Question

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

• A. 0.25 V
• B. 0.125 V
• C. 0.5 V
• D. zero

Answer 1

The Correct Option is B

To determine the electromotive force (EMF) generated between the center of the wheel and its rim, we can use the formula for EMF induced in a rotating disc in a magnetic field. This can be given by:

EMF = \dfrac{1}{2} B \omega r^2

Where:

• B is the magnetic field strength, B = 0.1 \, T.
• \omega is the angular velocity, \omega = 10 \, rad/s.
• r is the radius of the wheel, r = 0.5 \, m.

Let's substitute these values into the formula:

EMF = \dfrac{1}{2} \times 0.1 \, T \times 10 \, rad/s \times (0.5 \, m)^2

Simplifying inside the brackets:

= \dfrac{1}{2} \times 0.1 \times 10 \times 0.25

Further simplifying:

= \dfrac{1}{2} \times 0.25

= 0.125 \, V

Thus, the EMF generated between the center and the rim of the wheel is 0.125 \, V.

Hence, the correct answer is:

0.125 V