Q.Calculate the magnetic field inside and outside of a long solenoid using Ampere's circuital law.

Magnetic Field of a Solenoid

Imagine you take a long wire and wind it into a tight, helical coil — that's a solenoid. If you've ever seen a spring, it looks just like that. Now pass a current through this coil. Each loop of wire produces its own tiny magnetic field, and when you pack many loops close together, those individual fields add up.

The key insight is superposition: the total field at any point is the vector sum of the fields from every turn. Inside the coil, the fields from adjacent turns reinforce each other. Outside, they tend to cancel. The result is striking — a long solenoid behaves almost like a bar magnet, with a uniform field inside and a very weak field outside.

The Precise Statement

For an ideal solenoid — infinitely long, tightly wound, with negligible spacing between turns — the magnetic field is:

• Inside: uniform, directed along the axis of the solenoid, with magnitude

where $n = N/L$ is the number of turns per unit length, $I$ is the current, and $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

• Outside: nearly zero (strictly zero for an ideal solenoid; for a real one, it's very small and falls off rapidly with distance).

The direction of the field inside follows the right-hand rule: curl your fingers in the direction of the current around the coil, and your thumb points along the field inside.

Why Is the Field Uniform Inside?

Consider a rectangular Amperian loop that runs along the axis inside the solenoid, goes out radially, runs parallel to the axis outside, and returns. Ampère's law says:

The outside field is negligible, so only the inside segment contributes. The enclosed current is $n L I$ for a loop of axial length $L$. This gives:

$B L = \mu_0 (n L I) \quad \Rightarrow \quad B = \mu_0 n I$ …