A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 π rad s⁻¹ about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π=3.14)

Solution

Physics sir · 4 steps

Think of the disc as many spokes cutting the field; one spoke from centre to rim gives you the answer.

1. Disc as spokes. Picture the disc as many radial rods joined in parallel between the axis and the rim. Each rod rotates and cuts the field, so the EMF across the disc equals the EMF across one rod.
2. Speed varies. A small piece dr at distance r moves with speed v = ω r, so it gives dε = Bω r dr. Adding these from 0 to R gives the result below.

   ε = ∫₀^R Bω r dr = 1/2Bω R²

3. Put in numbers. With B = 0.4 T, ω = 10π rad s⁻¹ and R = 0.2 m: ε = 1/2 × 0.4 × 10π × 0.04 = 0.08π = 0.08 × 3.14.
4. So. The potential difference between the axis and the rim is ε = 0.2512 V, which is option (c).

The idea to keep. For a rod or disc rotating in a field, the speed grows from zero at the axis to ω R at the end. Use the average speed (ω R)/2, which gives ε = 1/2Bω R².

Answer from the NTA final answer key. Solution by PrepWiser.