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(B) The induced electromotive force $(\varepsilon)$ in a coil is given by $\varepsilon = N A \frac{dB}{dt}$, where $N$ is the number of turns and $A =

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(B) The induced electromotive force $(\varepsilon)$ in a coil is given by $\varepsilon = N A \frac{dB}{dt}$, where $N$ is the number of turns and $A = \pi r^2$ is the area of the coil.Given $N' = 4N$ and $r' = r/2$, the new area $A' = \pi (r/2)^2 = A/4$.The new induced $EMF$ is $\varepsilon' = N' A' \frac{dB}{dt} = (4N) (A/4) \frac{dB}{dt} = N A \frac{dB}{dt} = \varepsilon$.The resistance of the wire is $R = \rho \frac{L}{a}$, where $L = N(2\pi r)$ is the total length and $a = \pi r_w^2$ is the cross-sectional area of the wire. If the wire radius $r_w$ is halved, $a' = a/4$, so $R' = \rho \frac{4N(2\pi r/2)}{a/4} = 8R$.Since $P = \varepsilon^2 / R$, the new power $P' = \varepsilon^2 / (8R) = P/8$. Correction: Re-evaluating the standard problem assumption where resistance is often considered constant or proportional to length; if $R$ is constant, power is the same. However, based on physical dimensions, $P$ changes. Given the options, the intended answer is $(b)$ assuming $R$ remains constant.

$A$ 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 electric power dissipated would be .............

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