Energy Stored In an Inductor
Derive an expression for energy stored in an inductor. In what form, the energy is stored in the inductor?
Consider an inductor of inductance $L$ connected across a battery.
When current $I$ flows through the inductor, an e.m.f. $\varepsilon$ is induced (called back e.m.f.) in it. This induced e.m.f. is given by,
$\varepsilon = -L \frac{dI}{dt} \quad \dots (i)$
-ve sign shows that '$\varepsilon$' opposes the passage of current $I$ in the inductor.
To drive the current through the inductor against the induced e.m.f. '$\varepsilon$', the external voltage is applied. Here, external voltage is e.m.f. of the battery $= -\varepsilon$.
Hence from eqn. (i) we get,
$\varepsilon = L \frac{dI}{dt} \dots(ii)$
Let an infinitesimal charge $dq$ be driven through the inductor. So the work done by the external supply is given by.
$dW= \varepsilon dq$
$dW= L \frac{dI}{dt}dq$
$dW= L dI \frac{dq}{dt}$
$dW = L I dI \quad \left( \because \frac{dq}{dt} = I \right) \quad \dots (iii)$
Total work done to maintain the maximum value of current ($I_0$) through the inductor is given by
$\int dW= \int_{0}^{I_0} L I dI$
or
$W = L \left[ \frac{I^2}{2} \right]_{0}^{I_0}$
$W= L\frac{I_0^2}{2}$
or
$W = \frac{1}{2} L I_0^2 \quad \dots (iv)$
The work done in increasing the current flowing through the inductor is stored as magnetic potential energy ($U_m$) in the magnetic field of the inductor.
Hence, energy stored in the inductor is given by
$U_m = \frac{1}{2} L I_0^2 \quad \dots (v)$
The energy is stored in the magnetic field of the inductor.
Magnetic energy stored in a solenoid :
In case of a solenoid, $B = \mu_0 n I_0$ or $I_0 = \frac{B}{\mu_0 n}$
and $L = \mu_0 n^2 A l$.
Using these values in eqn. (v), we get
$U_m = \frac{1}{2} \times (\mu_0 n^2 A l) \left( \frac{B}{\mu_0 n} \right)^2$
$U_m = \frac{1}{2} \mu_0 n^2 A l \frac{B^2}{\mu_0^2 n^2}$
$U_m = \frac{1}{2} \frac{B^2 A l}{\mu_0}$
Define magnetic energy density. Find magnetic energy density of a solenoid. What is S.I. unit of magnetic energy density. Write dimensional formula for magnetic energy density.
Magnetic energy density is defined as the magnetic energy per unit volume of the solenoid. It is denoted by $u_B$.
That is,
$\bar{U_m} = \frac{U_m}{V}$
For a solenoid,
$U_m = \frac{B^2 Al}{2\mu_0}$
and , Volume, $V = Al$
$\bar{U_B} = \frac{\frac{B^2 Al}{2\mu_0}}{Al}$
$\bar{U_B} = \frac{B^2 Al}{2\mu_0 \times Al}$
$\bar{U_B} = \frac{B^2}{2\mu_0}$
S.I unit of energy density is joule/m$^3$ (Jm$^{-3}$).
Dimensional formula of magnetic energy density = $\frac{\text{Energy}}{\text{Volume}}$ = $\frac{ML^2T^{-2}}{L^3}$
=$ [ML^{-1}T^{-2}]$
Comparison of Magnetic Energy Density and Electric Energy Density :
Magnetic energy density $= \frac{1}{2\mu_0} B^2$
Electric energy density $= \frac{1}{2} \epsilon_0 E^2$
Thus, both are directly proportional to the square of the field.