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(D) The induced electromotive force $(EMF)$ is given by $\varepsilon = -N A \frac{dB}{dt}$.The resistance of the coil is $R = \rho \frac{\ell}{A_w}$,w

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Doubled

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(D) The induced electromotive force $(EMF)$ is given by $\varepsilon = -N A \frac{dB}{dt}$.The resistance of the coil is $R = ho \frac{\ell}{A_w}$,where $\ell = N(2\pi r)$ is the total length of the wire and $A_w = \pi r_w^2$ is the cross-sectional area of the wire.Power dissipated is $P = \frac{\varepsilon^2}{R} = \frac{(N A \frac{dB}{dt})^2}{ho \frac{N(2\pi r)}{\pi r_w^2}} = \frac{N^2 A^2 (dB/dt)^2 \cdot r_w^2}{2 ho N r} = \frac{N A^2 (dB/dt)^2 r_w^2}{2 ho r}$.Given: $N' = N/2$ and $r_w' = 2r_w$.$P' = \frac{(N/2) A^2 (dB/dt)^2 (2r_w)^2}{2 ho r} = \frac{(N/2) A^2 (dB/dt)^2 (4r_w^2)}{2 ho r} = 2 	imes \frac{N A^2 (dB/dt)^2 r_w^2}{2 ho r} = 2P$.Thus,the power dissipated is doubled.

$A$ coil is placed in a time-varying magnetic field. If the number of turns in the coil were to be halved and the radius of the wire doubled,the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short-circuited.)

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