The force points into the page.

E. The force points into the page.

Describe the force on the charged particles after they enter the magnetic fields shown in (Figure 2).

A. The force is zero.

B. The force points to the left.

C. The force points toward the bottom of the page.

D. The force points toward the top of the page.

E. The force points to the right.

A. The force is zero.

A proton is moving in a circular orbit in the earth's magnetic field directly above the north magnetic pole.

Viewed from above, is the rotation clockwise or counterclockwise?

A. counterclockwise

B. clockwise

B. clockwise

An electron and a proton are moving in circular orbits in the earth's magnetic field, high above the earth's atmosphere. The two particles move at the same speed.

Which particle takes more time to complete one orbit?

A. The proton has more mass and the larger orbit, so it takes longer to complete one orbit.

B. The proton has less mass and the larger orbit, so it takes longer to complete one orbit.

C. The electron has less mass and the larger orbit, so it takes longer to complete one orbit.

D. The electron has more mass and the larger orbit, so it takes longer to complete one orbit.

A. The proton has more mass and the larger orbit, so it takes longer to complete one orbit.

A proton moves in a region of uniform magnetic field, as shown in the figure. The velocity at one instant is shown.(Figure 1)

Will the subsequent motion be a clockwise or counterclockwise orbit?

A. clockwise

B. counterclockwise

A. clockwise

An unmagnetized metal sphere hangs by a thread. When the north pole of a bar magnet is brought near, the sphere is strongly attracted to the magnet, as shown in (Figure 1). Then the magnet is reversed and its south pole is brought near the sphere.

How does the sphere respond?

A. It is weakly repelled by the magnet.

B. It is weakly attracted to the magnet.

C. It is strongly repelled by the magnet.

D. It is strongly attracted to the magnet.

E. It does not respond.

D. It is strongly attracted to the magnet.

If a compass is placed above a current-carrying wire, as in the figure below, the needle will line up with the field of the wire.(Figure 1).

Which of the views shows the correct orientation of the needle for the noted current direction?

A. A

B. B

C. C

D. D

C

Two wires carry equal and opposite currents, as shown in the figure.(Figure 1)

At a point directly between the two wires, the field is

A. directed up, toward the top of the screen.

B. directed down, toward the bottom of the screen.

C. directed to the left.

D. directed to the right.

E. zero.

A. directed up, toward the top of the screen.

The figure below shows four particles moving to the right as they enter a region of uniform magnetic field, directed into the paper as noted. All particles move at the same speed and have the same charge.(Figure 1)

Which particle has the largest mass?

A. A

B. B

C. C

D. D

D

Four particles of identical charge and mass enter a region of uniform magnetic field and follow the trajectories shown in (Figure 1).

Which particle has the highest velocity?

A. A

B. B

C. C

D. D

D

If all of the particles shown in the figure are electrons, what is the direction of the magnetic field that produced the indicated deflection?(Figure 1)

A. Up (toward the top of the page)

B. Down (toward the bottom of the page)

C. Out of the plane of the paper

D. Into the plane of the paper

D. Into the plane of the paper

If two compasses are brought near enough to each other, the magnetic fields of the compasses themselves will be larger than the field of the earth, and the needles will line up with each other.(Figure 1)

Which of the arrangements of two compasses shown in the figure is a possible stable arrangement?

A. A

B. B

C. C

D. D

C

If you have a sample of unknown composition, a first step at analysis might be a determination of the masses of the atoms and molecules in the sample. A mass spectrometer to make such an analysis can take various forms, but for many years the best technique was to determine the masses of ionized atoms and molecules in a sample by observing their circular paths in a uniform magnetic field, as illustrated in (Figure 1). A sample to be analyzed is vaporized, then singly ionized. The ions are accelerated through an electric field, and ions of a known speed selected. These ions travel into a region of uniform magnetic field, where they follow circular paths. An exit slit allows ions that have followed a particular path to be counted by a detector, producing a record of the masses of the particles in the sample.

In the spectrometer shown in the figure, do the ions have positive or negative charge?

