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(B) Given: Number of turns $N = 500$,Area $A = 0.1\,m^2$,Magnetic field $B = 0.2\,T$,Resistance $R = 50\,\Omega$,Time $\Delta t = 0.1\,s$.Initial angl

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(B) Given: Number of turns $N = 500$,Area $A = 0.1\,m^2$,Magnetic field $B = 0.2\,T$,Resistance $R = 50\,\Omega$,Time $\Delta t = 0.1\,s$.Initial angle $	heta_1 = 0^o$ (since the coil is perpendicular to the field,the area vector is parallel to the field).Final angle $	heta_2 = 180^o$ (after rotation).The change in magnetic flux is $\Delta \phi = BA(\cos 	heta_2 - \cos 	heta_1)$.The induced charge is given by $\Delta Q = \frac{|\Delta \phi|}{R} = \frac{NBA}{R} |\cos 	heta_2 - \cos 	heta_1|$.Substituting the values: $\Delta Q = \frac{500 	imes 0.2 	imes 0.1}{50} |\cos 180^o - \cos 0^o|$.$\Delta Q = \frac{10}{50} |-1 - 1| = 0.2 	imes 2 = 0.4\,C$.

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