JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 3)
At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $$\omega $$ about an axis along their common diameter. Calculate the emf induced in their smaller coil after a time t of its start of rotation.
$${{{\mu _o}{\rm I}} \over {2\,R}}$$ $$\omega $$ $$\pi $$ r2 sin$$\omega $$ t
$${{{\mu _o}{\rm I}} \over {4\,R}}$$ $$\omega $$ $$\pi $$ r2 sin$$\omega $$ t
$${{{\mu _o}{\rm I}} \over {4\,R}}$$ $$\omega $$ r2 sin$$\omega $$ t
$${{{\mu _o}{\rm I}} \over {2\,R}}$$ $$\omega $$ r2 sin$$\omega $$ t
Explanation
We know that electric flux $$\phi = \overrightarrow B \,.\,\overrightarrow A $$
$$ \Rightarrow \phi = BA\cos \omega t$$
Now, $$B = {{{\mu _0}} \over 2}{I \over R}$$ is magnetic field due to circular coil of radius R and $$A = \pi {r^2}$$ is area of circular coil of radius r. Therefore,
$$\phi = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t$$
Now induced emf $$\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t} \right)$$
$$ \Rightarrow \varepsilon = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\sin \omega t$$
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