A. Positive

B. Negative

A. positive

If you have a sample of unknown composition, a first step at analysis might be a determination of the masses of the atoms and molecules in the sample. A mass spectrometer to make such an analysis can take various forms, but for many years the best technique was to determine the masses of ionized atoms and molecules in a sample by observing their circular paths in a uniform magnetic field, as illustrated in (Figure 1). A sample to be analyzed is vaporized, then singly ionized. The ions are accelerated through an electric field, and ions of a known speed selected. These ions travel into a region of uniform magnetic field, where they follow circular paths. An exit slit allows ions that have followed a particular path to be counted by a detector, producing a record of the masses of the particles in the sample. The moving ions can be thought of as a current loop, and it will produce its own magnetic field.

What is the direction of this field at the center of the particles' cicrcular orbit?

A. Opposite the direction of the spectrometer’s magnetic field.

B. In the same direction as the spectrometer’s magnetic field.

A. Opposite the direction of the spectrometer’s magnetic field.

If you have a sample of unknown composition, a first step at analysis might be a determination of the masses of the atoms and molecules in the sample. A mass spectrometer to make such an analysis can take various forms, but for many years the best technique was to determine the masses of ionized atoms and molecules in a sample by observing their circular paths in a uniform magnetic field, as illustrated in (Figure 1). A sample to be analyzed is vaporized, then singly ionized. The ions are accelerated through an electric field, and ions of a known speed selected. These ions travel into a region of uniform magnetic field, where they follow circular paths. An exit slit allows ions that have followed a particular path to be counted by a detector, producing a record of the masses of the particles in the sample.

Why is it important that the ions have a known speed?

A. If the ions are moving too fast, the magnetic field will not be able to bend their path to the detector.

B. The radius of the orbit depends on the mass, the charge, and the speed. If the charge and the speed are the same, the orbit depends on only the mass.

C. The orbit must be circular, and this is the case for only a certain range of speeds.

D. The ions are all accelerated by the same electric field, and so will all have the same speed anyway.

B. The radius of the orbit depends on the mass, the charge, and the speed. If the charge and the speed are the same, the orbit depends on only the mass.

If you have a sample of unknown composition, a first step at analysis might be a determination of the masses of the atoms and molecules in the sample. A mass spectrometer to make such an analysis can take various forms, but for many years the best technique was to determine the masses of ionized atoms and molecules in a sample by observing their circular paths in a uniform magnetic field, as illustrated in (Figure 1). A sample to be analyzed is vaporized, then singly ionized. The ions are accelerated through an electric field, and ions of a known speed selected. These ions travel into a region of uniform magnetic field, where they follow circular paths. An exit slit allows ions that have followed a particular path to be counted by a detector, producing a record of the masses of the particles in the sample. A mass spectrometer similar to the one in the figure is designed to analyze biological samples. Molecules in the sample are singly ionized, then they enter a 0.80 T uniform magnetic field at a speed of 2.3×10^5m/s. A molecule has a mass 255 times the mass of the proton.

What will be the approximate distance between the points where the ion enters and exits the magnetic field?

A. 150 cm

B. 300 cm

C. 225 cm

D. 75 cm

A. 150 cm

If you have a sample of unknown composition, a first step at analysis might be a determination of the masses of the atoms and molecules in the sample. A mass spectrometer to make such an analysis can take various forms, but for many years the best technique was to determine the masses of ionized atoms and molecules in a sample by observing their circular paths in a uniform magnetic field, as illustrated in (Figure 1). A sample to be analyzed is vaporized, then singly ionized. The ions are accelerated through an electric field, and ions of a known speed selected. These ions travel into a region of uniform magnetic field, where they follow circular paths. An exit slit allows ions that have followed a particular path to be counted by a detector, producing a record of the masses of the particles in the sample. A mass spectrometer similar to the one in the figure is designed to analyze biological samples. Molecules in the sample are singly ionized, then they enter a 0.80 T uniform magnetic field at a speed of 2.3×105m/s. A molecule has a mass 255 times the mass of the proton.

What will be the approximate distance between the points where the ion enters and exits the magnetic field?

A. 300 cm

B. 150 cm

C. 75 cm

D. 225 cm

B. 150 cm

The figures show one or more metal wires sliding on fixed metal rails in a magnetic field. For each, determine if the induced current is clockwise, counterclockwise, or is zero.

