Question

A metallic ring of mass m and radius r with a uniform metallic spoke of same mass m and length r is rotated about its axis with angular velocity $\omega$ in a perpendicular uniform magnetic field $B$ as shown. 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 centre 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 have zero resistance):

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

Answer 1

Answer: (D)

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

Answer 2

Step 1: Compute the motional emf between centre and rim along the spoke:

$$\mathcal E=\int_0^r B(\omega r')\,dr'=\tfrac12B\omega r^2.$$

Step 2: The current through the resistor is

$$I=\frac{\mathcal E}{R}=\frac{B\omega r^2}{2R}.$$

Step 3: The torque arises from the force on the radial spoke carrying current $I$. A small element $dr$ at radius $r'$ feels force $dF=I\,B\,dr'$, giving torque

$$\tau=\int_0^r (I\,B\,r')\,dr' =I\,B\,\frac{r^2}{2} =\frac{B\omega r^2}{2R}\,B\,\frac{r^2}{2} =\frac{B^2\omega r^4}{4R}.$$

Step 4: The required tangential force $F$ at radius $r$ is

$$F=\frac{\tau}{r}=\frac{B^2\omega r^3}{4R}.$$