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JEE Main 2024 (30 Jan, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2024 (30 Jan, Shift 1)

Q44JEE Main 2024 · 30 JAN · SHIFT 1Medium

Biot-Savart Law & Circular Coils

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

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Answer: (b)

Each loop gives $\dfrac{\mu_0 I}{2a}$ at the common centre.
The loops are in perpendicular planes, so the fields are perpendicular.
$B = \sqrt{2}\cdot\dfrac{\mu_0 I}{2a} = \dfrac{\mu_0 I}{\sqrt{2}a}$
So the answer is (b).

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