A metallic rod of length 'L' is rotated with an angular speed of 'ω' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :

Solution

Physics sir · 4 steps

Each bit of the rod moves at a different speed, so add up Bv dr along the length with v = ω r.

1. Speed varies along rod. A small piece dr at distance r from the pivot moves at v = ω r. The pivot end is at rest and the free end is the fastest, so you cannot use BLv with one single speed.
2. EMF of one piece. The velocity of the piece, the rod and B are mutually perpendicular. So that piece gives dε = B v dr = Bω r dr.

   ε = ∫₀^L Bω r dr = 1/2Bω L²

3. Quick check. The same result comes from the average speed (0 + ω L)/2, giving ε = BL × (ω L)/2. Or use flux: the rod sweeps area 1/2L²ω per second, so ε = B × 1/2L²ω.
4. So. The induced emf is 1/2BL²ω, option (b).

The idea to keep. For a rod rotating about one end, the emf is 1/2Bω L². Use the average speed of the rod, or the area it sweeps per second. EMF is always linear in B.

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