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(D) The induced electromotive force (emf) in a coil is given by Faraday's law of induction: $|\varepsilon| = N \frac{|\Delta \phi|}{\Delta t}$.Here,$N

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$150.0$

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(D) The induced electromotive force (emf) in a coil is given by Faraday's law of induction: $|\varepsilon| = N \frac{|\Delta \phi|}{\Delta t}$.Here,$N = 500$,$A = 0.15 \ m^2$,$B_1 = 0.2 \ T$,$B_2 = 1.0 \ T$,and $\Delta t = 0.4 \ s$.Since the magnetic field is perpendicular to the area,the magnetic flux $\phi = BA$.Therefore,$|\varepsilon| = N \frac{A(B_2 - B_1)}{\Delta t}$.Substituting the values:$|\varepsilon| = \frac{500 	imes 0.15 	imes (1.0 - 0.2)}{0.4}$.$|\varepsilon| = \frac{500 	imes 0.15 	imes 0.8}{0.4}$.$|\varepsilon| = 500 	imes 0.15 	imes 2 = 150 \ V$.

$A$ conducting coil having $500$ turns has a cross-sectional area of $0.15 \ m^2$. $A$ magnetic field of strength $0.2 \ T$ linked perpendicular to this area changes to $1.0 \ T$ in $0.4 \ s$. The induced emf produced in the coil will be . . . . . . $V$.

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