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 \(\dfrac{a}{2}\) from the axis of the wire will be

• A. Zero
• B. \(\dfrac{\mu_0 I}{2\pi a}\)
• C. \(\dfrac{\mu_0 I}{4\pi a}\)
• D. \(\dfrac{\mu_0 I}{6\pi a}\)

Hint:

For a point inside a wire carrying uniformly distributed current, \[ B=\frac{\mu_0 Ir}{2\pi a^2}, \] where \(r<a\). Substitute \(r=\dfrac{a}{2}\) directly to obtain \[ B=\frac{\mu_0 I}{4\pi a}. \]

Answer 1

The Correct Option is C

Concept: The magnetic field inside a current-carrying conductor is calculated using Ampere's Circuital Law. Ampere's law states that \[ \oint \vec B\cdot d\vec l=\mu_0 I_{\text{enc}} \] where \(I_{\text{enc}}\) is the current enclosed by the chosen Amperian loop. For a wire carrying uniformly distributed current, the current density is constant, and hence the current enclosed by a circle of radius \(r\) inside the conductor is proportional to the area enclosed.

Step 1: Determine the current enclosed within radius \(r=\dfrac{a}{2}\).
The total current density is \[ J=\frac{I}{\pi a^2} \] The area enclosed by radius \[ r=\frac{a}{2} \] is \[ A=\pi\left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{4} \] Therefore, \[ I_{\text{enc}} = J\times A = \frac{I}{\pi a^2} \times \frac{\pi a^2}{4} = \frac{I}{4} \]

Step 2: Apply Ampere's circuital law.
For a circular Amperian path of radius \(r=\dfrac{a}{2}\), \[ B(2\pi r)=\mu_0 I_{\text{enc}} \] Substituting, \[ B\left(2\pi\frac{a}{2}\right) = \mu_0\left(\frac{I}{4}\right) \] \[ B(\pi a)=\frac{\mu_0 I}{4} \] Hence, \[ B=\frac{\mu_0 I}{4\pi a} \] Therefore, \[ \boxed{B=\frac{\mu_0 I}{4\pi a}} \]