Self-Inductance & Magnetic Energy

A Coil Talks to Itself

A coil doesn't need a neighbour — changing its own current changes its own flux, inducing an emf in itself. This is self-induction.

Flux linkage is proportional to the current:

$$N\Phi_B = LI$$

where $L$ is the self-inductance (coefficient of self-induction). When the current varies:

$$\varepsilon = -\frac{d(N\Phi_B)}{dt} = -L\frac{dI}{dt}$$

The self-induced emf is called the back emf: it opposes any change - increase or decrease - of current in the circuit. Work must be done against it to establish or change a current.

Key Point: The minus sign works both ways. Rising current → back emf fights the rise (slows growth). Falling current → back emf fights the fall (sustains the current). The coil is a stubborn conservative: it likes its current exactly as it is.

[NEET Important] This is why switching OFF a highly inductive circuit produces a spark across the switch: the large $-L\,dI/dt$ from the rapid break drives the current through the air gap.

Self-Inductance of a Long Solenoid

For a long solenoid (cross-section $A$, length $l$, $n$ turns per unit length) carrying current $I$, the interior field is $B = \mu_0 n I$. The flux linkage of all $N = nl$ turns is:

$$N\Phi_B = (nl)(\mu_0 n I)(A) = \mu_0 n^2 A l\, I$$

Comparing with $N\Phi_B = LI$:

$$\boxed{L = \mu_0 n^2 A l}$$

Fill the interior with a material of relative permeability $\mu_r$ (e.g. soft iron):

$$L = \mu_r \mu_0 n^2 A l$$

Like $M$, the self-inductance depends only on geometry ($n$, $A$, $l$) and the medium ($\mu_r$) - never on the current.

[JEE Tip] Re-express via total turns $N = nl$: $L = \mu_0 N^2 A/l$. The $N^2$ (or $n^2$) dependence is the most-tested scaling in this topic: double the turns (same length) and $L$ quadruples.

Electrical Inertia & Stored Magnetic Energy

A beautiful analogy: self-inductance is the electromagnetic analogue of mass. Mass resists changes in velocity; $L$ resists changes in current. It is electrical inertia.

Because the back emf opposes the growth of current, the source must do work to establish a current. The rate of work is $\frac{dW}{dt} = |\varepsilon| I = LI\frac{dI}{dt}$ (ignoring resistive losses). Integrating from 0 to $I$:

$$W = \int_0^I LI'\,dI' = \boxed{\frac{1}{2}LI^2}$$

This energy is stored as magnetic potential energy - compare $\frac{1}{2}mv^2$, with $L \leftrightarrow m$ and $I \leftrightarrow v$.

Energy density: for the solenoid, $U_B = \frac{1}{2}LI^2 = \frac{1}{2}(\mu_0 n^2 Al)\left(\frac{B}{\mu_0 n}\right)^2 = \frac{B^2}{2\mu_0}Al$. Dividing by the volume $Al$:

$$u_B = \frac{B^2}{2\mu_0}$$

The perfect twin of the capacitor's electrostatic energy density $u_E = \frac{1}{2}\varepsilon_0 E^2$ - both proportional to the square of the field, and both completely general, valid in any region containing fields.

Two coils together: with currents in both, fluxes superpose: $N_1\Phi_1 = L_1 I_1 + M_{12}I_2$, so

$$\varepsilon_1 = -L_1\frac{dI_1}{dt} - M_{12}\frac{dI_2}{dt}$$

Solved Examples

Example 1: Energy in a solenoid

Obtain the magnetic energy stored in a solenoid in terms of $B$, $A$ and $l$.

Solution:

1. Stored energy: $U_B = \frac{1}{2}LI^2$ with $L = \mu_0 n^2 Al$.
2. From $B = \mu_0 n I$: $I = \frac{B}{\mu_0 n}$.
3. $U_B = \frac{1}{2}\mu_0 n^2 Al \cdot \frac{B^2}{\mu_0^2 n^2} = \frac{B^2}{2\mu_0}Al$.
4. Per unit volume: $u_B = B^2/2\mu_0$.

Example 2: The capacitor comparison

How does this compare with the electrostatic energy of a capacitor?

Solution:

1. Capacitor (Chapter 2): $u_E = \frac{1}{2}\varepsilon_0 E^2$.
2. Both energy densities are proportional to the square of the field strength.
3. Though derived for a solenoid and a parallel-plate capacitor, both results are general — valid for any region of space containing the fields.

