Question 89 of 153

A metallic rod of length $L$ is rotated with an angular speed of $\omega$ normal to a uniform magnetic field $B$ about an axis passing through one end of rod as shown in figure. The induced emf will be :
.

A$\dfrac{1}{2}{B}^{2}{L}^{2}\omega$B$\dfrac{1}{4}B{L}^{2}\omega$C$\dfrac{1}{2}B{L}^{2}\omega$D$\dfrac{1}{4}{B}^{2}L\omega$

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Answer: (C) $\dfrac{1}{2}B{L}^{2}\omega$

Linear velocity at the distance $x$ from the centre can be written as, $v=\omega x$.
Therefore, induced EMF due to small element will be,
$d\epsilon =Bvdx=B\left(\omega x\right)dx$
Integrating both sides, we get
$\int d\epsilon =\int B\left(\omega x\right)dx$
$\Rightarrow \epsilon =B\omega {\int }_{0}^{L}xdx=\dfrac{B\omega {L}^{2}}{2}$