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AP Physics C Electricity : Electricity and Magnetism Exam

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

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All AP Physics C Electricity Resources

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Example Questions

Example Question #191 : Ap Physics C

Two infinitely long wires having currents I_1 and I_2 are separated by a distance .

Ps0_twowiresbfield

The current I_1 is 6A into the page. The current I_2 is 9A into the page. The distance of separation is 1.5mm. The point lies 1.5mm away from I_1 on a line connecting the centers of the two wires.

What is the magnitude and direction of the net magnetic field at the point ?

$\mu_0=4\pi*10^{-7}\frac{T\cdot m}{A}$

Possible Answers:

$2.0*10^{-4}T\ \text{to the right}$

$1.4*10^{-3}T\ \text{to the left}$

$2.0*10^{-3}T\ \text{to the left}$

$1.4*10^{-3}T\ \text{to the right}$

Correct answer:

$1.4*10^{-3}T\ \text{to the right}$

Explanation:

At point , the magnetic field due to I_1 points right (via the right hand rule) with a magnitude given by:

$B_{1}=\left| \frac{\mu_{o} I_{1}}{2\pi r_{1}} \right|$

$B_{1}=\left| \frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(6A)}{2\pi (1.5mm)} \right|=8*10^{-4}T$

At point , the magnetic field due to I_2 points right (via the right hand rule) with a magnitude given by:

$B_{2}=\left| \frac{\mu_{o} I_{2}}{2\pi r_{2}} \right|$

$B_{2}=\left| \frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(9A)}{2\pi (3.0mm)} \right|=6*10^{-4}T$

The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude $1.4*10^{-3}T$ to the right.

$8*10^{-4}T+6*10^{-4}T=1.4*10^{-3}T$

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Example Question #1 : Magnetism

Ring

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

What is the direction of the magnetic field produced?

Possible Answers:

There is no resultant magnetic field

Into the screen

Out of the screen

Counterclockwise

Clockwise

Correct answer:

Into the screen

Explanation:

The correct answer is into the page. As the current is moving clockwise, we can use our right hand rule for magnetic fields produced by a current-carrying loop. Curl the fingers of your right hand in the direction of the current. This should result in your thumb pointing toward the screen, indicating the direction of the magnetic field.

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Example Question #192 : Ap Physics C

Ring

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

What would happen to the magnetic field at point if the radius was halved and current was multiplied by four?

Possible Answers:

The new magnetic field would be eight times weaker than the original

The new magnetic field would be four times as strong as the original

The new magnetic field would be eight times as strong as the original

The new magnetic field would reverse direction

The new magnetic field would be four times weaker than the original

Correct answer:

The new magnetic field would be eight times as strong as the original

Explanation:

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

Using out altered values, we can derive a ratio to determine the change in magnetic field.

$B_1=\frac{\mu_0(4I)}{2\pi(\frac{1}{2}R)}=\frac{8\mu_0I}{2\pi R}=8B$

The resulting field will be eight times stronger than the original.

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Example Question #1 : Magnetism

Ring

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

What is the magnitude of the magnetic field at point ?

Possible Answers:

$\frac{\mu_0I}{2\pi R}$

$\frac{\mu_{0}I}{R}$

There is no magnetic field at point

$\frac{\mu_{0}I}{\pi R^2}$

$\frac{\mu_0I}{2R}$

Correct answer:

$\frac{\mu_0I}{2\pi R}$

Explanation:

The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:

$B=\frac{\mu_{0}I}{2\pi R}$

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Example Question #193 : Ap Physics C

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

What is the magnitude of the magnetic field at a point equidistant from both wires?

Possible Answers:

$\frac{2\mu_0I}{\pi d}$

Zero

$\frac{4\mu_0I}{\pi d}$

$\frac{\mu_0I}{2\pi d}$

$\frac{\mu_0I}{\pi d}$

Correct answer:

$\frac{2\mu_0I}{\pi d}$

Explanation:

Using our right hand rule for magnetic fields produced by current-carrying wires, we know that the magnetic field produced by each wire is in the same direction within the distance between the wires. Therefore, we know that the total magnetic field is simply the magnetic field of one of the wires multiplied by two.

Use the equation for magnetic field:

$B=\frac{\mu_0I}{2\pi R}$

Multiply by two, since the magnetic field will be equal for each wire, and substitute the given variables:

$B=\frac{2\mu_0I}{2\pi (\frac{1}{2}d)}=\frac{2\mu_0I}{\pi d}$

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Example Question #1 : Calculating Magnetic Fields And Forces

An infinitely long wire has a current of running through it. Calculate the magnetic field at a distance 2cm away from the wire.

$\mu_0=4\pi *10^{-7}\frac{Tm}{A}$

Possible Answers:

$7.4*10^{-7}T$

$2*10^{-4}T$

$3*10^{-5}T$

$9*10^{-5}T$

$1.5*10^{-3}T$

Correct answer:

$3*10^{-5}T$

Explanation:

For infinitely long wires, the formula for the magnetic field is $B=\frac{\mu_0I}{2\pi r}$ , where is the current and is the distance from the wire.

The magnetic field is calculated using our given values.

$B=\frac{(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}})(3\text{A})}{2\pi(0.02\text{m})}$

$B=3*10^{-5}\ \text{T}$

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Example Question #1 : Calculating Magnetic Fields And Forces

A solenoid is 50cm long and it is made up of 800 turns of wire. How much current must run through the solenoid to generate a magnetic field of inside of the solenoid?

