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(B) The magnetic flux $\phi$ through the coil is given by $\phi = B \cdot A \cdot \cos(	heta)$. Since the plane of the coil is normal to the magnetic

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$\frac{3 AB}{4}$

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(B) The magnetic flux $\phi$ through the coil is given by $\phi = B \cdot A \cdot \cos(	heta)$. Since the plane of the coil is normal to the magnetic field,$	heta = 0^\circ$ and $\cos(0^\circ) = 1$,so $\phi = BA$. The initial magnetic flux is $\phi_i = B \cdot A$. The magnetic induction is reduced to $25 \%$ of its initial value,so the final magnetic induction is $B_f = 0.25 B = \frac{B}{4}$. The final magnetic flux is $\phi_f = \frac{B}{4} \cdot A = \frac{BA}{4}$. The change in magnetic flux is $\Delta \phi = \phi_i - \phi_f = BA - \frac{BA}{4} = \frac{3BA}{4}$. According to Faraday's law of induction,the magnitude of the induced e.m.f. is $e = \left| \frac{\Delta \phi}{\Delta t} ight|$. Given $\Delta t = 1 \ s$,we have $e = \frac{3BA/4}{1} = \frac{3AB}{4} \ V$.

$A$ coil having effective area $A$ is held with its plane normal to a magnetic field of induction $B$. The magnetic induction is quickly reduced to $25 \%$ of its initial value in $1 \ s$. The e.m.f. induced in the coil (in volt) will be

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