A wire is bent to form a semicircle of radius $a$ . The wire rotates about its one end with angular velocity $ω$ . Axis of rotation being perpendicular to the plane of the semicircle. In the space, a uniform magnetic field of induction $B$ exists along the axis of rotation, as shown in figure. Then

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Answer:

A

Solution:

$e = \frac{1}{2} B ω l^{2}$

$e = \frac{1}{2} B ω {(2 a)}^{2}$

$e = 2 B ω a^{2}$

TestHub · Accepted Answer

e = 1 2 B ω l 2 e = 1 2 B ω 2 a 2 e = 2 B ω a 2

Physics - EMI/AC Question with Solution | TestHub

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