$\mu_0=4\pi*10^{-7}\frac{Tm}{A}$

Possible Answers:

$274\ \text{A}$

$1237\ \text{A}$

$583\ \text{A}$

$497\ \text{A}$

$97\ \text{A}$

Correct answer:

$497\ \text{A}$

Explanation:

The formula for the magnetic field inside the solenoid is $B=\frac{N}{L}\mu_0I$ , where is the number of turns of wire, is the length of the solenoid, and is the current.

We want to find the current so we solve for .

$I=\frac{BL}{N\mu_0}$

Plug in the values.

$I=\frac{(1\text{T})(0.5\text{m})}{(800)(4\pi*10^{-7}\frac{\text{Tm}}{\text{A}{}})}$

$I=497\ \text{A}$

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Example Question #201 : Ap Physics C

Two parallel wires a distance apart each carry a current , and repel each other with a force per unit length. If the current in each wire is doubled to , and the distance between them is halved to $\frac{R}{2}$ , by what factor does the force per unit length change?

Possible Answers:

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

Explanation:

Relevant equations:

$\oint B\cdot dl = \mu _o I$

$F = I L \times B$

Step 1: Find the original and new magnetic fields created by wire 1 at wire 2, using Ampere's law with an Amperian loop of radius or $\frac{R}{2}$ , respectively.

Original

$B (2\pi R) = \mu _o I \rightarrow B = \frac{\mu _o I}{2 \pi R}$

New

$B (2\pi (\frac{R}{2})) = \mu _o (2I) \rightarrow B = \frac{\mu _o (2I)}{2\pi \frac{R}{2}} = \frac{4\mu _o I}{2\pi R}$

Since the wires are parallel to each other and wire 1's field is directed circularly around it, in each case wire 1's field is perpendicular to wire 2.

Step 2: Find the original and new magnetic forces per unit length on wire 2, due to the field created by wire 1.

$F = IL \times B \rightarrow \frac{F}{L} = IB$

Original

$\frac{F}{L} = I * \frac{\mu _o I }{2 \pi R} = \frac{\mu _o I^2}{2\pi R}\textup{}$

New

$\frac{F}{L} = 2I * \frac{4\mu _o I}{2\pi R} = \frac{8 \mu _o I^2}{2 \pi R}$

So, the new force per unit length is 8 times greater than the original.

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Example Question #1 : Calculating Magnetic Fields And Forces

A region of uniform magnetic field, , is represented by the grey area of the box in the diagram. The magnetic field is oriented into the page.

Ps0_movingqinbfield

A stream of protons moving at velocity $\vec{v}$ is directed into the region of the magnetic field, as shown. Identify the correct path of the stream of protons after they enter the region of magnetic field.

Possible Answers:

A semi-circular path oriented horizontally out of the page

A semi-circular path oriented vertically upward

A semi-circular path oriented horizontally into the page

A semi-circular path oriented vertically downward

Correct answer:

A semi-circular path oriented vertically upward

Explanation:

The magnetic force on a moving charged particle is given by the equation:

$\vec{F}=q\,\vec{v}\times \vec{B}$

Isolating the directional component of this equation yields the understanding that the resulting force on a moving charged particle is perpendicular to the plane of the velocity vector and magnetic field vector. Using the right-hand-rule on this cross-product shows that the velocity vector right-crossed into the magnetic field vector into the page yields a magnetic force vector upward on a positive charge. This will result in a semi-circular path oriented vertically upward.

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Example Question #1 : Electromagnetism

A metal ring is placed in a uniform magnetic field perpendicular to the plane of the ring. An emf of magnitude 15V is induced around the ring by increasing the field through it from zero to 5mT at a constant rate. If the area enclosed by the ring is $1 *10^{-2}m^2$ , what is the time interval over which the field is increased?

Possible Answers:

$15.0 \mu s$

$5.0 \mu s$

$3.3 \mu s$

$2.5 \mu s$

$1.0 \mu s$

Correct answer:

$3.3 \mu s$

Explanation:

Relevant equations:

$\epsilon = \frac{-\Delta \Phi}{\Delta t} = \frac {- \Delta (B\cdot A)}{\Delta t}$

Given:

$A = 1 * 10^{-2} m^2$

$\Delta B = 5mT = 5 * 10^{-3}T$

$\epsilon = 15 V$

Plug in (disregarding direction of emf):

$15 V= \frac{(5 * 10^{-3}T)(1 * 10^{-2}m^2)}{\Delta t}$

$\Delta t = \frac{(5*10^{-3}T)(1 * 10^{-2}m^2)}{15V}= 3.33 * 10^{-6}s = 3.33\mu s$

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AP Physics C Electricity : Electricity and Magnetism Exam

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

All AP Physics C Electricity Resources

Example Questions

Example Question #191 : Ap Physics C

Example Question #1 : Magnetism

Example Question #192 : Ap Physics C

Example Question #1 : Magnetism

Example Question #193 : Ap Physics C

Example Question #1 : Calculating Magnetic Fields And Forces

Example Question #1 : Calculating Magnetic Fields And Forces

Example Question #201 : Ap Physics C

Example Question #1 : Calculating Magnetic Fields And Forces

Example Question #1 : Electromagnetism

All AP Physics C Electricity Resources

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