Question

If a conducting rod of length $4l$ is rotated about at point $O$ in a uniform magnetic field $B$ directed into the paper and $DE=l$, $EA=3l$, then which of the folowing is true ?

• A. $V_A-V_E=\frac{9}{2}B\omega l^2$
• B. $V_E-V_A=\frac{9}{2}B\omega l^2$
• C. $V_D-V_E=\frac{B\omega l^2}{2}$
• D. $V_A-V_E=4B\omega l^2$

Answer 1

The correct option is D $V_A-V_E=4B\omega l^2$

Since, $V_E-V_D=\frac{B\omega l}{2}\times l$
$V_E-V_D=\frac{B\omega l^2}{2}$ ....(i)
$V_E-V_A=\frac{B\omega 3l}{2}\times 3l=\frac{9B\omega l^2}{2}$ ....(ii)
Subtracting equation (i) from equation (ii), we get,
$V_A-V_E=4B\omega l^2$
$\Rightarrow V_A>V_E$