Question

Derive an expression for energy stored in a magnetic field and also write the energy stored per-unit. volume?​

Answer 1

Answer:

The formula for the energy stored in a magnetic field is E = 1/2 LI2. The energy stored in a magnetic field is equal to the work needed to produce a current through the inductor. Energy is stored in a magnetic field. Energy density can be written as uB=B22μ u B = B 2 2 μ ...

Answer 2

The expression for energy stored in a magnetic field is  $\frac{1}{2}Li^{2}$.

The expression for energy stored in the magnetic field per unit volume is :

$B^{2}$/2μ

The derivation of the following is as follows:

We will consider a solenoid, the current through which is 'I', flowing because of switch 'S,' resistance R.

Initially, the switch is open, the current through the coil will be zero. When the switch is closed, it will develop a current, i. It will take 'dt' time to increase the current from '0' to 'i.' E.M.F. will start inducing.

Let induced e.m.f. in a coil be,

E= -L$\frac{di}{dt}$ ------------(i)

This opposes a supply voltage. Hence, we neglect the negative sign from equation (i)

⇒ Power supplied = Voltage x Current,

⇒ Power supplied = L$\frac{di}{dt}$ x i,

⇒ Power supplied in 'dt' time,

⇒ L$\frac{di}{dt}$ x i x dt,

⇒ Ldi x i,

For finding total energy we will integrate the above equation,

⇒   $\int\limits^i_0 {Li} \, di$

⇒   $L \int\limits^i_0 {i} \, di$

⇒   $L[\frac{i^{2} }{2}]$

⇒  Hence energy supplied in 'dt' time is $L*\frac{i^{2} }{2}$.

Now, energy stored per unit volume is also called energy density.

⇒  E= $L*\frac{i^{2} }{2}$

taking L= μ$N^{2}$A/L,

and, B = μNi / L

Hence,

⇒  We get energy density $U_{B}$= $B^{2}$/2μ.