Example 3: L of a solenoid

A solenoid has 1000 turns/m, cross-section $10^{-3}$ m$^2$, length 0.5 m. Find its self-inductance.

Solution:

1. $L = \mu_0 n^2 Al = 4\pi \times 10^{-7} \times (1000)^2 \times 10^{-3} \times 0.5$.
2. $= 4\pi \times 10^{-7} \times 10^6 \times 5 \times 10^{-4} = 6.3 \times 10^{-4}$ H.
3. Answer: $L \approx 0.63$ mH. (With a soft-iron core of $\mu_r = 1000$, this would leap to 0.63 H!)

Example 4: L from the back emf

The current in a coil falls steadily from 5.0 A to 0 in 0.1 s, inducing an average emf of 200 V. Estimate the self-inductance.

Solution:

1. $|\varepsilon| = L\left|\frac{\Delta I}{\Delta t}\right| \Rightarrow L = \frac{200}{5/0.1}$.
2. $\frac{\Delta I}{\Delta t} = 50$ A/s, so $L = 200/50$.
3. Answer: $L = 4$ H.

Example 5: Energy stored

How much energy is stored in that 4 H coil when it carries a steady 2 A?

Solution:

1. $W = \frac{1}{2}LI^2 = \frac{1}{2} \times 4 \times 2^2$.
2. Answer: 8 J — recoverable when the current is switched off (often as that spark!).

Example 6: Both directions of stubbornness

The current through an inductor is (a) increased, (b) decreased. Give the polarity of the back emf in each case.

Solution:

1. (a) Rising $I$: back emf opposes the rise — it acts like a battery opposing the driving source.
2. (b) Falling $I$: back emf reverses — now it acts with the original current direction, trying to sustain it.
3. Both follow from one sign: $\varepsilon = -L\,dI/dt$. Opposition is always to the change.

Example 7: Energy density numerical

Find the magnetic energy density in a 1.0 T field, and compare it with the electric energy density in air at its breakdown field, $E = 3 \times 10^6$ V/m.

Solution:

1. Magnetic: $u_B = \frac{B^2}{2\mu_0} = \frac{1}{2 \times 4\pi \times 10^{-7}} \approx 4.0 \times 10^5$ J/m$^3$.
2. Electric (at breakdown): $u_E = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times 9 \times 10^{12} \approx 40$ J/m$^3$.
3. Takeaway: an ordinary 1 T magnetic field stores about $10^4$ times more energy per unit volume than the strongest practical air-gap electric field — one reason magnetic storage of energy interests engineers.

Example 8: Doubling the turns

A solenoid's total turns are doubled, keeping its length and area fixed. What happens to $L$ and to the energy stored at the same current?

Solution:

1. $L = \mu_0 n^2 Al$ with $n = N/l$: doubling $N$ doubles $n$, so $L$ quadruples.
2. $W = \frac{1}{2}LI^2$ at the same $I$ also quadruples.
3. The $n^2$ scaling is the most-tested line of this section.

Example 9: Back emf from a time-dependent current [JEE Numerical]

The current through a 10 mH inductor varies as $I = (3t^2 + 2t)$ A. Find the back emf magnitude at $t = 2$ s.

Solution:

1. $\frac{dI}{dt} = 6t + 2 = 14$ A/s at $t = 2$ s.
2. $|\varepsilon| = L\frac{dI}{dt} = 10 \times 10^{-3} \times 14$.
3. Answer: 0.14 V, directed so as to oppose the rising current (the $L$-as-inertia picture: $|\varepsilon| = L\,dI/dt$ is the analogue of $F = m\,dv/dt$).

Example 10: Two coils at once

Coil 1 ($L_1 = 10$ mH) carries a current rising at 5 A/s, while a neighbour ($M = 2$ mH) carries one rising at 10 A/s. Find the magnitude of the total emf induced in coil 1.

Solution:

1. Superpose the fluxes: $\varepsilon_1 = -L_1\frac{dI_1}{dt} - M\frac{dI_2}{dt}$.
2. $= -(10 \times 10^{-3} \times 5) - (2 \times 10^{-3} \times 10) = -(0.05 + 0.02)$.
3. Answer: $|\varepsilon_1| = 0.07$ V = 70 mV.