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AP Physics C Electricity : Understanding Electromagnetic Induction

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

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Example Questions

Example Question #50 : Electricity And Magnetism Exam

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 $1 *10^{-2}m^2$ $\displaystyle 1 *10^{-2}m^2$, what is the time interval over which the field is increased?

Possible Answers:

$5.0 \mu s$ $\displaystyle 5.0 \mu s$

$15.0 \mu s$ $\displaystyle 15.0 \mu s$

$1.0 \mu s$ $\displaystyle 1.0 \mu s$

$3.3 \mu s$ $\displaystyle 3.3 \mu s$

$2.5 \mu s$ $\displaystyle 2.5 \mu s$

Correct answer:

$3.3 \mu s$ $\displaystyle 3.3 \mu s$

Explanation:

Relevant equations:

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

Given:

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

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

$\epsilon = 15 V$ $\displaystyle \epsilon = 15 V$

Plug in (disregarding direction of emf):

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

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

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Understanding Electromagnetic Induction

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

All AP Physics C Electricity Resources

Example Questions

Example Question #50 : Electricity And Magnetism Exam

All AP Physics C Electricity Resources

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