JEE Physics 2026 Question with Solution

Given: Magnetic field at the centre of the circular loop is $B_0 = 16\,\mu\text{T}$ and the point on the axis is at $x=\sqrt{3}R$.

Find: The magnetic field at that axial point and hence the correct option.

For a circular current-carrying loop, the magnetic field on its axis at a distance $x$ from the centre is

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

At the centre of the loop, $x=0$, so

$$B_0 = \frac{\mu_0 I}{2R}$$

Using this, the axial field can be written as

$$B = B_0\left(\frac{R^2}{R^2+x^2}\right)^{3/2}$$

Substitute $x=\sqrt{3}R$:

$$B = 16\left(\frac{R^2}{R^2+3R^2}\right)^{3/2}$$ $$B = 16\left(\frac{1}{4}\right)^{3/2}$$

Now,

$$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$$

Therefore,

$$B = 16 \times \frac{1}{8} = 2\,\mu\text{T}$$

the solution marks option B, but the working gives $2\,\mu\text{T}$, which matches option D. Hence, based on the extracted working, the defensible answer is D.

Given: $B_0 = 16\,\mu\text{T}$ at the centre and $x=\sqrt{3}R$.

Find: Verify the numerical value carefully.

Starting from

$$B = B_0\left(\frac{R^2}{R^2+x^2}\right)^{3/2}$$

put $x^2 = 3R^2$:

$$B = 16\left(\frac{R^2}{R^2+3R^2}\right)^{3/2} = 16\left(\frac{R^2}{4R^2}\right)^{3/2} = 16\left(\frac{1}{4}\right)^{3/2}$$

Now,

$$\left(\frac{1}{4}\right)^{3/2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$$

So,

$$B = 16 \times \frac{1}{8} = 2\,\mu\text{T}$$

Thus the intermediate algebra on the page leads to $2\,\mu\text{T}$. The printed conclusion $8\,\mu\text{T}$ is inconsistent with the shown substitution and simplification.