A magnetic field, confined in a cylindrical region of radius R, is changing at the rate of $4 {Ts}^{- 1}$ . A conducting rod PQ of length $\sqrt{\frac{3}{2}} R$ is placed in the region as shown. The induced emf across the rod will be
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\frac{π R^{2}}{6}

\frac{R^{2}}{6} (π + 6)

\frac{R^{2}}{24} (π + 6)

\frac{π R^{2}}{24}

answer is D.

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Detailed Solution

$\tan 60 ° = \frac{PQ}{CP}$

$\Rightarrow \sqrt{3} CP = \sqrt{\frac{3}{2}} R$

$\Rightarrow CP = \frac{R}{\sqrt{2}} and CQ = \sqrt{\frac{{3R}^{2}}{2} + \frac{R^{2}}{2}} = \sqrt{2} R$

$∴ cos ϕ = \frac{CP}{CS} = \frac{R}{\sqrt{2} R} = \frac{1}{\sqrt{2}}, ϕ = 45 °$

Total flux, $ϕ = B (area of CPS + CST)$

$Σ \frac{dB}{dt} [\frac{R^{2}}{4} + \frac{π R^{2}}{24}] = 4 (\frac{π + 6}{24}) R^{2} = \frac{R^{2}}{6} π + 6$

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