Question

State and prove Ampere's circuital law.

Hint:

Ampere's law is especially useful for calculating the magnetic field produced by symmetric current distributions, like wires, loops, and solenoids.

Answer 1

Ampere's Circuital Law:
Ampere's circuital law states that the line integral of the magnetic field \( \vec{B} \) around any closed loop is proportional to the total current \( I \) passing through the loop. Mathematically, it is expressed as: \[ \oint_{\mathcal{C}} \vec{B} . d\vec{l} = \mu_0 I_{\text{enc}} \] where:
- \( \oint_{\mathcal{C}} \vec{B} . d\vec{l} \) is the line integral of the magnetic field \( \vec{B} \) around a closed loop \( \mathcal{C} \),
- \( I_{\text{enc}} \) is the current enclosed by the loop,
- \( \mu_0 \) is the permeability of free space, which is a constant \( \mu_0 = 4\pi \times 10^{-7}~\text{T} . \text{m/A} \).
Proof:
Consider a steady current \( I \) flowing through a long straight conductor. Using the Biot-Savart law, we can compute the magnetic field \( \vec{B} \) around the conductor. The magnetic field at a distance \( r \) from the wire is given by: \[ B = \frac{\mu_0 I}{2 \pi r} \] Now, consider a circular path of radius \( r \) around the wire. The magnetic field \( \vec{B} \) is tangent to the circular path and its magnitude is constant along the path. Thus, the line integral of \( \vec{B} \) around the loop is: \[ \oint_{\mathcal{C}} \vec{B} . d\vec{l} = B . 2\pi r \] Substitute the expression for \( B \): \[ \oint_{\mathcal{C}} \vec{B} . d\vec{l} = \frac{\mu_0 I}{2\pi r} . 2\pi r = \mu_0 I \] This proves Ampere's circuital law.