A rectangular coil of single turn,having area | vedclass

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Question

(D) The magnetic flux through the coil is given by $\phi = BA \cos 	heta$,where $	heta$ is the angle between the area vector and the magnetic field.

Vedclass · Accepted Answer

$\frac{2\omega BA}{\pi}$

Vedclass · Answer

(D) The magnetic flux through the coil is given by $\phi = BA \cos 	heta$,where $	heta$ is the angle between the area vector and the magnetic field.Initially,the plane of the coil is perpendicular to the field,so the area vector is parallel to the field. Thus,$	heta = 0^{\circ}$ and $\phi_i = BA \cos 0^{\circ} = BA$.After rotating through $90^{\circ}$,the plane of the coil is parallel to the field,so the area vector is perpendicular to the field. Thus,$	heta = 90^{\circ}$ and $\phi_f = BA \cos 90^{\circ} = 0$.The change in flux is $\Delta \phi = |\phi_f - \phi_i| = BA$.The time taken to rotate through $90^{\circ}$ is $\Delta t = \frac{\pi/2}{\omega} = \frac{\pi}{2\omega}$.The average induced $e.m.f.$ is given by $\varepsilon_{avg} = \frac{\Delta \phi}{\Delta t} = \frac{BA}{\pi / (2\omega)} = \frac{2BA\omega}{\pi}$.

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