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JEE Main 2013 (9 Apr, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2013 (9 Apr, Shift 1)

Q97JEE Main 2013 · 9 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

An electric current is flowing through a circular coil of radius R. The ratio of the magnetic field at the centre of the coil and that at a distance $2\sqrt{2}R$ from the centre of the coil and on its axis is:

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Answer: (b)

$\dfrac{B_c}{B_{axis}} = \dfrac{(R^2+x^2)^{3/2}}{R^3}$ with $x = 2\sqrt{2}R$ .
$R^2+8R^2 = 9R^2 \Rightarrow (9R^2)^{3/2} = 27R^3$
So the answer is (b) $27$ .

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