A coil of resistance 8 Omega,number of turns | vedclass

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Question

(B) The magnetic flux $\phi$ through a coil is given by $\phi = NBA \cos 	heta$,where $	heta$ is the angle between the area vector and the magnetic

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(B) The magnetic flux $\phi$ through a coil is given by $\phi = NBA \cos 	heta$,where $	heta$ is the angle between the area vector and the magnetic field. Given that the plane of the coil makes an angle of $\frac{\pi}{6}$ with the magnetic field,the angle $	heta_1$ between the area vector and the magnetic field is $	heta_1 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} = 60^\circ$. Initial flux $\phi_1 = NBA \cos(60^\circ) = 250 	imes 2 	imes (120 	imes 10^{-4}) 	imes 0.5 = 3 \ Wb$. When the plane of the coil is parallel to the magnetic field,the area vector is perpendicular to the magnetic field,so $	heta_2 = 90^\circ$. Final flux $\phi_2 = NBA \cos(90^\circ) = 0 \ Wb$. The change in flux is $\Delta \phi = |\phi_2 - \phi_1| = 3 \ Wb$. The induced $EMF$ is $\varepsilon = \frac{\Delta \phi}{\Delta t} = \frac{3}{100 	imes 10^{-3}} = 30 \ V$. The induced current is $I = \frac{\varepsilon}{R} = \frac{30}{8} = 3.75 \ A$.

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