Question

Define self-inductance of a coil. Derive the expression for the energy required to build up a current \(I\) in a coil of self-inductance \(L\).

Hint:

The energy required to build up a current in a coil is proportional to the square of the current and the self-inductance. This energy is stored as magnetic potential energy in the coil.

Answer 1

Self
-inductance of a coil is a property of the coil that describes its ability to oppose changes in current flowing through it. When a current \(I\) flows through a coil, it creates a magnetic flux. The self
-inductance \(L\) is defined as the ratio of the induced emf (electromotive force) in the coil to the rate of change of current. Mathematically, it is given by: \[ L = \frac{N \Phi}{I} \] where: 
- \(N\) is the number of turns in the coil, 
- \(\Phi\) is the magnetic flux through each turn, 
- \(I\) is the current flowing through the coil. Now, let’s derive the expression for the energy required to build up a current \(I\) in a coil of self-inductance \(L\). Energy required to build up a current: The work \(dW\) required to increase the current \(I\) by an infinitesimal amount \(dI\) in the coil is given by the product of the induced emf \( \mathcal{E} \) and the infinitesimal current change \(dI\): \[ dW = \mathcal{E} \cdot dI \] From Faraday's law, the induced emf is: \[ \mathcal{E} = -L \frac{dI}{dt} \] Substituting this in the above equation: \[ dW = -L \frac{dI}{dt} \cdot dI \] Since \(dI/dt\) is the rate of change of current, we need to integrate to find the total work required to build up the current from 0 to \(I\). The total work done (or energy) to establish the current is: \[ W = \int_0^I L \, I \, dI \] Solving the integral: \[ W = \frac{1}{2} L I^2 \] Thus, the energy required to build up a current \(I\) in a coil of self-inductance \(L\) is: \[ \boxed{W = \frac{1}{2} L I^2} \] This is the expression for the energy stored in the magnetic field of the coil as the current increases.