A round flat metal coil has 180 turns and negligible resistance. It is connected in a series circuit with a 17- $\Omega$ resistor, with nothing else in the circuit. You measure that a $6.0 - \mathrm { A }$ current flows through the resistor when a magnetic field through the coil, perpendicular to its area, is changing at $3.0 \mathrm {~T} / \mathrm { s }$. What is the radius of the coil?
A) $0.25 \mathrm {~m}$
B) $0.014 \mathrm {~m}$
C) $0.043 \mathrm {~m}$
D) $0.54 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The magnetic flux through the coil is given by:$$\Phi = BA = \pi r^2 B$$The emf induced in the coil is given by:$$\varepsilon = -\frac{d\Phi}{dt} = -\pi r^2 \frac{dB}{dt}$$The current in the circuit is given by:$$I = \frac{\varepsilon}{R} = -\frac{\pi r^2}{R} \frac{dB}{dt}$$Solving for the radius of the coil, we get:$$r = \sqrt{\frac{IR}{\pi \frac{dB}{dt}}} = \sqrt{\frac{(6.0 \mathrm {~A})(17 \mathrm {~\Omega})}{\pi (3.0 \mathrm {~T} / \mathrm {~s})}} = 0.25 \mathrm {~m}$$

A Round Flat Metal Coil Has 180 Turns and Negligible Ω\OmegaΩ

A Round Flat Metal Coil Has 180 Turns and Negligible $\Omega$