Question

A conducting rod AB of length L is rotating with a constant angular
velocity in the plane of the paper about an axis perpendicular to the plane and passing through a point ‘C’ as shown in the diagram. A uniform magnetic field of magnitude B is perpendicular to the plane of the paper and into the plane. The velocities of A and B are perpendicular to the length of the rod AB. The emf induced between the ends of the rod is $\frac{1}{2}BVl\times \frac{1}{n}$. Then n is ___

Answer 1

$v=v_B=\frac{L}{3}\omega and2v=v_A=\frac{2L\omega }{3}{V}_A-v_0=\frac{1}{2}B\omega {\left(\frac{2L}{3}\right)}^2{V}_B-V_0=\frac{1}{2}B\omega {\left(\frac{L}{3}\right)}^2{V}_A-V_B=\frac{1}{2}B\omega L^2\times \frac{3}{9}=\frac{1}{2}B\times \frac{3v}{L}\times L^2\times \frac{1}{3}=\frac{1}{2}BvL\therefore n=1$