**Energy Stored in an Inductor**

Consider a coil connected to a battery through switch S. When the switch is turned on vol voltage Vis applied across the and of the coil current through rises from zero to its maximum value I.

Due to the change of current, an emf is induced which is opposite to that of the battery. The work done by the battery in moving a small charge ∆q is given by,

**W = ∆qεl→eq1**

where εl is the magnitude of induced emf and given by,

**εl=L ∆I/∆t→eq2**

Putting the values of El from eq2 in eq1.

**W= ∆qL ∆I/∆t**

**W=∆q/∆t L ∆I →eq3**

**As current rises from 0 – I, then average current = ∆q/∆t=o+I/2=1/2I**

**Change in current = ∆I=I-O=I**

**The total work = W = (1/2I)**

**W=1/(2LI^2 )→eq4**

This work is done stored as potential energy in the inductor. Hence the energy stored in an inductor is

**Um = 1/2LI² eq 5**

For the inductor, the energy stored in terms of magnetic field strength, B = μₒnIfor the solenoid which has n turns per unit length and area of cross-section A. Tflux through the coil, ∅ =AB

**⇒ ∅= μₒ nI A →eq 6**

**From self-induction, N∅=LI**

Putting eq 6 in eq 7. Then

**L = ((μₒIA))/I=N μₒ nA**

**L = N μₒ nA →eq 8**

If l is the length of the solenoid, then put N = nl in eq 8. Then self-inductance of the coil is given by

**L = (nl) μₒ n² Al = μₒn²Al→eq 9**

**Putting volue of L from eq 9 in the eq 5**

**Um = 1/2 (μₒn² Al) I²→eq 10**

**From the eq of solenoid, B= μₒnI**

**⇒I=B/μₒn putting in eq 10**

**Um = 1/2 (μₒn²Al) (B/μₒn)²**

**Um = 1/2 ×μₒ n^2 Al ×B^2/(μₒ^2 n^2 )=(1/2 B^2)/μₒ Al**

**Um = ½ (B²/μₒ) Al →eq 11**

This is the expression for energy stored in the induced magnetic field

**Energy Density**

The energy stored in the inductor(solenoid) per unit value is called energy density.

**Um= Um/Vol = (1/2 B^2/μₒ Al)/Al =1/2 B^2/μₒ**

**Um =1/2 B²/μₒ**