Question

A disc of radius 0.1 m is rotating with a frequency 10 rev/s in a normal magnetic field of strength 0.1 T. Net induced emf is

• A. $2\pi\times10^{-2}V$
• B. $\pi\times10^{-2}V$
• C. $\frac{\pi}{2}\times10^{-2}V$
• D. None of these

Hint:

A disc of radius 0.1 m is rotating with a frequency 10 rev/s in a normal magnetic field of strength 0.1 T. Net induced emf is

Answer 1

The Correct Option is B

To find the net induced emf in the rotating disc, we must first understand the concept of motional emf in a magnetic field.

Here, the disc rotates in a uniform magnetic field. The induced emf (\(E\)) can be calculated using the formula for motional emf in a rotating disc:

\(E = \frac{1}{2} B \omega R^2\)

where:

• \(B\) is the magnetic field strength (0.1 T in this case).
• \(\omega\) is the angular speed in radians per second (rad/s).
• \(R\) is the radius of the disc (0.1 m).

First, we need to convert the frequency from revolutions per second (rev/s) to angular speed (rad/s):

\(\omega = 2\pi \times \text{frequency}\)

Substituting the frequency in the given formula:

\(\omega = 2\pi \times 10 = 20\pi \text{ rad/s}\)

Now we can substitute these values into the formula for the emf:

\(E = \frac{1}{2} \times 0.1 \times 20\pi \times (0.1)^2\)

This simplifies to:

\(E = \frac{1}{2} \times 0.1 \times 20\pi \times 0.01\)

\(E = 0.01\pi \text{ V}\)

Therefore, the correct induced emf is \(\pi \times 10^{-2} \text{ V}\), which matches the option "

$\pi\times10^{-2}V$