AP Physics C E and M Unit 13 Induction

[Maximum number: 15]

A wire is connected to a resistor of resistance R to form a rigid rectangular loop of width L and height 2 L. An external force is exerted on the loop so that the loop always moves with constant speed v in the +x-direction, as shown in Figure 1. The loop then enters Region 1 of external uniform magnetic field of magnitude B that is directed in the -z-direction. Region 1 has boundaries x=L and x=2.5 L. The loop later enters Region 2 with two external, uniform magnetic fields, each of magnitude B, that are parallel but are directed in opposite z-directions. Region 2 has boundaries x=2.5 L and x=3.5 L. Point S is the midpoint of the leading edge of the loop and is aligned with the horizontal boundary in Region 2 that separates the two magnetic fields.

(a)

On the following axes, sketch a graph of the magnetic flux $\Phi$ through the rectangular loop as a function of the position x of Point S from x=0 to x=4.5 L. The +z-direction indicated in Figure 1 corresponds to $+\Phi$ .

Figure for Question (a) — AP Physics C: Electricity & Magnetism

[ 4 ]

(b)

Consider the instant when Point S reaches x=1.5 L.

Indicate whether the current $I_{\mathrm{R}}$ that is induced in the rectangular loop when Point S reaches x=1.5 L is clockwise, counterclockwise, or zero.

Clockwise Counterclockwise Zero
Briefly justify your answer.

ii. Derive an expression for $I_{\mathrm{R}}$ when Point S reaches x=1.5 L. If $I_{\mathrm{R}}=0$ , indicate how the derived expression shows that $I_{\mathrm{R}}=0$ . Express your answer in terms of R, L, v, B, and physical constants, as appropriate.
iii. Derive an expression for the power P dissipated by the resistor when Point S reaches x=1.5 L. Express your answer in terms of R, L, v, B, and physical constants, as appropriate.

The total energy dissipated by the resistor in the rectangular loop as Point S moves from x=0 to x=4.5 L is $E_{\text {original }}$ .

The vertical boundary between regions 1 and 2 is now shifted to x=1.5 L. After the boundary is shifted, the rectangular loop again moves with speed v in the +x-direction, as shown in Figure 2. The total energy dissipated by the resistor as Point S moves from x=0 to x=4.5 L is $E_{\text {new }}$ .

Figure 2

[ 7 ]

(i)

For selecting Counterclockwise with an attempt at a relevant justification

1 point

For indicating the magnetic flux through the loop is increasing in the -z-direction
OR
For indicating the magnetic field due to the induced current will be directed in the + z-direction

1 point

Example Response
Counterclockwise. The magnetic flux through the loop is increasing in the -z-direction.
Therefore, a magnetic field is induced by the current in the loop to oppose the increasing
magnetic flux. To establish this field, the current must be counterclockwise.

Scoring Note: The point can be earned if a negative sign is not included.
Example Response

\mathcal{E}=-\frac{d \Phi_{B}}{d t}=-\frac{d}{d t}(B A)

For indicating $\frac{d A}{d t}$ is 2 L v
1 point

Scoring Note: The point can be earned if a negative sign is included.
Example Response

\mathcal{E}=B \frac{d A}{d t}=2 B L v

For using Ohm's law, resulting in an expression for $I_{\mathrm{R}}$ that is consistent with the expression
1 point determined for the emf $\mathcal{E}$

Scoring Note: The point can be earned if a negative sign is included.
Example Response

I_{\mathrm{R}}=\frac{2 B L v}{R}

Example Solution

\begin{aligned} & \mathcal{E}=-\frac{d \Phi_{B}}{d t} \\ & \mathcal{E}=-\frac{d}{d t}\left(\int \vec{B} \cdot d \vec{A}\right) \\ & \mathcal{E}=-\frac{d}{d t}(B A)=-\frac{d}{d t}(-B(2 L) x) \\ & \mathcal{E}=2 B L \frac{d x}{d t} \\ & \mathcal{E}=2 B L v \end{aligned}

\begin{aligned} & I=\frac{\Delta V}{R} \\ & I_{\mathrm{R}}=\frac{\mathcal{E}}{R} \\ & I_{\mathrm{R}}=\frac{2 B L v}{R} \end{aligned}

(iii) For using a correct general expression for P 1 point
Example Responses
$P=I^{2} R \quad$ OR $\quad P=\frac{(\Delta V)^{2}}{R} \quad$ OR $\quad P=I \Delta V$
For an expression for P that is consistent with the response in part (b)(ii) that is in terms of 1 point the provided quantities only
Example Response
$P=\frac{4 B^{2} L^{2} v^{2}}{R}$
Example Solution
$P=I^{2} R$ $P=\frac{(2 B L v)^{2}}{R}$ $P=\frac{4 B^{2} L^{2} v^{2}}{R}$

(c)

Indicate whether $E_{\text {new }}$ is greater than, less than, or equal to $E_{\text {original }}$ . $E_{\text {new }}>E_{\text {original }}$ $E_{\text {new }}<E_{\text {original }}$ $E_{\text {new }}=E_{\text {original }}$
Briefly justify your answer.

