Question

A copper rod $ab$ 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. $\frac{3B\omega L^2}{8}$
• B. $\frac{B\omega L^2}{8}$
• C. $\frac{B\omega L^2}{4}$
• D. $\frac{3B\omega L^2}{4}$

Answer 1

The correct option is A $\frac{3B\omega L^2}{8}$

$Solution:$ Motional EMF induced in a conductor of length L moving in magnetic field is given by equation

$EMF=vBL$

now lets take a small part of length "dx"which is at distance"x" from the hinge point

the speed of the point is given by

$v=x\omega$

now the motional emf in that small part of the rod will be

$EMF={}_{x=L/2}^{x=L}\int vBdx$

$EMF={}^{x=L}{}_{x=L/2}\int x\omega Bdx$

$EMF=\frac{(L^2-(L/2{)}^2)}{2}\omega B$

$EMF=\frac{3B\omega L^2}{8}$

so above is the induced EMF from centre of

rod to end of rod.

$So,the$ $correct$ $option:A$