A circular loop made of thin copper wire of m | vedclass

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(B) Let $r$ be the radius of the loop and $A_w$ be the cross-sectional area of the wire.The mass of the loop is $m = (2 \pi r) A_w d$,so $A_w = \frac{

Vedclass · Accepted Answer

$\frac{m}{4 \pi \rho d}\left(\frac{dB}{dt}\right)$

Vedclass · Answer

(B) Let $r$ be the radius of the loop and $A_w$ be the cross-sectional area of the wire.The mass of the loop is $m = (2 \pi r) A_w d$,so $A_w = \frac{m}{2 \pi r d}$.The resistance of the loop is $R = \rho \frac{l}{A_w} = \rho \frac{2 \pi r}{A_w} = \rho \frac{2 \pi r}{m / (2 \pi r d)} = \frac{4 \pi^2 r^2 \rho d}{m}$.The magnetic flux through the loop is $\phi = B \cdot \pi r^2$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = -\pi r^2 \frac{dB}{dt}$.The induced current is $I = \frac{|\varepsilon|}{R} = \frac{\pi r^2 (dB/dt)}{4 \pi^2 r^2 \rho d / m} = \frac{m}{4 \pi \rho d} \left(\frac{dB}{dt}\right)$.

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