JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 22)
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s$$-$$1 in a uniform horizontal magnetic field of 3.0 $$\times$$ 10$$-$$2 T. The maximum emf induced the coil will be ................. $$\times$$ 10$$-$$2 volt (rounded off to the nearest integer)
Answer
60
Explanation
Maximum emf $$\varepsilon = N\,\omega AB$$
N = 20, $$\omega$$ = 50, B = 3 $$\times$$ 10$$-$$2 T
$$\varepsilon $$ = 20 $$\times$$ 50 $$\times$$ $$\pi$$ $$\times$$ (0.08)2 $$\times$$ 3 $$\times$$ 10$$-$$2 = 60.28 $$\times$$ 10$$-$$2
Rounded off to nearest integer = 60
N = 20, $$\omega$$ = 50, B = 3 $$\times$$ 10$$-$$2 T
$$\varepsilon $$ = 20 $$\times$$ 50 $$\times$$ $$\pi$$ $$\times$$ (0.08)2 $$\times$$ 3 $$\times$$ 10$$-$$2 = 60.28 $$\times$$ 10$$-$$2
Rounded off to nearest integer = 60
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