Question

An equilateral triangular conducting frame is rotated with angular velocity $\omega$ in uniform magnetic field $B$ as shown. Side of triangle is $l$. Choose the correct options

• A. $V_a-V_c=0$
• B. $V_a-V_c=\frac{B\omega l^2}{2}$
• C. $V_a-V_b=\frac{B\omega l^2}{2}$
• D. $V_c-V_b=-\frac{B\omega l^2}{2}$

Answer 1

Answer: (A), (C)

$V_a-V_b=\frac{B\omega l^2}{2}$
$V_a-V_c=0$

Velocity of each element between points a and c has components both along the element and perpendicular to it. However the direction of perpendicular components of velocity reverses for elements in the other half of the segment ac. Thus, the direction of potential difference created reverses for other half of the segment ac with equal magnitude. Hence, $V_a=V_c$.

Consider small elements of length $dl$ in the segment ab. Velocity of each element perpendicular to the element $=I\omega$

Thus, $V_a-V_b={\int }_0^{l}B\left(l\omega \right)dl=\frac{1}{2}B\omega l^2$