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(B) According to Faraday's law of electromagnetic induction,the magnitude of induced electromotive force (emf) is given by:$|\varepsilon| = N \frac{\D

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$30$

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(B) According to Faraday's law of electromagnetic induction,the magnitude of induced electromotive force (emf) is given by:$|\varepsilon| = N \frac{\Delta \phi}{\Delta t}$Since the magnetic field is perpendicular to the area,the magnetic flux $\phi = B \cdot A$. Thus,the change in flux is $\Delta \phi = A(B_2 - B_1)$.Substituting the given values:$N = 200$,$A = 0.15 \ m^2$,$B_1 = 0.2 \ T$,$B_2 = 0.6 \ T$,$\Delta t = 0.4 \ s$$|\varepsilon| = \frac{N \cdot A \cdot (B_2 - B_1)}{\Delta t}$$|\varepsilon| = \frac{200 	imes 0.15 	imes (0.6 - 0.2)}{0.4}$$|\varepsilon| = \frac{200 	imes 0.15 	imes 0.4}{0.4}$$|\varepsilon| = 200 	imes 0.15 = 30 \ V$Therefore,the induced emf is $30 \ V$.

$A$ coil having $200$ turns has a surface area of $0.15 \ m^2$. $A$ magnetic field of strength $0.2 \ T$ applied perpendicular to this changes to $0.6 \ T$ in $0.4 \ s$,then the induced emf in the coil is . . . . . . $V$.

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