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 \sqrtx: 1 . The value of x is \underline\hspace2cm

Magnetics PYQ — JEE Main 2023 · 8 Apr · Shift 2 · Q65 with Solution

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Correct answer: 8

For a circular current-carrying coil of radius $r$ , the magnetic field at the centre is

B_c = \frac{\mu_0 I}{2r}

At a distance $x$ from the centre on the axis of a circular coil, the magnetic field is

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

Here, the point is at distance $r$ from the centre, so put $x=r$ :

B_a = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}}

Now,

(2r^2)^{3/2} = 2^{3/2}r^3 = 2\sqrt{2}\,r^3

So,

B_a = \frac{\mu_0 I r^2}{2 \cdot 2\sqrt{2} \, r^3} = \frac{\mu_0 I}{4\sqrt{2} \, r}

\frac{B_c}{B_a} = \frac{\mu_0 I}{2r} \div \frac{\mu_0 I}{4\sqrt{2}r}

= \frac{\mu_0 I}{2r} \cdot \frac{4\sqrt{2}r}{\mu_0 I} = 2\sqrt{2}

Thus,

B_c : B_a = 2\sqrt{2} : 1

Given,

\sqrt{x} : 1 = 2\sqrt{2} : 1

So,

\sqrt{x} = 2\sqrt{2}

Squaring both sides,

x = (2\sqrt{2})^2 = 8

\boxed{8}