Question

Along straight and solid metal wire of radius 2mm carries a current uniformly distributed over its circular cross section. The magnetic field at a distance 2 mm from its axis is B. Magnetic field at a distance 1 mm from the axis of the wire is :-

• A. B
• B. 4B
• C. 2B
• D. B/2

Answer 1

Answer: (D)

B/2

Answer 2

The magnetic field inside a current-carrying wire with uniform current distribution is given by $B_{in} = \frac{\mu_0 I r}{2\pi R^2}$, where R is the wire's radius, I is the total current, and r is the distance from the axis. The magnetic field at the surface (r=R) is $B_{surface} = \frac{\mu_0 I}{2\pi R}$.

Given:

• Radius of wire R = 2 mm.
• Magnetic field at r = 2 mm (which is R) is B.

So, $B = \frac{\mu_0 I}{2\pi R} = \frac{\mu_0 I}{2\pi (2 \text{ mm})} = \frac{\mu_0 I}{4\pi \text{ mm}}$.

Required: Magnetic field B' at r = 1 mm.

Since 1 mm < 2 mm, this point is inside the wire.

$B' = \frac{\mu_0 I r}{2\pi R^2} = \frac{\mu_0 I (1 \text{ mm})}{2\pi (2 \text{ mm})^2} = \frac{\mu_0 I (1 \text{ mm})}{2\pi (4 \text{ mm}^2)} = \frac{\mu_0 I}{8\pi \text{ mm}}$.

Comparing B' with B:

$B' = \frac{1}{2} \left( \frac{\mu_0 I}{4\pi \text{ mm}} \right) = \frac{1}{2} B$.

The magnetic field at a distance 1 mm from the axis is B/2.