The original magnetic fields are modified so that the region L<x<3.5 L contains an external uniform magnetic field of magnitude B that is directed in the -z-direction.

A new wire is connected to a resistor of resistance R to form a rigid triangular loop with base length L and height 2 L. An external force is exerted on the loop so that the loop always moves with speed v in the +x-direction, as shown in Figure 3. Point S represents the lower-leading corner of the loop.

Figure 3

[ 2 ]

For selecting $E_{\text {new }}<E_{\text {original }}$ with an attempt at a relevant justification For indicating one of the following: 1 point

- The change in magnetic flux is greater for the original scenario.

- The induced current occurs for a greater amount of time in the original scenario.

- The induced emf occurs for a greater amount of time in the original scenario.

Example Response

The change in magnetic flux is greater in the original scenario, which produces an emf and current for a longer time. Therefore, $E_{\text {new }}<E_{\text {original }}$ .

Total for part (c) 2 points

(d)

On the following axes, sketch a graph of the induced current $I_{\mathrm{T}}$ in the triangular loop as Point S moves from x=L to x=3 L.

[ 2 ]

[Maximum number: 7]

A rotating, circular, conducting loop of area A and resistance R is in an external uniform

magnetic field of magnitude B that is directed in the -z-direction. At time t=0, the magnetic

field is perpendicular to the plane of the loop, as shown in Figure 1. The loop is rotating with

constant angular speed $\omega$ and period T about the dashed line that is along the diameter of the

loop. The value of the magnetic flux through the loop as a function of time t is $\Phi=B A \cos (\omega t)$ .

Figure 1

(a)

The absolute value of the induced emf in the loop is $|\varepsilon|$ . The partially completed bar chart in

Figure 2 shows a bar that represents $|\varepsilon|$ at $t=\frac{3}{4} T$ . In Figure 2, draw bars to represent $|\varepsilon|$ at

times $t=0, \frac{1}{4} T$ , and $\frac{1}{2} T$ relative to $|\varepsilon|$ shown at $\frac{3}{4} T$ . If $|\varepsilon|=0$ , write a " 0 " in that column.

Figure 2

[ 3 ]

A	Point A1
	For drawing a bar at $\frac{1}{4} T$ that has a height of 3 units	Point A2
	For indicating that $\|\mathcal{E}\|$ is zero at times t=0 and $\frac{1}{2} T$	Point A3
	Example Response

(b)

Derive an expression for the maximum induced current in the loop in terms of A, R, B,

$\omega$ , and physical constants, as appropriate. Begin your derivation by writing a fundamental

physics principle or an equation from the reference information.

[ 4 ]

\multirow[t]{4}{*}{B} & Point B1 \\ \hline For a correct expression for the induced emf (e.g., $\mathcal{E}=B A \omega \sin (\omega t)$ ) & Point B2 \\ \hline For using Ohm's law to relate current and $\operatorname{emf}\left(\right.$ e.g., $\left.I=\frac{\mathcal{E}}{R}\right)$ & Point B3 \\ \hline For a correct expression for the absolute value of the maximum induced current (e.g., $I=\frac{B A \omega}{R}$ ) & Point B4 \\ \hline \end{tabular}

Example Response $\begin{aligned} & \mathcal{E}=-\frac{d \Phi_{\mathrm{B}}}{d t} & \mathcal{E}=-\frac{d}{d t}[B A \cos (\omega t)]=B A \omega \sin (\omega t) & \mathcal{E}_{\max }=B A \omega \end{aligned}$ $\begin{aligned} & I=\frac{\Delta V}{R}=\frac{\mathcal{E}}{R} & I_{\max }=\frac{\mathcal{E}_{\max }}{R}=\frac{B A \omega}{R} \end{aligned}$

[Maximum number: 10]

E\&M.3.
A conducting bar of mass M, length L, and negligible resistance is connected to two long vertical conducting rails of negligible resistance. The two rails are connected by a resistor of resistance R at the top. The entire apparatus is located in a magnetic field of magnitude B directed into the page, as shown in the figure above. The bar is released from rest and slides without friction down the rails.

