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A rod of length $l$ rotates in the form of a conical pendulum with an angular velocity w about its axis as shown in figure. The rod makes an angle $θ$ with the axis. The magnitude of the motional e.m.f. developed the two ends of the rod is;

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$\frac{1}{2} B w l^{2}$

$\frac{1}{2} B w l^{2} \tan^{2} θ$

$\frac{1}{2} B w l^{2} \cos^{2} θ$

$\frac{1}{2} B w l^{2} \sin^{2} θ$

answer is D.

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Detailed Solution

$e = \int (\vec{V} \times \vec{B}) \vec{d l} \to e = \frac{1}{2} B L^{2} ω \sin^{2} θ$

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