Q. A rod of length $\ell $ rotates with a uniform angular velocity $\omega $ about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction $B$ with its direction parallel to the axis of rotation. The induced emf between the two ends of the rods is-

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A

$\frac{B \ell ^{2} \omega }{2}$

B

zero

C

$\left(B \ell ^{2} \omega / 8\right)$

D

$2B\ell ^{2}\omega $

Solution:

Length of the rod between the axis of rotation
and one end of the rod
$=L/2$ Area swept out in one rotation
$=\pi \left(\right.L/2\left(\left.\right)^{2}=\left(\pi L^{2} / 4\right)$
Angular velocity $=\omega $ radian/sec
Frequency $n$ of revolution $=\frac{\omega }{2 \pi }$
Area swept out per second $=\frac{\pi L^{2}}{4}\left(\frac{\omega }{2 \pi }\right)=\frac{L^{2} \omega }{8}$
Magnetic Induction $=B$
Rate of change of magnetic flux $=\left(B L^{2} \omega / 8\right)$
Numerical value of induced $emf=BL^{2}\omega /8.$
Numerical value of induced emf between the axis and the other end is also $\left(B L^{2} \omega / 8\right).$ These two emf are in opposite directions. Hence, the potential difference between the two ends of the rod is zero.
Hence, the correct answer is option $\left(\right.2\left.\right).$

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