Question

A rod of length $10cm$ made up of conducting and non-conducting material (shaded part is non-conducting). The rod is rotated with constant angular velocity $10rad/sec$ about point $O$, in constant and uniform magnetic field of $2$ Tesla as shown in figure. The induced emf between the point $A$ and point $B$ of rod will be

• A. $0.029v$
• B. $0.1v$
• C. $0.051v$
• D. $0.064v$

Answer 1

The correct option is C $0.051v$

Here,

$\omega =10rad/s$

$L=10cm$

Let us consider an elemental length of the conductor at a distance r from o. Thus the motional emf experienced by the elementary length of the conductor will be

$d\epsilon =Bvdr$

OA= L= 0.1 m

OB= L-3= 0.07 m

But. $v=r\omega$

$d\epsilon =B\omega rdr$

Thus the total potential difference between A and B is given by

$d\epsilon$= ${\int }_{0B}^{OA}B\omega rdr$

$\epsilon$ = ${\int }_{0.1}^{0.007}B\omega rdr$

on solving above integration we will get

$\epsilon =0.051V$

Hence option (c) is correct answer