Question

A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance (a/2) from the axis of the wire will be ______.

पर्याय

• Zero

• `(mu_0 I)/(2 pi a)`

• `(mu_0 I)/(4 pi a)`

• `(mu_0 I)/(6 pi a)`

Answer 1

A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance (a/2) from the axis of the wire will be `bbunderline((mu_0 I)/(4 pi a))`.

Explanation:

Given: r = `a/2`

Magnetic field inside a current-carrying conductor is found using Ampere’s Law.

For a uniformly distributed current:

B = `(mu_0 I r)/(2 pi a^2)`, (r < a)

= `(mu_0 I(a//2))/(2 pi a^2)`

= `(mu_0 I)/(4 pi a)`