Question

Derive the expression for the magnetic field (near its centre) due to an air-filled long solenoid carrying a current $I$ and having $n$ number of turns per unit length.

Hint:

Apply Ampere's law to a rectangular loop with one side inside the solenoid. The enclosed current is nLI, which gives B = mu0 n I.

Answer 1

Step 1: Understanding the Concept:
A long air-filled solenoid has \(n\) turns per unit length and carries current \(I\). Inside it near the centre the field is uniform and runs along the axis. Outside it the field is very small and can be taken as zero. We use Ampere's circuital law to find the field.

Step 2: Key Formula or Approach:
Ampere's law:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]

Step 3: Choose the Amperian loop:
Take a rectangular loop abcd. Side ab (length \(L\)) is inside the solenoid and parallel to the axis. Side cd is outside, far from the solenoid. Sides bc and da are perpendicular to the axis.

Step 4: Evaluate the line integral:
On ab, \(\vec{B}\) is along \(d\vec{l}\), so the contribution is \(BL\).
On cd, the field is zero outside, so the contribution is zero.
On bc and da, \(\vec{B}\) is perpendicular to \(d\vec{l}\) inside, and zero outside, so the contribution is zero.
\[ \oint \vec{B} \cdot d\vec{l} = BL \]

Step 5: Find the enclosed current:
The loop encloses \(nL\) turns, each carrying current \(I\).
\[ I_{enc} = nLI \]

Step 6: Apply Ampere's law:
\[ BL = \mu_0 nLI \quad \Rightarrow \quad B = \mu_0 n I \]

Final Answer:
The field near the centre of the solenoid is uniform and directed along its axis. \[ \boxed{B = \mu_0 n I} \]