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(A) When a conducting rod of length '$l$' rotates about one of its ends in a uniform magnetic field '$B$' with an angular velocity '$\omega$',the indu

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$\frac{e}{B \pi l^2}$

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(A) When a conducting rod of length '$l$' rotates about one of its ends in a uniform magnetic field '$B$' with an angular velocity '$\omega$',the induced e.m.f. '$e$' is given by the formula:$e = \frac{1}{2} B \omega l^2$We know that the angular velocity '$\omega$' is related to the frequency of rotation '$n$' (revolutions per second) by the relation:$\omega = 2 \pi n$Substituting this value of '$\omega$' into the e.m.f. equation:$e = \frac{1}{2} B (2 \pi n) l^2$$e = B \pi n l^2$To find the number of revolutions per second '$n$',we rearrange the formula:$n = \frac{e}{B \pi l^2}$

$A$ metal rod of length '$l$' rotates about one of its ends in a plane perpendicular to a magnetic field of induction '$B$'. If the e.m.f. induced between the ends of the rod is '$e$',then the number of revolutions made by the rod per second is

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