A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s⁻¹. If the horizontal component of earth's magnetic field is 0.2 × 10⁻⁴ T, then emf induced between the two ends of the conductor is :

Solution

Physics sir · 4 steps

A rod spinning about one end gives 1/2Bω L². Just pick the field component that cuts it.

1. Which field counts. The rod turns in a vertical east-west plane, so the normal to that plane points north-south. The horizontal component of the earth's field also points north-south, so it is perpendicular to the plane and the rod cuts it. The vertical component lies in the plane and induces nothing.
2. Why the half. A point at distance r from the pivot moves with speed ω r. The speed grows from 0 at the pivot to ω L at the tip, so you add up Bω r dr along the rod.

   ε = ∫₀^L Bω r dr = 1/2Bω L²

3. Put in numbers. ε = 1/2 × (0.2 × 10⁻⁴) × 5 × 1² = 0.5 × 10⁻⁴ V = 5 × 10⁻⁵ V.
4. So. The emf between the ends is 50 μV, option (b).

The idea to keep. For a rod rotating about one end, ε = 1/2Bω L². Use only the field component perpendicular to the plane of rotation.

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