Question

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.

Hint:

Think of the induced EMF as coming from the average velocity of the rod. The tip moves at \(v = L\omega\) and the pivot at \(v=0\). The average velocity is \(\frac{0 + L\omega}{2} = \frac{1}{2}L\omega\). So the EMF is just \(B \times L \times v_{avg} = B L (\frac{1}{2}L\omega) = \frac{1}{2}B\omega L^2\).

Answer 1

Consider a conducting bar of length \(L\) rotating with a constant angular velocity \(\omega\) about a pivot at one end. The rotation occurs in a uniform magnetic field \(B\) which is perpendicular to the plane of rotation.
Consider a small element of the bar of length \(dr\) at a distance \(r\) from the pivot.
The linear velocity of this element is \(v = r\omega\).
The motional e.m.f. induced in this small element \(dr\) is given by: \[ d\mathcal{E} = B v \, dr \] Substituting \(v = r\omega\): \[ d\mathcal{E} = B (r\omega) \, dr \] To find the total e.m.f. induced across the entire length of the bar, we integrate this expression from the pivot (\(r=0\)) to the other end (\(r=L\)): \[ \mathcal{E} = \int_{0}^{L} d\mathcal{E} = \int_{0}^{L} B\omega r \, dr \] Since \(B\) and \(\omega\) are constant: \[ \mathcal{E} = B\omega \int_{0}^{L} r \, dr \] \[ \mathcal{E} = B\omega \left[ \frac{r^2}{2} \right]_0^L \] \[ \mathcal{E} = B\omega \left( \frac{L^2}{2} - 0 \right) \] \[ \mathcal{E} = \frac{1}{2} B \omega L^2 \] This is the expression for the rotational e.m.f. induced in the conducting bar.