JEE Physics 2025 Question with Solution

Given: A long straight wire of radius $a$ carries a steady current $I$ uniformly distributed over its cross-section.

Find: The distances from the centre where the magnetic field is half of the maximum possible magnetic field, once inside the wire and once outside the wire.

For a uniformly distributed current, the magnetic field inside the wire is

$$B = \frac{\mu_0 I r}{2\pi a^2}$$

The maximum magnetic field occurs at the surface, that is at $r=a$:

$$B_{\max} = \frac{\mu_0 I}{2\pi a}$$

Half of the maximum value is

$$\frac{B_{\max}}{2} = \frac{\mu_0 I}{4\pi a}$$

Set the inside-field expression equal to this value:

$$\frac{\mu_0 I r}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$$

So,

$$r = \frac{a}{2}$$

Thus, inside the wire, the required distance is $\frac{a}{2}$.

For points outside the wire,

$$B = \frac{\mu_0 I}{2\pi r}$$

Again, taking half of the maximum value,

$$\frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{4\pi a}$$

This gives

$$r = 2a$$

Thus, outside the wire, the required distance is $2a$.

Therefore, the distances are $\left[ \frac{a}{2}, 2a \right]$ and the correct option is B.

Note: The solution states "The Correct Option is C", but its working and final value clearly give $\left[ \frac{a}{2}, 2a \right]$, which matches option B.

Given: Radius of wire $= a$, current $= I$, and current distribution is uniform.

Find: Where the magnetic field becomes half of its maximum value inside and outside the wire.

1. Inside the wire, the field varies linearly with distance $r$ from the centre:

$$B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2}$$

2. The maximum field anywhere due to the wire occurs at the surface $r=a$:

$$B_{\max} = \frac{\mu_0 I}{2\pi a}$$

3. Half of this maximum is

$$\frac{1}{2}B_{\max} = \frac{\mu_0 I}{4\pi a}$$

4. Equating inside field to half maximum,

$$\frac{\mu_0 I r}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$$

Cancelling common terms,

$$\frac{r}{a^2} = \frac{1}{2a}$$

Hence,

$$r = \frac{a}{2}$$

5. Outside the wire, the field is

$$B_{\text{outside}} = \frac{\mu_0 I}{2\pi r}$$

6. Set this equal to half maximum:

$$\frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{4\pi a}$$

So,

$$\frac{1}{r} = \frac{1}{2a}$$

Therefore,

$$r = 2a$$

Thus, the required pair is $\left[ \frac{a}{2}, 2a \right]$, so the correct option is B.