Clockwise: C

Counterclockwise: A, D

Zero: B, E, F

A circular loop rotates at constant speed about an axle through the center of the loop. The figure shows an edge view and defines the angle ϕ, which increases from 0∘ to 360∘ as the loop rotates.(Figure 1)

At what angle or angles is the magnetic flux a maximum?

0, 180, 360

A circular loop rotates at constant speed about an axle through the center of the loop. The figure shows an edge view and defines the angle ϕ, which increases from 0∘ to 360∘ as the loop rotates.(Figure 1)

At what angle or angles is the magnetic flux a minimum?

90, 270

A circular loop rotates at constant speed about an axle through the center of the loop. The figure shows an edge view and defines the angle ϕ, which increases from 0∘ to 360∘ as the loop rotates.(Figure 1)

At what angle or angles is the magnetic flux changing most rapidly?

90, 270

There is a counterclockwise induced current in the conducting loop shown in the figure.(Figure 1)

Is the magnetic field inside the loop increasing in strength, decreasing in strength, or steady?

A. increasing in strength

B. decreasing in strength

C. steady

increasing in strength

The conducting loop in the figure is moving into the region between the magnetic poles shown.(Figure 1)

Is the induced current (viewed from above) clockwise or counterclockwise?

A. clockwise

B. counterclockwise

A. clockwise

The conducting loop in the figure is moving into the region between the magnetic poles shown.(Figure 1)

Is there an attractive magnetic force that tends to pull the loop in, like a magnet pulls on a paper clip? Or do you need to push the loop in against a repulsive force?

A. There is an attractive magnetic force that tends to pull the loop in.

B. You need to push the loop in against a repulsive force.

B. You need to push the loop in against a repulsive force.

(Figure 1) shows two concentric, conducting loops. We will define a counterclockwise current (viewed from above) to be positive, a clockwise current to be negative. The graph shows the current in the outer loop as a function of time.

Choose the correct graph of the induced current in the inner loop.

A

(Figure 1) shows conducting loops next to each other. We will define a counterclockwise current (viewed from above) to be positive, a clockwise current to be negative. The graph shows the current in the left loop as a function of time.

Choose the correct graph of the induced current in the right loop.

C

A loop of wire is horizontal. A bar magnet is pushed toward the loop from below, along the axis of the loop, as shown in (Figure 1).

In what direction is the current in the loop? Select the correct answer and explanation.

A. As the north magnetic pole of the magnet approaches the coil, there is an increasing upward magnetic flux through the coil. So the induced current is counterclockwise (as viewed from above).

B. As the north magnetic pole of the magnet approaches the coil, there is an increasing upward magnetic flux through the coil. So the induced current is clockwise (as viewed from above).

C. As the north magnetic pole of the magnet approaches the coil, there is an increasing downward magnetic flux through the coil. So the induced current is counterclockwise (as viewed from above).

D. As the north magnetic pole of the magnet approaches the coil, there is an increasing downward magnetic flux through the coil. So the induced current is clockwise (as viewed from above).

B. As the north magnetic pole of the magnet approaches the coil, there is an increasing upward magnetic flux through the coil. So the induced current is clockwise (as viewed from above).

A loop of wire is horizontal. A bar magnet is pushed toward the loop from below, along the axis of the loop, as shown in (Figure 1).

Is there a magnetic force on the loop? If so, in which direction? Select the correct answer and explanation. Hint: Recall that a current loop is a magnetic dipole.

A. The magnetic field at the site of the coil may be resolved into a vertical and a horizontal component. The current is clockwise. As the induced current interacts with the magnetic field of the approaching magnet, the vertical component of the magnetic field results in force that tries to collapse the coil, and the horizontal component of the magnetic field results in force that pushes the coil vertically downward.

B. The magnetic field at the site of the coil may be resolved into a vertical and a horizontal component. The current is clockwise, and forces caused by the vertical and horizontal components compensate each other. So there is no magnetic force on the loop.

C. The magnetic field at the site of the coil may be resolved into a vertical and a horizontal component. The current is clockwise. As the induced current interacts with the magnetic field of the approaching magnet, the vertical component of the magnetic field results in a force that tries to collapse the coil, and the horizontal component of the magnetic field results in force that pushes the coil vertically upward.

D. The magnetic field at the site of the coil may be resolved into a vertical and a horizontal component. However, they don't cause any force so there is no magnetic force on the loop.

