Question

A 1 m metallic rod is rotated with an angular velocity of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5T parallel to axis exists everywhere. Derive the expression of emf developed between the centre of the ring and also calculate its value.

Answer 1

Expression for induced emf:

$$\begin{gathered} \epsilon ={\int }_0^{L}BvdL \\ ={\int }_0^{L}B\omega LdL \\ =B\omega \frac{L^2}{2} \\ =\frac{B\omega L^2}{2} \end{gathered}$$

Calculation of the value of emf:

$$\begin{gathered} \epsilon =\frac{1}{2}\times 0.5\times 400\times 1^2 \\ =100\mathrm{V} \end{gathered}$$