A metallic rod of length ‘L’ is rotated with an angular speed of ‘ $ω$ ’ 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 :

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$\frac{1}{4} B^{2} L ω$

$\frac{1}{2} B L^{2} ω$

$\frac{1}{4} B L^{2} ω$

$\frac{1}{2} B^{2} L^{2} ω$

answer is A.

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Detailed Solution

$\begin{array}{l} de = B \cdot dL \cdot v \\ de = B \cdot dx \cdot xω \\ \int de = \int_{0}^{L} Bωx \cdot dx \\ e = Bω \frac{L^{2}}{2} \\ e = \frac{1}{2} {BL}^{2} ω \end{array}$

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