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The diagram given below shows a solenoid carrying time varying current $l = l_{0} t$ . On the axis of the solenoid, a ring has been placed. The mutual inductance of the ring and the solenoid is M and the self inductance of the ring is L. If the resistance of the ring is R then maximum current which can flow through the ring is

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$\frac{M I_{0}}{R}$

$\frac{(M + L) l_{0}}{R}$

$\frac{(2 M + L) l_{0}}{R}$

$\frac{(2 M - L) I_{0}}{R}$

answer is B.

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Detailed Solution

$ϕ = M I = M I_{0} t$

Hence, induced emf $e = \frac{d ϕ}{d t} = M I_{0}$

Current through ring $= \frac{e}{R} = \frac{M I_{0}}{R}$

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