C. The magnetic field at the site of the coil may be resolved into a vertical and a horizontal component. The current is clockwise. As the induced current interacts with the magnetic field of the approaching magnet, the vertical component of the magnetic field results in a force that tries to collapse the coil, and the horizontal component of the magnetic field results in force that pushes the coil vertically upward.

A loop of wire is horizontal. A bar magnet is pushed toward the loop from below, along the axis of the loop, as shown in (Figure 1).

Is there a magnetic force on the magnet? If so, in which direction? Select the correct answer and explanation.

A. By Newton's third law, if the magnet exerts an upward force on the loop, the loop must exert a downward force on the magnet.

B. The magnet exerts an upward force on the loop, but the loop doesn't exert any force on the magnet.

C. By Newton's third law, if the magnet exerts an upward force on the loop, the loop must exert an upward force on the magnet.

D. The loop exerts an upward force on the magnet, but it's compensated by the magnet's force. So there is no force exerted on the magnet.

A. By Newton's third law, if the magnet exerts an upward force on the loop, the loop must exert a downward force on the magnet.

A metal wire is resting on a U-shaped conducting rail, as shown in (Figure 1). The rail is fixed in position, but the wire is free to move.

If the magnetic field is increasing in strength, which way does the wire move? Select the correct answer and explanation.

A. The original magnetic field B is into the page and increasing, so Binduced is into the page. This leads to a clockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the right, so the wire moves to the right. The field extends beyond the rail so the wire will break contact. At that point the force on the wire will drop to zero.

B. The original magnetic field B is into the page and increasing, so Binduced is out of the page. This leads to a counterclockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the right, so the wire moves to the right. The field extends beyond the rail so the wire will break contact. At that point the force on the wire will drop to zero.

C. The original magnetic field B is into the page and increasing, so Binduced is into the page. This leads to a clockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the left, so the wire moves to the left.

D. The original magnetic field B is into the page and increasing, so Binduced is out of the page. This leads to a counterclockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the left, so the wire moves to the left.

D. The original magnetic field B is into the page and increasing, so Binduced is out of the page. This leads to a counterclockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the left, so the wire moves to the left.

A metal wire is resting on a U-shaped conducting rail, as shown in (Figure 1). The rail is fixed in position, but the wire is free to move.

If the magnetic field is decreasing in strength, which way does the wire move? Select the correct answer and explanation.

A. The original magnetic field B is into the page and decreasing, so Binduced is out of the page. This leads to a counterclockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the right, so the wire moves to the right. The field extends beyond the rail so the wire will break contact. At that point the force on the wire will drop to zero.

B. The original magnetic field B is into the page and decreasing, so Binduced is out of the page. This leads to a counterclockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the left, so the wire moves to the left.

C. The original magnetic field B is into the page and decreasing, so Binduced is into the page. This leads to a clockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the left, so the wire moves to the left.

D. The original magnetic field B is into the page and decreasing, so Binduced is into the page. This leads to a clockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the right, so the wire moves to the right. The field extends beyond the rail so the wire will break contact. At that point the force on the wire will drop to zero.

D. The original magnetic field B is into the page and decreasing, so Binduced is into the page. This leads to a clockwise induced current. By the right-hand rule, the force exerted by the field on the wire is to the right, so the wire moves to the right. The field extends beyond the rail so the wire will break contact. At that point the force on the wire will drop to zero.

Although sunlight is unpolarized, the light that reflects from smooth surfaces may be partially polarized in the direction parallel to the plane of the reflecting surface.

How should the axis of the polarizers in sunglasses be oriented—vertically or horizontally—to reduce the glare from a horizontal surface such as a road or a lake? Select the correct answer and explanation.

A. The light we are interested in blocking is partially polarized parallel to the plane of the reflecting surface; hence the electric field vector we wish to block is oscillating parallel to the horizontal surface. If the long axis of the polarizing molecules is also parallel to the horizontal surface, the electrons in the molecules will be forced into oscillations along the long molecules by the electric field, so the electric field does not pass through the sunglasses.

B. The light we are interested in blocking is partially polarized parallel to the plane of the reflecting surface; hence the electric field vector we wish to block is oscillating perpendicular to the horizontal surface. If the long axis of the polarizing molecules is also perpendicular to the horizontal surface, the electrons in the molecules will be forced into oscillations along the long molecules by the electric field, so the electric field does not pass through the sunglasses.

