Question

A rod of length $L$ rotates in the form of a conical pendulum with an angular velocity $\omega$ about its axis as shown in figure. The rod makes an angle $\theta$ with the axis. The magnitude of the motional emf developed across the two ends of the rod is

• A. $\frac{1}{2}B\omega L^2$
• B. $\frac{1}{2}B\omega L^2{\tan }^2\theta$
• C. $\frac{1}{2}B\omega L^2{\cos }^2\theta$
• D. $\frac{1}{2}B\omega L^2{\sin }^2\theta$
• E. answer required

Answer 1

Answer: (D)

$\frac{1}{2}B\omega L^2{\sin }^2\theta$

Figure shows the Top view, in which the rod of length $r$ is revolving in a circle about an axis passing through its one end.

From top view, effective length $=r=l\sin \theta$

Induced emf $e=\frac{1}{2}B\omega r^2=\frac{1}{2}B\omega l^2{\sin }^2\theta$