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JEE Main 2023 (8 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2023 (8 Apr, Shift 2)

Q61JEE Main 2023 · 8 APR · SHIFT 2Medium

Biot-Savart Law & Circular Coils

The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of coil on its axis is $\sqrt{x} : 1$ . The value of $x$ is __________

Numerical answer type. Work it out, then open the answer.

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Answer: 8

$\dfrac{B_c}{B_{axis}} = \dfrac{(r^2+x^2)^{3/2}}{r^3}$ with $x = r$ .
$(2r^2)^{3/2} = 2\sqrt{2}\,r^3 \Rightarrow$ ratio $= 2\sqrt{2} = \sqrt{8}$
So $x = 8$ .

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