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(B) The induced electromotive force $(e)$ is given by Faraday's law: $e = -N \frac{d\phi}{dt} = -NA \frac{dB}{dt}$.Since $A = \pi r_c^2$ (where $r_c$

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(B) The induced electromotive force $(e)$ is given by Faraday's law: $e = -N \frac{d\phi}{dt} = -NA \frac{dB}{dt}$.Since $A = \pi r_c^2$ (where $r_c$ is the coil radius),$e \propto N$.The resistance of the wire is $R = ho \frac{L}{A_w} = ho \frac{N(2\pi r_c)}{\pi r^2}$,where $r$ is the wire radius.Thus,$R \propto \frac{N}{r^2}$.The power dissipated is $P = \frac{e^2}{R} \propto \frac{N^2}{N/r^2} = N r^2$.Given $N_2 = 4N_1$ and $r_2 = r_1/2$,the new power $P_2$ is:$P_2 \propto (4N_1) 	imes (r_1/2)^2 = 4N_1 	imes (r_1^2/4) = N_1 r_1^2 = P_1$.Therefore,the electrical power dissipated remains the same.

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

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