Question

On the basis of Ampere's circuital law, find the expression for the magnetic field produced by an infinitely long straight current-carrying conductor.

Hint:

For a long straight conductor carrying current \( I \), the magnetic field at a distance \( r \) from the conductor is given by \( B = \frac{\mu_0 I}{2\pi r} \).

Answer 1

Step 1: Ampere's Circuital Law.
Ampere's circuital law states that the line integral of the magnetic field \( B \) around a closed loop is proportional to the total current \( I_{\text{enc}} \) passing through the loop. The law is given by: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \] where: - \( \mu_0 \) is the permeability of free space, - \( I_{\text{enc}} \) is the enclosed current.
Step 2: Magnetic Field due to a Straight Conductor.
Consider a long straight conductor carrying a current \( I \). We choose a circular path of radius \( r \) around the wire. Since the magnetic field due to a straight conductor is radially symmetric, the magnetic field at a distance \( r \) from the wire is tangential to the circle. The line integral of the magnetic field around the circular path is: \[ B \times 2\pi r = \mu_0 I \] Solving for \( B \), we get the magnetic field at a distance \( r \) from the conductor: \[ B = \frac{\mu_0 I}{2\pi r} \]