AP Physics 2 : Electricity and Magnetism

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #8 : Magnetic Fields

An infinitely long wire carries a current of  determine the magnitude of the magnetic field  away.

Possible Answers:

None of these

Correct answer:

Explanation:

Magnetic field of an infinitely long wire:

Where 

Plugging in values:

Example Question #9 : Magnetic Fields

A circuit contains a  battery and a  resistor in series. Determine the magnitude of the magnetic field outside of the loop  away from the wire.

Possible Answers:

None of these

Correct answer:

Explanation:

Using

Converting  to  and plugging in values

Determining current:

Example Question #1 : Magnetic Fields

How strong would a magnetic field need to be in order to make a particle with a mass of  and a charge of  move in a circular path with a speed of  and a radius of ?

Possible Answers:

Correct answer:

Explanation:

For this question, we are being asked to determine the magnetic field necessary to make a particle of a given mass and charge to move in a circular path with a given speed and radius.

To begin with, we can realize that the particle will be moving in a circular path. Thus, there is going to be a centripetal force associated with this circular motion. Moreover, because we know the particle will be present in a magnetic field, we can infer that the magnetic force will be the source of the centripetal force. Thus, we can start by writing out the expression for each of these forces, and then setting them equal to one another.

Rearranging the above expression to isolate the term for magnetic field, we arrive at the following expression.

Now, we can plug in the values given to us in the question stem to solve for the magnetic field strength.

Example Question #11 : Magnetic Fields

Radius of moon: 

A loop of current carrying wire runs along the equator of the moon. Determine the magnetic field at the center of the loop if  are traveling through it

Possible Answers:

None of these

Correct answer:

Explanation:

Using

Plugging in values

Example Question #1 : Magnetic Force

A particle with a charge of  is moving at perpendicularly through a magnetic field with a strength of . What is the magnitude of the force on the particle?

Possible Answers:

There is no force on the particle

Correct answer:

Explanation:

The equation for finding the force on a moving charged particle in a magnetic field is as follows:

Here,  is the force in Newtons,  is the charge in Coulombs,  is the velocity in , and  is the magnetic field strength in Teslas.

Another way to write the equation without the cross-product is as follows:

Here,  is the angle between the particles velocity and the magnetic field.

For our problem, because theta is  evaluates to 1, so we just need to perform multiplication.

Therefore, the force on the particle is 0.3N.

Example Question #2 : Magnetic Force

An  wire with a current of  is oriented  from parallel to a magnetic field with a strength of . What is the force on the wire?

Possible Answers:

The wire doesn't experience any force

Correct answer:

Explanation:

The equation for the force on a current carrying wire in a magnetic field is as follows:

 is the force in Newtons,  is the current in amperes,  is the magnetic field strength in Teslas, and  is the angle from parallel to the magnetic field.

Because our wire is not fully perpendicular to the magnetic field, it does not experience the full possible force. Instead, it experiences  times the maximum force value.

Example Question #1 : Magnetic Force

A proton traveling at  in a horizontal plane passes through an opening into a mass spectrometer with a uniform  magnetic field directed upward. The particle then moves in a circular path through  and crashes into the wall of the spectrometer adjacent to the entrance opening. How far down from the entrance is the proton when it crashes into the wall?

The proton’s mass is  and its electric charge is .

Possible Answers:

Correct answer:

Explanation:

Mass_spec_fig

A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:

 is the charge,  is the particle speed, and  is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:

We can set these two equations equal to one another, allowing us to solve for the radius of the arc.

Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).

Twice this value is the lateral offset of its crash point from the entrance:

Example Question #1 : Magnetic Force

What is the force experienced by a  charge moving at  through a magnetic field with strength  at  from perpendicular to the field?

Possible Answers:

There is no net force on the charge

Correct answer:

Explanation:

To find the force experienced by the charge, we use this equation:

Because the charge is moving at an angle from perpendicular, we need to take the cross product into account. 

Theta is the angle from perpendicular, which is . Plug in known values and solve.

Example Question #1 : Magnetic Force

An  charge is moving through a  magnetic field at a speed of  perpendicular to the direction of the field. What is the force on the charge?

Possible Answers:

There is no force on the charge

Correct answer:

Explanation:

The equation for force on a charge moving through a magnetic field is:

Because the velocity is perpendicular to the field, the cross product doesn't matter, and we can do simple multiplication.

Therefore, the force on the charge is

Example Question #1 : Magnetic Force

 charge is moving through a  magnetic field at a speed of  from parallel to the magnetic field. What is the force on the charge?

Possible Answers:

The charge experiences no force

Correct answer:

Explanation:

The equation for force on a charge moving through a magnetic field is:

.

The cross product is:

Above,  is the degree from parallel the charge is moving. The charge is moving at  from parallel, so the equation, once we plug in our numbers, is:

The force on the charge is about .

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