Question

A metallic disc of radius 0.3 m is rotating with a constant angular speed of \( 60 \, \text{rad s}^{-1} \) in a plane perpendicular to a uniform magnetic field of \( 5 \times 10^{-2} \, \text{T} \). The emf induced between a point on the rim and the centre of the disc is

• A. 0.06 V
• B. 0.612 V
• C. 1.35 V
• D. 0.135 V

Hint:

The emf induced in a conducting rod of length \(L\) rotating with angular speed \( \omega \) about one end in a uniform magnetic field \(B\) perpendicular to the plane of rotation is \( \mathcal{E} = \frac{1}{2}B\omega L^2 \). A rotating disc can be considered as an infinite number of such rods (radii) rotating. The emf between the centre and the rim is the same as for a single rod of length R (the radius of the disc). Ensure all units are in SI.

Answer 1

The Correct Option is D

Radius of the disc \( R = 0.
3 \) m.
Angular speed \( \omega = 60 \, \text{rad s}^{-1} \).
Magnetic field strength \( B = 5 \times 10^{-2} \, \text{T} \).
The plane of the disc is perpendicular to the magnetic field, so the field lines are parallel to the axis of rotation.
The emf induced between the centre and a point on the rim of a rotating metallic disc in a uniform magnetic field perpendicular to its plane is given by: \[ \mathcal{E} = \frac{1}{2} B \omega R^2 \] Substitute the given values: \[ \mathcal{E} = \frac{1}{2} (5 \times 10^{-2} \, \text{T}) (60 \, \text{rad s}^{-1}) (0.
3 \, \text{m})^2 \] \[ \mathcal{E} = \frac{1}{2} \times 5 \times 10^{-2} \times 60 \times (0.
09) \, \text{V} \] \[ \mathcal{E} = \frac{1}{2} \times 5 \times 60 \times 0.
09 \times 10^{-2} \, \text{V} \] \[ \mathcal{E} = 1 \times 5 \times 30 \times 0.
09 \times 10^{-2} \, \text{V} \] \[ \mathcal{E} = 150 \times 0.
09 \times 10^{-2} \, \text{V} \] \( 150 \times 0.
09 = 150 \times \frac{9}{100} = \frac{15 \times 9}{10} = \frac{135}{10} = 13.
5 \).
\[ \mathcal{E} = 13.
5 \times 10^{-2} \, \text{V} = 0.
135 \, \text{V} \] This matches option (4).