Question

Define the coefficient of self-induction and find an expression for it for a solenoid.

Hint:

The self-inductance of a solenoid depends on the number of turns, the cross-sectional area, and the length of the solenoid. Increasing the number of turns or the area increases the inductance.

Answer 1

Step 1: Coefficient of self-induction.
The coefficient of self-induction \( L \) is a property of a coil (or solenoid) that quantifies the ability of the coil to induce an emf in itself due to a change in current. It is defined as the ratio of the induced emf to the rate of change of current. Mathematically, it is given by: \[ L = \frac{\text{Induced emf}}{\text{Rate of change of current}} = \frac{V_{\text{induced}}}{\frac{dI}{dt}} \] Step 2: Expression for self-induction in a solenoid.
For a solenoid, the self-induction \( L \) depends on the number of turns \( N \), the length \( l \), the area of cross-section \( A \), and the permeability of free space \( \mu_0 \). The expression for the self-inductance of a solenoid is given by: \[ L = \mu_0 \frac{N^2 A}{l} \] where:
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns,
- \( A \) is the cross-sectional area,
- \( l \) is the length of the solenoid.
Step 3: Conclusion.
Thus, the self-inductance of a solenoid is given by \( L = \mu_0 \frac{N^2 A}{l} \), and it quantifies the solenoid's ability to induce an emf in itself.