QJEEM24Ph2065: JEE Main 2024 Physics - Class 12 JEEM24Ph2065 Solution: A long straight wire of radius a carries a steady current I.…

To find the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ distances from the axis of a long straight wire, we use Ampère's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.

Ampère’s Law is given by:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$

where:

$\vec{B}$ is the magnetic field,

$\mu_0$ is the permeability of free space,

$I_{\text{enc}}$ is the enclosed current.

For a point inside the wire (at radius $r=a/2$ ):

The current enclosed by a radius $r$ is proportional to the area of the cross-section at radius $r$ .

The area of the cross-section at radius $r$ is given by:

$\pi \left( \frac{a}{2} \right)^2 = \frac{\pi a^2}{4}$

The total current $I$ is uniformly distributed, thus the current enclosed $I_{\text{enc}}$ at radius $r = \frac{a}{2}$ is:

$I_{\text{enc}} = I \times \frac{\text{Area enclosed}}{\text{Total area}} = I \times \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4}$

Applying Ampère’s Law inside the conductor, we get:

$B \cdot 2 \pi \left( \frac{a}{2} \right) = \mu_0 \left( \frac{I}{4} \right)$

So,

$B \cdot \pi a = \frac{\mu_0 I}{4}$

Therefore, the magnetic field inside the wire at $r = \frac{a}{2}$ is:

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

For a point outside the wire (at radius $r = 2a$ ):

The total current enclosed by a radius $r = 2a$ is the entire current $I$ .

Applying Ampère’s Law outside the conductor, we get:

$B \cdot 2 \pi (2a) = \mu_0 I$

So,

$B \cdot 4 \pi a = \mu_0 I$

Therefore, the magnetic field outside the wire at $r = 2a$ is:

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

Hence, the ratio of the magnetic field at $\frac{a}{2}$ and $2a$ is:

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

So, the correct option is:

Option C: $1:1$