Question

A copper rod $A B$ of length $l$, pivoted at one end $A$, rotates at constant angular velocity $\omega$, at right angles to a uniform magnetic field of induction $B$. The emf, developed between the mid point $C$ of the rod and end $B$ is

• A. $\frac{B \omega l^{2}}{8}$
• B. $\frac{3}{4} B \omega l^{2}$
• C. $\frac{B \omega l^{2}}{4}$
• D. $\frac{3}{8} B \omega l^{2}$

Answer 1

Answer: (D)

$\varepsilon=\int\limits_{1 / 2}^{1} B \omega x d x$
$=B \omega\left(\frac{x^{2}}{2}\right)_{1 / 2}^1$
$=\frac{B \omega}{2}\left(l^{2}-l^{2} / 4\right)$
$=\frac{3}{8} B \omega l^{2}$