AP Physics C Electricity : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #31 : Electricity And Magnetism Exam

Ap physics c e m potential problems  2 6 16   3

A thin bar of length L lies in the xy plane and carries linear charge density , where  ranges from 0 to . Calculate the potential at the point  on the y-axis.

Possible Answers:

Correct answer:

Explanation:

Use the linear charge density  and length element , where each point is  from the point . The potential is therefore

Example Question #11 : Calculating Electric Potential

Ap physics c e m potential problems  2 6 16   2

A uniformly charged ring of radius  carries a total charge . Calculate the potential a distance  from the center, on the axis of the ring.

Possible Answers:

Correct answer:

Explanation:

Use the linear charge density  and length element . The distance from each point on the ring to the point on the axis is . Lastly, integrate over  from  to  to obtain

Example Question #31 : Electricity And Magnetism Exam

Ap physics c e m potential problems  2 6 16

A uniformly charged square frame of side length  carries a total charge . Calculate the potential at the center of the square.

You may wish to use the integral:

Possible Answers:

Correct answer:

Explanation:

Calculate the potential due to one side of the bar, and then multiply this by  to get the total potential from all four sides. Orient the bar along the x-axis such that its endpoints are at , and use the linear charge density . The potential is therefore

 

 

Example Question #32 : Electricity And Magnetism Exam

Consider a spherical shell with radius  and charge . What is the magnitude of the electric field at the center of this spherical shell?

Possible Answers:

Zero

Correct answer:

Zero

Explanation:

According to the shell theorem, the total electric field at the center point of a charged spherical shell is always zero. At this point, any electric field lines will result in symmetry, canceling each other out and creating a net field of zero at that point.

Example Question #33 : Electricity And Magnetism Exam

A charge of unknown value is held in place far from other charges. Its electric field lines and some lines of electric equipotential (V1 and V2) are shown in the diagram.

Onecharge_negative_fieldandpotentiallines

A second point charge, known to be negative, is placed at point A in the diagram. In which direction will the second, negative charge freely move?

Possible Answers:

Toward the original charge Q

Toward point C

It will remain stationary

Toward point A

Correct answer:

Toward point C

Explanation:

The original charge Q is negative, as indicated by the direction of the electric field lines in the diagram. A negative point charge of any value placed at point A will cause both charges to feel a mutually repelling force on each other. Therefore, the second charge will be repelled by the original negative charge with a force pointing radially along a line connecting the center of Q and the point A, resulting in free movement toward C. Additionally, point B is on a line of equipotential to point A; negative point charges will move from regions of lower electric potential to higher electric potential (against the direction of the electric field lines), so point B is not a viable answer.

Example Question #31 : Electricity And Magnetism Exam

A proton traveling  enters a uniform magnetic field and experiences a magnetic force, causing it to travel in a circular path. Taking the magnetic field to be , what is the radius of this circular path (shown in red)?

Uniform_field

Possible Answers:


 

Correct answer:

Explanation:

To calculate the magnetic force of a single charge, we use , where  is the charge of the proton,  is its velocity,  is the uniform magnetic field.

Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.

 is the mass of the proton and  is the radius of the circular path. Solve for .

Using the values given in the question, we can solve for the radius.

Example Question #31 : Electricity And Magnetism Exam

Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?

Possible Answers:

Negative only if the north pole of the magnet is within the surface

Zero regardless of the orientation of the magnet

Zero only if the magnet is completely enclosed within the surface

More than one of the other options is true

Positive only if the north pole of the magnet is within the surface

Correct answer:

Zero regardless of the orientation of the magnet

Explanation:

The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."

Example Question #35 : Electricity And Magnetism Exam

A particle of charge  and mass  moves with a speed of  perpendicular to a uniform magnetic field, . What is the period of the particle's orbit in the field?

Possible Answers:

Correct answer:

Explanation:

Relevant equations:

Set the magnetic force equal to the centripetal force, since the magnetic force is directed towards the center of the particle's circular path and centripetal force is defined as the net force towards the center of a circular path.

Rearrange to isolate  the velocity:

Determine the distance, , traveled in one revolution, which is the circumference of a circle of radius :

Plug this distance and velocity into , to solve for the period :

Example Question #36 : Electricity And Magnetism Exam

Ring

Consider a current-carrying loop with current , radius , and center 

A particle with charge  flies through the center and into the page with velocity . What is the total electromagnetic force on the particle at the instant that it flies through the loop, in terms of the variables given?

Possible Answers:

Correct answer:

Explanation:

The correct answer is zero. To calculate the force of a magnetic field on a moving charged particle, we use the cross product. We know that if the magnetic field is parallel to the velocity vector of the particle, then the force produced is zero.

Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.

Example Question #33 : Electricity And Magnetism Exam

Consider two long, straight, current-carrying wires at distance  from each other, each with a current of magnitude  going in opposite directions.

If the two wires described are not held in place, what motion will result from the magnetic fields produced?

Possible Answers:

The wires will move toward each other

The wires will remain in place

The wires will move away from each other

The wires will rotate counterclockwise

The wires will rotate clockwise

Correct answer:

The wires will move away from each other

Explanation:

The answer is that the wires will move from each other. Using our right hand rule, we know that the magnetic fields produced by each wire are in the same direction, as long as their currents oppose. Using the right hand rule again to determine the direction of the force exerted on each wire by the magnetic field with which they are interacting yields a force in the direction away from the other wire for each wire.

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