Question

Derive an expression for the self-inductance of an air-filled long solenoid of length \(l\) and cross-sectional area \(A\) having \(N\) turns.

Hint:

For a long air-cored solenoid, \[ L=\frac{\mu_0N^2A}{l} \] Thus, self-inductance is directly proportional to: \[ N^2 \] and cross-sectional area \(A\), while it is inversely proportional to the length \(l\) of the solenoid.

Answer 1

Concept: Whenever the current flowing through a coil changes, the magnetic flux linked with the coil also changes. Due to this change in magnetic flux, an emf is induced in the same coil itself. This phenomenon is called self-induction and the property of the coil by virtue of which it opposes any change in current is called self-inductance. The self-inductance \(L\) is defined by \[ \boxed{ L=\frac{N\phi_B}{I} } \] where \[ N\phi_B \] is the total magnetic flux linkage and \(I\) is the current through the coil.

Step 1: Magnetic field inside a long solenoid. Consider an air-cored solenoid having \[ N=\text{number of turns}, \] \[ l=\text{length of solenoid}, \] \[ A=\text{cross-sectional area}, \] \[ I=\text{current flowing through it}. \] The magnetic field inside a long solenoid is \[ \boxed{ B=\mu_0 nI } \] where \[ n=\frac{N}{l} \] is the number of turns per unit length. Therefore, \[ B = \mu_0 \frac{N}{l}I. \] \[ \boxed{ B=\frac{\mu_0 NI}{l} } \]

Step 2: Calculate magnetic flux through one turn. Magnetic flux through one turn is \[ \phi_B=BA. \] Substituting the value of \(B\), \[ \phi_B = \left(\frac{\mu_0 NI}{l}\right)A. \] Hence, \[ \boxed{ \phi_B=\frac{\mu_0 NIA}{l} } \]

Step 3: Determine total flux linkage. Since the solenoid contains \(N\) turns, \[ N\phi_B = N\left(\frac{\mu_0 NIA}{l}\right). \] Therefore, \[ N\phi_B = \frac{\mu_0 N^2 IA}{l}. \] \[ \boxed{ N\phi_B = \frac{\mu_0 N^2 IA}{l} } \]

Step 4: Apply the definition of self-inductance. Using \[ L=\frac{N\phi_B}{I}, \] we get \[ L = \frac{1}{I} \left( \frac{\mu_0 N^2 IA}{l} \right). \] Cancelling \(I\), \[ \boxed{ L=\frac{\mu_0 N^2 A}{l} } \] Final Result: Hence the self-inductance of a long air-filled solenoid is \[ \boxed{ L=\frac{\mu_0 N^2A}{l} } \] where \[ \mu_0 = 4\pi\times10^{-7}\ \text{H m}^{-1}. \]