C. The light we are interested in blocking is partially polarized parallel to the plane of the reflecting surface; hence the electric field vector we wish to block is oscillating perpendicular to the horizontal surface. If the long axis of the polarizing molecules is parallel to the horizontal surface, the electrons in the molecules will be forced into oscillations along the long molecules by the electric field, so the electric field does not pass through the sunglasses.

D. The light we are interested in blocking is partially polarized parallel to the plane of the reflecting surface; hence the electric field vector we wish to block is oscillating parallel to the horizontal surface. If the long axis of the polarizing molecules is perpendicular to the horizontal surface, the electrons in the molecules will be forced into oscillations along the long molecules by the electric field, so the electric field does not pass through the sunglasses.

A. The light we are interested in blocking is partially polarized parallel to the plane of the reflecting surface; hence the electric field vector we wish to block is oscillating parallel to the horizontal surface. If the long axis of the polarizing molecules is also parallel to the horizontal surface, the electrons in the molecules will be forced into oscillations along the long molecules by the electric field, so the electric field does not pass through the sunglasses.

An AM radio detects the oscillating magnetic field of the radio wave with an antenna consisting of a coil of wire wrapped around a ferrite bar, as shown in (Figure 1). Ferrite is a magnetic material that “amplifies” the magnetic field of the wave. The radio antenna broadcasts waves with the electric field vertical, the magnetic field horizontal.

If a radio antenna is located due north of you, how must the ferrite bar be oriented for best reception? Select the correct answer and explanation.

A. The coil and ferrite bar need to be horizontal since the magnetic field will be oscillating in a horizontal plane (perpendicular to the vertical plane of the electric field). If the station is north of you, then the magnetic field will oscillate in an east-west direction, and that is the way you should orient your ferrite bar to detect the largest changes in flux.

B. The coil and ferrite bar need to be horizontal since the magnetic field will be oscillating in a horizontal plane (perpendicular to the vertical plane of the electric field). If the station is north of you, then the magnetic field will oscillate in an east-west direction, and you should orient your ferrite bar in a north-south direction to detect the largest changes in flux.

C. The coil and ferrite bar need to be vertical since the magnetic field will be oscillating in a horizontal plane (perpendicular to the horizontal plane of the electric field), and you should orient your ferrite bar vertically irrespective of the location of the station to detect the largest changes in flux.

D. The coil and ferrite bar need to be vertical since the magnetic field will be oscillating in a horizontal plane (perpendicular to the horizontal plane of the electric field). If the station is north of you, then the magnetic field will oscillate in an east-west direction, and you should orient your ferrite bar in a north-south direction to detect the largest changes in flux.

A. The coil and ferrite bar need to be horizontal since the magnetic field will be oscillating in a horizontal plane (perpendicular to the vertical plane of the electric field). If the station is north of you, then the magnetic field will oscillate in an east-west direction, and that is the way you should orient your ferrite bar to detect the largest changes in flux.

A circular loop of wire has an area of 0.25 m^2 . It is tilted by 43 ∘ with respect to a uniform 0.45 T magnetic field. What is the magnetic flux through the loop?

A. 8.2×10⁻² T⋅m^2

B. 0.20 T⋅m^2

C. 0.40 T⋅m^2

D. 0.75 T⋅m^2

E. 1.5 T⋅m^2

A. 8.2×10⁻² T⋅m^2

A circular loop of wire has an area of 0.27 m^2 . It is tilted by 42 ∘ with respect to a uniform 0.38 T magnetic field.

What is the magnetic flux through the loop?

A. 7.6×10⁻² T⋅m2

B. 0.11 T⋅m^2

C. 0.38 T⋅m^2

D. 0.72 T⋅m^2

E. 1.4 T⋅m^2

A. 7.6×10⁻² T⋅m2

In the figure, a square loop is rotating in the plane of the page around an axis through its center. A uniform magnetic field is directed into the page.(Figure 1)

What is the direction of the induced current in the loop?

