A rod of length l rotates with a small but uniform angular velocity $ω$ about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The PD between the centre of the rod and an end is :

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Zero

$\frac{1}{8} {ωBl}^{2}$

$\frac{1}{2} {ωBl}^{2}$

${ωBl}^{2}$

answer is B.

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Detailed Solution

$\begin{aligned} e = \int d e \\ = \int_{0}^{l / 2} B ω x d x \\ = B ω \int_{0}^{\frac{l}{2}} x d x \\ = B ω {[\frac{x^{2}}{2}]}_{0}^{\frac{l}{2}} \\ e = \frac{1}{8} ω B l^{2} \end{aligned}$

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