An elliptical loop having resistance R,semi-m | vedclass

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(A) The magnetic flux through the loop at time $t$ is $\phi = BA \cos(\omega t)$,where $A = \pi ab$ is the area of the elliptical loop.The induced ele

Vedclass · Accepted Answer

$\frac{\pi^{2} a^{2} b^{2} B^{2} \omega^{2}}{2 R}$

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(A) The magnetic flux through the loop at time $t$ is $\phi = BA \cos(\omega t)$,where $A = \pi ab$ is the area of the elliptical loop.The induced electromotive force $(EMF)$ is given by Faraday's law: $\epsilon = -\frac{d\phi}{dt} = AB\omega \sin(\omega t)$.The instantaneous power loss due to Joule heating is $P = \frac{\epsilon^{2}}{R} = \frac{(AB\omega \sin(\omega t))^{2}}{R} = \frac{A^{2}B^{2}\omega^{2}}{R} \sin^{2}(\omega t)$.The average power loss over a complete cycle is $P_{avg} = \langle P \rangle = \frac{A^{2}B^{2}\omega^{2}}{R} \langle \sin^{2}(\omega t) \rangle$.Since the average value of $\sin^{2}(\omega t)$ over a cycle is $\frac{1}{2}$,we have $P_{avg} = \frac{A^{2}B^{2}\omega^{2}}{R} \cdot \frac{1}{2}$.Substituting $A = \pi ab$,we get $P_{avg} = \frac{(\pi ab)^{2} B^{2} \omega^{2}}{2R} = \frac{\pi^{2} a^{2} b^{2} B^{2} \omega^{2}}{2R}$.

An elliptical loop having resistance $R$,semi-major axis $a$,and semi-minor axis $b$ is placed in a magnetic field as shown in the figure. If the loop is rotated about the $x$-axis with angular frequency $\omega$,the average power loss in the loop due to Joule heating is:

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