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Energy conservation - class-XII
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A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of 40 rad $s^{-1}$ in a uniform horizontal magnetic field of magnitude $2 \times 10^{-2}$ T. If the coil form a closed loop of resistance 8 $\Omega$, then the average power loss
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A
$2.07\times 10^{-3}$W
💡 Explanation:
Here, $r = 6 cm = 6 \times 10^{-2}m,$
$N = 20, \omega = 40 \, rad \, s^{-1}$
$B = 2 \times 10^{-2} T, R = 8 \Omega$
Maximum emf induced, $\epsilon \, = \, NAB \omega$
$ = \, N(\pi r^2)B \omega$
$ = 20 \times \pi \times (6 \times 10^{-2})^2 \times 2 \times 10^{-2} \times 40 \, = \, 0.18 V$
Average value of emf induced over a full cycle $\epsilon _{av}\, = \, 0$
Maximum value of current in the coil, $ I \, = \, \dfrac{\varepsilon }{R} = \dfrac{0.18}{8} = 0.023 \,A$
Average power dissipated, $ P \, = \, \dfrac{\varepsilon I}{2} \, = \, \dfrac{0.18 \times 0.023}{2} \, = \, 2.07 \times 10^{-3} W$