In a uniform magnetic field of induction B,a | vedclass

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Question

(B) The area of the semicircle is $A = \frac{1}{2} \pi r^{2}$.The magnetic flux linked with the loop at time $t$ is $\phi = B A \cos(\omega t) = B \le

Vedclass · Accepted Answer

$\frac{(B \pi r^{2} \omega)^{2}}{8 R}$

Vedclass · Answer

(B) The area of the semicircle is $A = \frac{1}{2} \pi r^{2}$.The magnetic flux linked with the loop at time $t$ is $\phi = B A \cos(\omega t) = B \left(\frac{1}{2} \pi r^{2}\right) \cos(\omega t)$.The induced electromotive force $(EMF)$ is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt} = \frac{1}{2} B \pi r^{2} \omega \sin(\omega t)$.The instantaneous power generated is $P(t) = \frac{\varepsilon^{2}}{R} = \frac{(\frac{1}{2} B \pi r^{2} \omega \sin(\omega t))^{2}}{R} = \frac{B^{2} \pi^{2} r^{4} \omega^{2} \sin^{2}(\omega t)}{4 R}$.The mean power over a full period is $\langle P \rangle = \frac{B^{2} \pi^{2} r^{4} \omega^{2}}{4 R} \langle \sin^{2}(\omega t) \rangle$.Since the mean value of $\sin^{2}(\omega t)$ over a period is $\frac{1}{2}$,we have $\langle P \rangle = \frac{B^{2} \pi^{2} r^{4} \omega^{2}}{4 R} \cdot \frac{1}{2} = \frac{(B \pi r^{2} \omega)^{2}}{8 R}$.

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