JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
The magnitude of magnetic induction at mid point O due to current arrangement as shown in Fig will be
- (a) (μ₀ I)/(π a)
- (b) (μ₀ I)/(4 π a)
- (c) (μ₀ I)/(2 π a)
- (d) 0
Answer and solution
Answer (a) (μ₀ I)/(π a)
Solution
Physics sir · 4 steps
The horizontal pieces point straight at O and give nothing, and the two vertical half-wires add up.
- Horizontal parts. Wires AB and DE both lie on the line through O. For every element, dl⃗ is parallel to r⃗, so dl⃗ × r⃗ = 0 and they give zero field at O.
- Vertical half-wires. BC and ET are semi-infinite wires. Each one starts at the foot of the perpendicular from O. O is the mid point of BE, so its distance from each wire is a/2.
Bₒₙₑ = (μ₀ I)/4π(a/2) = (μ₀ I)/(2π a)
- Check directions. In BC the current flows down, and O is to its right. In ET the current flows up, and O is to its left. The right-hand rule gives out of the page for both, so the two fields add.
- So. B_O = 2 × (μ₀ I)/(2π a) = (μ₀ I)/(π a), which is option (a).
The idea to keep. A straight wire whose line passes through the point gives zero field there. A semi-infinite wire seen from its end gives half of (μ₀ I)/(2π d). Always fix d and the direction before adding.
Answer from the NTA final answer key. Solution by PrepWiser.