A. clockwise

B. counterclockwise

E. There is no induced current

C. There is no induced current.

A diamond-shaped loop of wire is pulled at a constant velocity through a region where the magnetic field is directed into the paper in the left half and is zero in the right half, as shown in the figure.(Figure 1)

As the loop moves from left to right, which graph best represents the induced current in the loop as a function of time? Let a clockwise current be positive and a counterclockwise current be negative.

A. A

B. B

C. C

D. D

E. E

B

The figure shows a triangular loop of wire in a uniform magnetic field.(Figure 1)

If the field strength changes from 0.45 to 0.10 Tin 40 ms , what is the induced emf in the loop?

A. 0.18 V

B. 0.26 V

C. 0.35 V

D. 0.53 V

E. 0.79 V

A. 0.18 V

The figure shows a triangular loop of wire in a uniform magnetic field.(Figure 1)

If the field strength changes from 0.35 to 0.20 Tin 35 ms , what is the induced emf in the loop?

A. 8.6×10⁻² V

B. 0.13 V

C. 0.17 V

D. 0.26 V

E. 0.39 V

A. 8.6×10⁻² V

The Metal Detector Metal detectors use induced currents to sense the presence of any metal-not just magnetic materials such as iron. A metal detector, shown in the figure, consists of two coils: a transmitter coil generates an oscillating magnetic field along the axis and an oscillating induced current in the receiver coil.

If a piece of metal is placed between the transmitter and the receiver, the oscillating magnetic field in the metal induces eddy currents in a plane parallel to the transmitter and receiver coils. The receiver coil then responds to the superposition of the transmitter's magnetic field and the magnetic field of the eddy currents. Because the eddy currents attempt to prevent the flux from changing, in accordance with Lenz's law, the net field at the receiver decreases when a piece of metal is inserted between the coils. Electronic circuits detect the current decrease in the receiver coil and set off an alarm. (Figure 1)

Why won't the metal detector detect insulators?

A. Insulators block magnetic fields.

B. No eddy current can be produced in an insulator.

C. No emf can be produced in an insulator.

D. An insulator will increase the field at the receiver.

B. No eddy current can be produced in an insulator.

The Metal Detector Metal detectors use induced currents to sense the presence of any metal-not just magnetic materials such as iron. A metal detector, shown in the figure, consists of two coils: a transmitter coil generates an oscillating magnetic field along the axis and an oscillating induced current in the receiver coil.

If a piece of metal is placed between the transmitter and the receiver, the oscillating magnetic field in the metal induces eddy currents in a plane parallel to the transmitter and receiver coils. The receiver coil then responds to the superposition of the transmitter's magnetic field and the magnetic field of the eddy currents. Because the eddy currents attempt to prevent the flux from changing, in accordance with Lenz's law, the net field at the receiver decreases when a piece of metal is inserted between the coils. Electronic circuits detect the current decrease in the receiver coil and set off an alarm. (Figure 1)

A metal detector can detect the presence of metal screws used to repair a broken bone inside the body. What does this tell us?

A. The screws are made of magnetic materials.

B. The tissues of the body are conducting.

C. The magnetic fields of the device can penetrate the tissues of the body.

D. The screws must be perfectly aligned with the axis of the device.

C. The magnetic fields of the device can penetrate the tissues of the body.

The Metal Detector Metal detectors use induced currents to sense the presence of any metal-not just magnetic materials such as iron. A metal detector, shown in the figure, consists of two coils: a transmitter coil generates an oscillating magnetic field along the axis and an oscillating induced current in the receiver coil.

If a piece of metal is placed between the transmitter and the receiver, the oscillating magnetic field in the metal induces eddy currents in a plane parallel to the transmitter and receiver coils. The receiver coil then responds to the superposition of the transmitter's magnetic field and the magnetic field of the eddy currents. Because the eddy currents attempt to prevent the flux from changing, in accordance with Lenz's law, the net field at the receiver decreases when a piece of metal is inserted between the coils. Electronic circuits detect the current decrease in the receiver coil and set off an alarm. (Figure 1)

Which of the following changes would not produce a larger eddy current in the metal?

A. Increasing the frequency of the oscillating current in the transmitter coil.

B. Increasing the magnitude of the oscillating current in the transmitter coil.

C. Increasing the resistivity of the metal.

D. Decreasing the distance between the metal and the transmitter

C. Increasing the resistivity of the metal.