1. Magnetic field at the center of a circular loop

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

$B_1 = \frac{\mu_0 I}{2R}$

2. Magnetic field on the axis of a circular loop

At a point on the axis at distance $x$ from the center, the magnetic field is

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

Here, we are given

$x = \sqrt{3}R$

So,

$B_2 = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}}$

$= \frac{\mu_0 I R^2}{2(4R^2)^{3/2}}$

Now,

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

Therefore,

$B_2 = \frac{\mu_0 I R^2}{2 \cdot 8R^3} = \frac{\mu_0 I}{16R}$

3. Find the ratio $B_1/B_2$

$\frac{B_1}{B_2} = \frac{\mu_0 I/(2R)}{\mu_0 I/(16R)}$

$= \frac{16R}{2R} = 8$

Hence,

$B_1 : B_2 = 8 : 1$

4. Option check

• A: $9:4$ ❌
• B: $12:\sqrt{5}$ ❌
• C: $8:1$ ✅
• D: $5:\sqrt{3}$ ❌

So the correct answer is Option C.