A cycle wheel of radius $0.5 \mathrm{~m}$ is rotated with constant angular velocity of $10 \mathrm{rad} / \mathrm{s}$ in a region of magnetic field of $0.1 \mathrm{~T}$ which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is

• A

  $0.25 \mathrm{~V}$
• Correct answer
  B

  $0.125 \mathrm{~V}$
• C

  $0.5 \mathrm{~V}$
• D

  zero

Solution

When a conducting disc or wheel of radius $r$ rotates with constant angular velocity of $\omega$ about its axis in a uniform magnetic field perpendicular to its plane and parallel to its axis of rotation, then,
Induced emf is given by, $\mathrm{e}=\frac{1}{2}{\mathrm{~B} \omega \mathrm{r}^2}^2$ Here, $B=0.1 \mathrm{~T}, \omega=10 \mathrm{rad} / \mathrm{s}, \mathrm{r}=0.5 \mathrm{~m}$ Substituting these values in Eq. (i), we get $$ \therefore \quad \mathrm{e}=\frac{1}{2} \times 0.1 \times 10 \times(0.5)^2=\frac{1}{8} \mathrm{~V}=0.125 \mathrm{~V} $$

Hence, the correct answer is option: 'B'.