AP Physics C E and M 13.3: Induced Currents and Magnetic Forces

Analyze magnetic forces on conductors carrying induced current and connect changing flux, loop resistance, velocity, and acceleration.

[Maximum number: 9]

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)

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 $\frac{d v}{d t}=g-\frac{B^{2} L^{2} v}{M R}$

(b)

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}}

(c)

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.

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Covered syllabus points

13.3 Induced Currents and Magnetic Forces question 1

Question (a)

Question (b)

Question (c)