Question

A rod AB of length L is placed on a frictionless horizontal surface and pivoted at a point P which lies at a distance L/3 from the end A. A uniform magnetic field of intensity B directed along the vertical direction is set up. If the rod is made to rotate about a vertical axis passing through P, the potential difference that develops between the ends A and B is:

• A. $\frac{1}{3}B\omega L^2$
• B. $\frac{2}{3}B\omega L^2$
• C. $\frac{1}{6}B\omega L^2$
• D. $\frac{5}{6}B\omega L^2$

Answer 1

Answer: (C)

$\frac{1}{6}B\omega L^2$

Answer 2

The rod AB rotates about pivot P. The induced EMF in a rotating rod segment of length $l$ about its end in a perpendicular magnetic field $B$ is $\frac{1}{2}B\omega l^2$.

Segment PA has length $L_A = L/3$. EMF across PA is $\varepsilon_{PA} = \frac{1}{2}B\omega (L/3)^2 = \frac{1}{18}B\omega L^2$.

Segment PB has length $L_B = 2L/3$. EMF across PB is $\varepsilon_{PB} = \frac{1}{2}B\omega (2L/3)^2 = \frac{1}{2}B\omega (4L^2/9) = \frac{2}{9}B\omega L^2$.

The end further from the pivot is at a higher potential. So, $V_A - V_P = \varepsilon_{PA}$ and $V_B - V_P = \varepsilon_{PB}$.

The potential difference between A and B is $V_A - V_B = (V_A - V_P) - (V_B - V_P) = \frac{1}{18}B\omega L^2 - \frac{2}{9}B\omega L^2 = \frac{1}{18}B\omega L^2 - \frac{4}{18}B\omega L^2 = -\frac{3}{18}B\omega L^2 = -\frac{1}{6}B\omega L^2$.

The magnitude of the potential difference is $\frac{1}{6}B\omega L^2$.