A uniform circular loop of radius a and resistance R is pulled at a constant velocity v out of a region of a uniform magnetic field whose magnitude is B . The plane of the loop and the velocity are bo

Question

A uniform circular loop of radius a and resistance $R$ is pulled at a constant velocity $v$ out of a region of a uniform magnetic field whose magnitude is $B$ . The plane of the loop and the velocity are both perpendicular to $B$ . Then the electrical power in the circular loop at the instant when the arc (of the circular loop) outside the region of magnetic field subtends an angle $\frac{π}{3}$ at the centre of the loop is

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Answer:

A

Solution:

The induced emf in the circular loop $= B (PQ) v = Bav$

Therefore electrical power consumed by the circular loop

$= \frac{{(emf)}^{2}}{R} = \frac{{(Bav)}^{2}}{R}$

TestHub · Accepted Answer

The induced emf in the circular loop = B PQ v = Bav Therefore electrical power consumed by the circular loop = emf 2 R = Bav 2 R

Physics - EMI/AC Question with Solution | TestHub

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