A circular coil of 500 turns encloses an area | vedclass

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Question

(C) The magnetic flux linked with the coil is given by $\phi = NBA \cos 	heta$.Initially,the magnetic field is perpendicular to the plane of the coil

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(C) The magnetic flux linked with the coil is given by $\phi = NBA \cos 	heta$.Initially,the magnetic field is perpendicular to the plane of the coil,so the angle between the area vector and the magnetic field is $	heta_1 = 0^o$. Thus,$\phi_1 = NBA \cos 0^o = NBA$.After rotating by $90^o$ about its diameter,the plane of the coil becomes parallel to the magnetic field,so the angle between the area vector and the magnetic field is $	heta_2 = 90^o$. Thus,$\phi_2 = NBA \cos 90^o = 0$.The change in magnetic flux is $\Delta \phi = |\phi_2 - \phi_1| = NBA$.The induced charge $q$ is given by $q = \frac{\Delta \phi}{R} = \frac{NBA}{R}$.Substituting the given values: $N = 500$,$B = 0.25 \,Wb/m^2$,$A = 0.04 \,m^2$,and $R = 25 \,\Omega$.$q = \frac{500 	imes 0.25 	imes 0.04}{25} = \frac{5}{25} = 0.2 \,C$.

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