Question

A metal rod of length l, moving with an angular velocity $\omega$ has velocity of its centre as v. A uniform magnetic field of strength B exists perpendicular to plane of paper. Potential difference between points A and B at the instant shown in figure is:

• A. $Bvl$
• B. $Bvl+\frac{1}{2}B\omega l^2$
• C. $B\omega l-\frac{1}{2}B\omega l^2$
• D. $Bvl+B\omega (\frac{l}{2}{)}^2$

Answer 1

The correct option is A $Bvl$

P is labelled on the rod as shown in the figure.

Its linear velocity is given by:

$v_P=v_{COM}-d\omega$

$v_P=v-\frac{v}{\omega }\omega =0$

Hence, points above P are moving rightwards while points below P are moving leftwards. Hence, polarities of induced emf above and below are as shown in the figure.

Now,

Motional emf due to rotating rod is given by:
${\epsilon }_1=V_A-V_P=\frac{1}{2}B\omega {(\frac{l}{2}+\frac{v}{\omega })}^2{\epsilon }_2=V_B-V_P=\frac{1}{2}B\omega {(\frac{l}{2}-\frac{v}{\omega })}^2|V_A-V_B|={\epsilon }_1-{\epsilon }_2=\frac{1}{2}B\omega \left(\frac{2lv}{\omega }\right)=Blv$