Self induction, mutual induction, the choke coil, energy in an inductor and the LR circuit

Self induction and self inductance

When the current in a coil changes, the flux through the same coil changes. So an emf is induced in the coil itself. This is self induction. The total flux linked with the coil is proportional to its current: φ = LI. The constant L is the coil's self inductance, and ε = −L dI/dt. By Lenz's law this emf opposes the change in current, so it is called a back emf. The SI unit of L is the henry (H): 1 H = 1 V s A⁻¹ = 1 ohm second. The millihenry (mH) and microhenry (μH) are also used. L depends only on the coil's shape, size, number of turns and core. An inductor acts like inertia in a circuit: the current through it cannot change suddenly. That is why a spark can jump at a switch when a fan or a mixer is switched off.

Derivations: inductance of a solenoid, and the energy stored

1. 1
   Inductance: a long solenoid has N turns, length l and cross-section A, and carries a current I. Inside it, B = μ₀nI = μ₀NI/l.
2. 2
   Flux through one turn: BA = μ₀NIA/l. Total flux through N turns: φ = μ₀N²AI/l.
3. 3★
   Compare with φ = LI: L = μ₀N²A/l. Doubling the turns makes L four times as large.
4. 4
   Energy: while the current grows, the source pushes charge against the back emf, of size L dI/dt.
5. 5
   Work done in a short time dt: dW = (L dI/dt) × I dt = LI dI.
6. 6
   Total work to raise the current from 0 to I: W = ∫ LI dI (from 0 to I) = ½LI².
7. 7★
   So, the energy stored in the inductor's magnetic field is U = ½LI².

★ key step

Mutual induction, and two coaxial solenoids

When the current in one coil changes, the flux through a **nearby coil** changes and an emf is induced in it. This is **mutual induction**. If φ₂ is the flux linked with coil 2 due to a current I₁ in coil 1, then φ₂ = MI₁ and **ε₂ = −M dI₁/dt**. M is the **mutual inductance**, also measured in henry.

1. 1
   Take two long solenoids of the same length l, one inside the other. The outer one has n₁ turns per metre. The inner one has N₂ turns in all and radius r₂.
2. 2
   Let a current I₁ flow in the outer solenoid. The field inside it is B₁ = μ₀n₁I₁.
3. 3
   This field passes through each turn of the inner solenoid, of area πr₂². Flux linked with the inner solenoid: φ₂ = N₂ × μ₀n₁I₁ × πr₂².
4. 4★
   Compare with φ₂ = MI₁: M = μ₀n₁N₂πr₂². With N₂ = n₂l, this is M = μ₀n₁n₂πr₂²l.
5. 5
   M is larger for more turns, a soft iron core, and coils close together on the same axis. Coils at right angles to each other have almost no M.

★ key step

Growth and decay of current in an LR circuit, and how to draw it

An inductor L and a resistance R are joined in series with a battery of emf E and a switch.

1. 1
   Close the switch. The current starts to rise, but the back emf opposes the rise. So the current grows gradually, not at once.
2. 2★
   Growth: I = I₀(1 − e^(−Rt/L)), where I₀ = E/R is the final steady current.
3. 3
   The time constant τ = L/R is the time for the current to reach about 63% (nearly two-thirds) of I₀.
4. 4
   Now take the battery out of the loop, leaving L and R joined. The current does not stop at once; the back emf keeps it flowing.
5. 5★
   Decay: I = I₀e^(−Rt/L). In a time L/R the current falls to about 37% of I₀.
6. 6
   So, a larger L or a smaller R makes the changes slower. A larger L also gives a larger back emf when a switch is opened.
7. 7
   How to draw: a battery, a switch, a coil symbol for L and a resistor R in series. Beside it, plot I against t: a curve rising from zero and levelling off at I₀, with 0.63 I₀ marked at t = L/R.
8. 8
   For decay, draw a curve falling from I₀ towards zero, with 0.37 I₀ marked at t = L/R. Label both axes and the curves 'growth' and 'decay'.

★ key step

The choke coil and other uses of inductance

• A choke coil is copper wire wound on a laminated soft iron core. It has a large inductance but a small resistance.
• In an AC circuit, a choke cuts down the current through its large reactance. The current lags the voltage by nearly 90°, so the choke wastes very little energy as heat.
• A resistor used for the same job would waste energy as heat. So chokes are used in old tube-light fittings and in some fan regulators.
• An ignition coil in a petrol engine uses mutual induction. The current in a coil of few turns is suddenly broken. This induces a very high emf in a coil of many turns, which makes the spark.
• In resistance boxes, the wire is folded back on itself before winding. The two halves carry opposite currents, so the coil has almost no inductance: a non-inductive winding.

Worked example: back emf, energy and an LR circuit