Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is \underline\hspace2cm .

Magnetics PYQ — JEE Main 2021 · 25 Feb · Shift 1 · Q50 with Solution

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Correct answer: D

For a circular coil of radius $R$ , the magnetic field at a point on its axis at distance $x$ from the centre is

B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}

So,

B \propto \frac{1}{(R^2 + x^2)^{3/2}}

At $x_1 = 0.05\,\text{m}$ and $x_2 = 0.2\,\text{m}$ ,

\frac{B_1}{B_2} = 8

Using the proportionality,

\frac{B_1}{B_2} = \frac{(R^2 + x_2^2)^{3/2}}{(R^2 + x_1^2)^{3/2}} = 8

Substitute values:

\frac{(R^2 + 0.2^2)^{3/2}}{(R^2 + 0.05^2)^{3/2}} = 8

Take both sides to the power $2/3$ :

\frac{R^2 + 0.04}{R^2 + 0.0025} = 8^{2/3}

Since

8^{2/3} = (2^3)^{2/3} = 2^2 = 4

So,

\frac{R^2 + 0.04}{R^2 + 0.0025} = 4

R^2 + 0.04 = 4(R^2 + 0.0025)

R^2 + 0.04 = 4R^2 + 0.01

0.03 = 3R^2

R^2 = 0.01

R = 0.1\,\text{m}

$0.1\,\text{m}$ corresponds to Option D.

Stored correct answer: D

Derived answer: D

They match.