(a)

What is the direction of the current in the resistor? Left Right

[ 1 ]

(b)

Write, but do NOT solve, a differential equation that could be used to determine the velocity of the falling bar as a function of time t.

[ 4 ]

4 points
For correctly applying Newton's second law to the motion of the bar $F_{\text {net }}=M g-F_{M}=M g-B I L$ $I=\mathcal{E} / R$
For attempting to use Faraday's law to obtain an expression for the emf in the bar $\boldsymbol{\varepsilon}=d \boldsymbol{\Phi} / d t=B L(d x / d t)$
For correctly using the expression for emf to obtain an expression for the current $I=\frac{B L(d x / d t)}{R}=\frac{B L v}{R}$ $M a=M g-B \frac{B L v}{R} L$ $a=g-\frac{B^{2} L^{2} v}{M R}$
For writing the acceleration as d v / d t $\frac{d v}{d t}=g-\frac{B^{2} L^{2} v}{M R}$

(c)

Determine an expression for the terminal velocity $v_{T}$ of the bar.

Express your answers to parts (e) and (f) in terms of $v_{T}, M, L, R, B$ , and physical constants, as appropriate.

[ 2 ]

2 points
For setting the net force at terminal velocity equal to zero 1 point

\begin{aligned} & 0=M g-\frac{B^{2} L^{2} v_{T}}{R} \\ & \frac{B^{2} L^{2} v_{T}}{R}=M g \end{aligned}

For an answer consistent with part (c) 1 point

v_{T}=\frac{M g R}{B^{2} L^{2}}

(d)

Using your differential equation from part (c), derive an expression for the speed of the falling bar v(t) as a function of time t.

[ 3 ]

3 points

Using the equation from part (c)

\begin{aligned} \frac{d v}{d t} & =g-\frac{B^{2} L^{2} v}{M R} \\ \frac{d v}{d t} & =\frac{-B^{2} L^{2}}{M R}\left(v-\frac{M R g}{B^{2} L^{2}}\right) \end{aligned}

For attempting separation of variables 1 point

For attempting to integrate with the correct limits or the correct constant of 1 point integration

\begin{aligned} & \int_{v^{\prime}=0}^{\Rightarrow} \frac{1}{\left(v^{\prime}-\frac{M R g}{B^{2} L^{2}}\right)} d v^{\prime}=\int_{t^{\prime}=0}^{t^{\prime}=t} \frac{-B^{2} L^{2}}{M R} d t^{\prime} \\ & \ln \left[v^{\prime}-\frac{M R g}{B^{2} L^{2}}\right]_{v^{\prime}=0}^{v^{\prime}=v(t)}=\frac{-B^{2} L^{2}}{M R} t \\ & \ln \left(\frac{v(t)-\frac{M R g}{B^{2} L^{2}}}{\frac{M R g}{B^{2} L^{2}}}\right)=\frac{-B^{2} L^{2}}{M R} t \end{aligned}

For a correct answer 1 point
$v(t)=\frac{M R g}{B^{2} L^{2}}\left(1-e^{-\frac{B^{2} L^{2}}{M R} t}\right)$
Using a trial solution in the differential equation and verifying its correctness is also acceptable.

[Maximum number: 1]

A long, cylindrical-shaped solenoid has length $\ell$ and N loops. The diameter of one loop is d. Which of the following expressions is equal to the inductance of the solenoid?

A

$\frac{1}{4} \frac{N \mu_{0} \pi d^{2}}{\ell}$

B

$\frac{1}{2} \frac{N \mu_{0} \pi d^{2}}{\ell}$

C

$\frac{1}{4} \frac{N^{2} \mu_{0} \pi d^{2}}{\ell}$

D

$\frac{1}{2} \frac{N^{2} \mu_{0} \pi d^{2}}{\ell}$

E

$\frac{N^{2} \mu_{0} \pi d^{2}}{\ell}$

Figure for Question 13 Electromagnetic Induction question 4 — AP Physics C: Electricity & Magnetism

AP Physics C E and M Unit 13: Electromagnetic Induction

Covered syllabus points

Covered syllabus points

13 Electromagnetic Induction question 1

Question (a)

Question (b)

Question (c)

Question (d)

13 Electromagnetic Induction question 2

Question (a)

Question (b)

13 Electromagnetic Induction question 3

Question (a)

Question (b)

Question (c)

Question (d)

13 Electromagnetic Induction question 4