JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Magnetic Effects of Current

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Two insulated circular loop A and B radius ' a ' carrying a current of ' I ' in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be :

(A) $\frac{2 μ _{0} I}{a}$
(B) $\frac{μ _{0} I}{2 a}$
(C) $\frac{μ _{0} I}{2 a}$ ✓ Correct
(D) $\frac{2 μ _{0} I}{a}$

Answer: (C)

Solution

Each loop produces a field of magnitude

B = \frac{μ _{0} I}{2 a}

at the common centre, directed perpendicular to its own plane. The two loops are set at right angles to each other, so the two field contributions are mutually perpendicular:

B_{n e t} = B^{2} + B^{2} = 2 \frac{μ _{0} I}{2 a} = \frac{μ _{0} I}{2 a}

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option

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