Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 1/√2 Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm / s, induced emf between points A and E is ________ mV .

Solution

Physics sir · 3 steps

Join A to E with a straight line: only its length perpendicular to the velocity makes emf.

1. Effective length. For any bent wire moving in a uniform field, the emf between its ends equals Bvℓ. Here ℓ is the length of the straight line from A to E, measured perpendicular to v⃗. Arms AB and DE lie along v⃗, so they add nothing.
2. Find the separation. BC and CD each make 45° with the velocity, so each has a vertical part of 10 sin 45° = 10/√2 cm. A and E are therefore 20/√2 = 10√2 cm = 0.1√2 m apart, perpendicular to v⃗.

   ε = Bvℓ_⊥ = 1/√2 × 0.1 × 0.1√2 = 0.01 V

3. So. The induced emf between A and E is 0.01 V, that is 10 mV.

The common slip: 14. Using the full 20 cm of arms BC and CD instead of their components perpendicular to v⃗ gives about 14 mV, not 10.

The idea to keep. For a bent conductor in a uniform field, ignore the shape. Join the two ends with a straight line and use its length perpendicular to the velocity in ε = Bvℓ.

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