JEE Main 2025 · 8 April · evening · Single correct · +4 / −1
Figure shows a current carrying square loop ABCD of edge length is ‘a’ lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is
- (a) (2μ₀ I)/(3π a)
- (b) (√2μ₀ I)/(3π a)
- (c) (3πμ₀ I)/√2a
- (d) (μ₀ I)/(2π a)
Answer and solution
Answer (b) (√2μ₀ I)/(3π a)
Solution
Physics sir · 4 steps
The two halves carry unequal currents round the centre in opposite senses, so only their difference counts.
- Split the current. ABC and ADC are in parallel, so the current divides inversely as resistance. ABC (resistance r) carries i₁ = 2I/3 and ADC (resistance 2r) carries i₂ = I/3.
- Field of one side. Each side is a finite wire at perpendicular distance a/2 from the centre. Its ends subtend 45° on either side, so the field from a side carrying current i is:
B_(side) = (μ₀ i)/(4π (a/2))(sin 45° + sin 45°) = (√2 μ₀ i)/(2π a)
- Fix the directions. Along A to B to C the current goes clockwise round the centre, so its field points into the page. Along A to D to C it goes anticlockwise, so its field points out of the page. The straight leads at A and C lie on a line through the centre and give no field there.
- So. Each path has two sides, so B = 2 × (√2 μ₀)/(2π a)(2I/3 - I/3) = (√2 μ₀ I)/(3π a), which is option (b).
The idea to keep. When current splits around a loop, the two branches circle the centre in opposite senses. Find each branch current from the resistances, then subtract their fields.
Answer from the NTA final answer key. Solution by PrepWiser.