JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1

A wheel with 30 metallic spokes each of

0.7 \mathrm{~m}

long is rotated with a speed of

120 \mathrm{~rev} / \mathrm{min}

, in a plane normal to the horizontal component of earth's magnetic field

\mathrm{H_E}

at a place. If

\mathrm{H_E}=0.8\,G

at the place, what is the induced emf between the axle and the rim of the wheel?
Given

1 \mathrm{~G}=10^{-4} \mathrm{~T}

View answer and explanation

Correct option: A

2.46 \times 10^{-4}

volt

Option A (

2.46 \times 10^{-4}

volt) is correct because the induced emf is calculated using the formula

E = B \pi R^2 v

, where

B

is the magnetic field,

R

is the length of the spoke, and

v

is the frequency of rotation.

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