A circular loop of radius r is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is :

Magnetics PYQ — JEE Main 2023 · 24 Jan · Shift 1 · Q52 with Solution

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Correct answer: C

For a circular loop of radius $r$ carrying current $I$ , the magnetic field at its center is

B_\text{center} = \frac{\mu_0 I}{2r}

At a distance $x$ from the center on the axis of the loop,

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

Here, the point is at a distance $r$ from the center on the axis, so $x=r$ .

Thus,

B_{x=r} = \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_{x=r} = \frac{\mu_0 I r^2}{2 \cdot 2\sqrt{2} \, r^3} = \frac{\mu_0 I}{4\sqrt{2} \, r}

We need

B_\text{center} : B_{x=r}

Substitute the values:

\frac{\mu_0 I}{2r} : \frac{\mu_0 I}{4\sqrt{2}r}

Cancel common terms $\mu_0 I/r$ :

\frac{1}{2} : \frac{1}{4\sqrt{2}}

Multiply both terms by $4\sqrt{2}$ :

2\sqrt{2} : 1

The correct option is:

\boxed{\text{C: } 2\sqrt{2} : 1}

Stored correct answer: C

Our derived answer is also C, so they agree.