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A conducting disc of conductivity $σ$ has radius a and thickness $δ$ . A uniform magnetic field $\vec{B}$ is applied in a direction perpendicular to the plane of the disc. The
cylindrical region of the magnetic field is coaxial with the disc. If the magnetic field changes with time at the rate of dB/dt, then the power dissipated in the disc due to the
induced current is?

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$\frac{π δ σ a^{4}}{8} {(\frac{d B}{d t})}^{2}$

$\frac{π δ σ a^{4}}{12} {(\frac{dB}{dt})}^{2}$

$\frac{π δ σ a^{4}}{4} {(\frac{dB}{dt})}^{2}$

$\frac{π δ σ a^{4}}{6} {(\frac{d B}{d t})}^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} ε = π r^{2} (\frac{dB}{dt}) \\ dR = \frac{2 π r}{σ δ dr} \\ dP = \frac{ε^{2}}{dR} = \frac{π δ σ}{2} {(\frac{dB}{dt})}^{2} r^{3} dr \\ P = \frac{π δ σ a^{4}}{8} {(\frac{dB}{dt})}^{2} \end{array}$

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