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(A) Magnetic flux is given by $\phi = BA \cos(	heta)$.The time taken for one revolution is $T = 8 \ s$. The time taken for a $90^{\circ}$ rotation is

Vedclass · Accepted Answer

$\frac{3}{2} \ V; \frac{3}{2} \ V; -\frac{3}{2} \ V; -\frac{3}{2} \ V$

Vedclass · Answer

(A) Magnetic flux is given by $\phi = BA \cos(	heta)$.The time taken for one revolution is $T = 8 \ s$. The time taken for a $90^{\circ}$ rotation is $\Delta t = \frac{T}{4} = \frac{8}{4} = 2 \ s$.The average emf is $e = -\frac{\Delta \phi}{\Delta t} = -\frac{\phi_2 - \phi_1}{\Delta t}$.$i. 0^{\circ}$ to $90^{\circ}$: $e = -\frac{BA \cos(90^{\circ}) - BA \cos(0^{\circ})}{2} = -\frac{0 - (0.3 	imes 10)}{2} = \frac{3}{2} \ V$.$ii. 90^{\circ}$ to $180^{\circ}$: $e = -\frac{BA \cos(180^{\circ}) - BA \cos(90^{\circ})}{2} = -\frac{-3 - 0}{2} = \frac{3}{2} \ V$.$iii. 180^{\circ}$ to $270^{\circ}$: $e = -\frac{BA \cos(270^{\circ}) - BA \cos(180^{\circ})}{2} = -\frac{0 - (-3)}{2} = -\frac{3}{2} \ V$.$iv. 270^{\circ}$ to $360^{\circ}$: $e = -\frac{BA \cos(360^{\circ}) - BA \cos(270^{\circ})}{2} = -\frac{3 - 0}{2} = -\frac{3}{2} \ V$.

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