Q.21 A conducting bar is rotating with constant angular speed around a pivot at one end in a uniform magnetic field perpendicular to its plane of rotation. Obtain an expression for the rotational e.m.f. induced between the ends of the bar.
Best answer
Answer: \(e = \dfrac{1}{2}B\omega l^2\)
Let a conducting bar OP of length \(l\) rotate about one end O with constant angular speed \(\omega\) in a plane perpendicular to a uniform magnetic field \(B\).
Method (flux swept): In one rotation (time \(T = 2\pi/\omega\)) the bar sweeps a circle of area \(\pi l^2\), so the flux it cuts is \(\Phi = B\pi l^2\). The magnitude of induced e.m.f. is
\[e = \frac{\Phi}{T} = \frac{B\pi l^2}{2\pi/\omega} = \frac{1}{2}B\omega l^2\]Alternative (motional e.m.f.): A small element \(dr\) at distance \(r\) from O moves with speed \(v = r\omega\) perpendicular to \(B\), so its e.m.f. is \(de = Bv\,dr = B\omega r\,dr\). Integrating along the bar,
\[e = \int_0^l B\omega r\,dr = \frac{1}{2}B\omega l^2\]If the bar makes \(f\) rotations per second, \(\omega = 2\pi f\) and \(e = \pi B f l^2\).
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