JEE Main 2024 · 5 April · morning · Single correct · +4 / −1
Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is:
- (a) μₒ/(2 π) · b²/a
- (b) (μₒ π a²)/(2 b)
- (c) (μ₀ π b²)/(2 a)
- (d) μ₀/(2 π) · a²/b
Answer and solution
Answer (b) (μₒ π a²)/(2 b)
Solution
Physics sir · 3 steps
Pass current through the big loop: its field is uniform over the small loop, so flux is field times area.
- Choose the easy direction. Mutual inductance is the same whichever loop carries the current. Pass a current I through the large loop B, because its field at the centre is easy to find.
- Field over loop A. At the centre of B the field is (μ₀ I)/2b. Since b ≫ a, the small loop A stays very close to the centre, so this field is almost uniform over all of A.
φ_A = (μ₀ I)/2b × π a² ⇒ M = φ_A/I = (μ₀ π a²)/2b
- So. The mutual inductance is (μ₀ π a²)/2b, which is option (b).
The idea to keep. M is the same either way round, so send current through the coil whose field is uniform over the other coil. Then the flux is just field times area.
Answer from the NTA final answer key. Solution by PrepWiser.