JEE Main 2023 · 8 April · evening · Single correct · +4 / −1
An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when moves with a velocity of:
- (a) 20 ms⁻¹
- (b) 2 ms⁻¹
- (c) 3.2 ms⁻¹
- (d) 0.5 ms⁻¹
Answer and solution
Answer (b) 2 ms⁻¹
Solution
Physics sir · 3 steps
Motional emf is just Blv: put the length in metres and solve for v.
- Motional emf. A rod of length l moving with speed v through a field B, with rod, field and velocity all at right angles, has an emf e = Blv across its ends.
v = e/Bl
- Put in numbers. Convert the length first: l = 10 cm = 0.1 m. Then Bl = 0.4 × 0.1 = 0.04 T m, and v = 0.08/0.04 = 2 m s⁻¹.
- So. The rod must move at 2 m s⁻¹, which is option (b).
The idea to keep. For a rod moving across a field, e = Blv. Always change cm to m before you substitute, because a slip there shifts the answer by a factor of 10.
Answer from the NTA final answer key. Solution by PrepWiser.