NCERT · Class 12 · Physics · Moving Charges and MagnetismState Ampere's Circuital Law. Using this law, derive an expression for the magnetic field at a distance $r$ from a long, straight, current-carrying conductor.

Statement of Ampere's Circuital Law

\nAmpere's Circuital Law states that the line integral of the magnetic field vector $\vec{B}$ around any closed loop is equal to $\mu_0$ times the total net current $I_{\text{enclosed}}$ passing through the surface bounded by the closed loop. \nMathematically, it is expressed as: $$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$$\nwhere:

• $\vec{B}$ is the magnetic field.
• $d\vec{l}$ is a small element of length on the closed loop (Amperian loop).
• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7};\text{T}\cdot\text{m}/\text{A}$).
• $I_{\text{enclosed}}$ is the total current enclosed by the loop.

Derivation of Magnetic Field due to a Long Straight Conductor

\nConsider a long, straight conductor carrying a steady current $I$. We need to find the magnetic field $B$ at a perpendicular distance $r$ from the conductor.

1. Choice of Amperian Loop: Due to symmetry, the magnetic field lines around a long straight current-carrying conductor are concentric circles centered on the conductor. We choose a circular Amperian loop of radius $r$, centered on the wire and lying in a plane perpendicular to the wire.

2. Applying Ampere's Law: Consider a small length element $d\vec{l}$ on this circular loop. At any point on the loop, the magnetic field $\vec{B}$ is tangent to the circle, and the length element $d\vec{l}$ is also along the tangent. Therefore, the angle $\theta$ between $\vec{B}$ and $d\vec{l}$ is $0^\circ$.

3. Evaluating the Line Integral: $$\oint \vec{B} \cdot d\vec{l} = \oint B , dl \cos 0^\circ = \oint B , dl$$

4. Constant Magnetic Field Magnitude: The magnitude of the magnetic field $B$ is the same at every point on the circular loop because every point is equidistant ($r$) from the current-carrying wire. Thus, $B$ can be taken outside the integration sign: $$\oint \vec{B} \cdot d\vec{l} = B \oint dl$$

5. Summing the Path: The integral $\oint dl$ represents the total circumference of the circular Amperian loop, which is equal to $2\pi r$: $$\oint \vec{B} \cdot d\vec{l} = B (2\pi r)$$

6. Equating with Ampere's Law: According to Ampere's Circuital Law, this line integral is equal to $\mu_0 I$: $$B (2\pi r) = \mu_0 I$$

7. Final Expression: Solving for $B$, we get: $$B = \frac{\mu_0 I}{2\pi r}$$