A 20 cm long metallic rod is rotated with 210 rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.2 T parallel to the axis exists everywhere. The emf developed between the centre and the ring is ____________ mV.
Take π=22/7

Solution

Physics sir · 4 steps

Each bit of the rod moves at a different speed, so the average speed of half of ωL gives ½BωL².

1. Why a half. A small piece of rod at distance r from the axis moves with speed v = ω r, so its emf is Bω r dr. Adding these from r = 0 to r = L gives the full emf between the centre and the ring.

   ε = ∫₀^L Bω r dr = 1/2Bω L²

2. Angular speed. ω = 210 × 2π/60 = 7π = 7 × 22/7 = 22 rad s⁻¹. The length is L = 0.2 m, so L² = 0.04 m².
3. Put in numbers. ε = 1/2 × 0.2 × 22 × 0.04 = 0.088 V.
4. So. The emf between the centre and the ring is 0.088 V, that is 88 mV.

The common slip: 176. Leaving out the factor 1/2 and using Bω L² gives 176 mV, not 88. The tip moves at ω L, but the centre does not move at all.

The idea to keep. For a rod spinning about one end in a field along the axis, use ε = 1/2Bω L². Always convert rpm to rad/s and cm to m first.

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