Question

A metallic ring of radius $r$ with a uniform metallic spoke of negligible mass and length $r$ is rotated about its axis with angular velocity $\omega$ in a perpendicular uniform magnetic field $B$ as shown in figure. The central end of the spoke is connected to the rim of the wheel through a resistor $R$ as shown. The resistor does not rotate, its one end is always at the center of the ring and the other end is always in contact with the ring. A force $F$ as shown is needed to maintain constant angular velocity of the wheel. $F$ is equal to (the ring and the spoke has zero resistance)

• A. $\frac{B^2{\omega r}^2}{8R}$
• B. $\frac{B^2{\omega r}^2}{2R}$
• C. $\frac{B^2{\omega r}^3}{2R}$
• D. $\frac{B^2{\omega r}^3}{4R}$
• E. answer required

Answer 1

Answer: (D)

$\frac{B^2{\omega r}^3}{4R}$

Emf induced across the metallic spoke=${\int }_0^{r}B\left(\omega r\right)dr=\frac{1}{2}B\omega r^2$

This is the potential difference across the resistor. Thus the current in the resistor and the spoke=$\frac{B\omega r^2}{2R}$.

Hence force on the spoke due to external magnetic field=$Bir=\frac{B^2\omega r^3}{2R}$.This force acts on the mid point of the spoke.

Thus to balance the torque due to this force, a force at the end of the spoke must be $\frac{B^2\omega r^3}{4R}$.