AP Physics C E and M 13.2: Electromagnetic Induction

Relate changing magnetic flux to induced emf and current using Faraday’s law, Lenz’s law, and right-hand rules for direction.

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

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Covered syllabus points

13.2 Electromagnetic Induction question 1

Question (a)

Question (b)