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(C) The initial magnetic flux through the coil is $\Phi_i = B \cdot A$. Since the magnetic field is reduced by $25 \%$,the final magnetic field is $B_

Vedclass · Accepted Answer

$\frac{AB}{8}$

Vedclass · Answer

(C) The initial magnetic flux through the coil is $\Phi_i = B \cdot A$. Since the magnetic field is reduced by $25 \%$,the final magnetic field is $B_f = B - 0.25B = 0.75B = \frac{3}{4}B$. The final magnetic flux is $\Phi_f = \frac{3}{4}B \cdot A$. The change in magnetic flux is $\Delta \Phi = \Phi_f - \Phi_i = \frac{3}{4}BA - BA = -\frac{1}{4}BA$. The induced e.m.f. is given by Faraday's Law: $\varepsilon = -\frac{\Delta \Phi}{\Delta t}$. Given $\Delta t = 2 \ s$,we have $\varepsilon = -\left( \frac{-\frac{1}{4}BA}{2} \right) = \frac{BA}{8}$. Thus,the magnitude of the induced e.m.f. is $\frac{AB}{8}$.

$A$ coil having effective area $A$ is held with its plane normal to a magnetic field of induction $B$. The magnetic induction is quickly reduced by $25 \%$ of its initial value in $2 \ s$. The e.m.f. induced across the coil will be:

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