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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 magnetic field and the area vector

Vedclass · Accepted Answer

$2: 3$

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(D) The magnetic flux through the coil is given by $\Phi = BA \cos(	heta)$,where $	heta$ is the angle between the magnetic field and the area vector. Initially,the field is perpendicular to the plane,so $	heta_0 = 0^{\circ}$.In step $A$,the coil rotates by $60^{\circ}$,so the final angle is $	heta_A = 60^{\circ}$. The change in flux is $\Delta \Phi_A = BA(\cos 60^{\circ} - \cos 0^{\circ}) = BA(0.5 - 1) = -0.5 BA$. The induced emf is $\varepsilon_A = -\frac{\Delta \Phi_A}{t} = \frac{0.5 BA}{t}$.In step $B$,the coil rotates by another $120^{\circ}$ in the same sense,so the final angle is $	heta_B = 60^{\circ} + 120^{\circ} = 180^{\circ}$. The change in flux is $\Delta \Phi_B = BA(\cos 180^{\circ} - \cos 60^{\circ}) = BA(-1 - 0.5) = -1.5 BA$. The induced emf is $\varepsilon_B = -\frac{\Delta \Phi_B}{2t} = \frac{1.5 BA}{2t} = \frac{0.75 BA}{t}$.The ratio of the induced emf is $\frac{\varepsilon_A}{\varepsilon_B} = \frac{0.5 BA / t}{0.75 BA / t} = \frac{0.5}{0.75} = \frac{2}{3}$.

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