Energy in Inductors & Magnetic Fields

Push the current in a coil upward and the coil pushes back: Faraday's law gives an induced emf of size L dI/dt opposing the increase, so the source must supply that voltage on top of any resistive drop. The extra power it delivers is P = L I (dI/dt), so in a time dt the coil receives dU = L I dI. Integrating from zero to the final current, with L constant, gives U = ½LI². Notice what dropped out: dt. A 45 mH coil taken to 3.2 A stores 0.23 J whether the ramp lasts 12 ms or 12 s, because the energy depends only on the current reached. What the timing changes is the power. Squeeze the same ramp into a thousandth of the time and both the induced emf and the average power climb a thousandfold, while the 0.23 J waiting in the field at the end of it does not move at all. So the energy rating of a coil is set by the current it is taken to, and its voltage rating by how fast that current is allowed to change.

½LI² says how much, not where. Take a long air-cored solenoid of n turns per metre, cross-section A and length l, whose inductance is L = μ₀n²Al. Its stored energy is ½μ₀n²Al I², and that can be regrouped as ((μ₀nI)²/(2μ₀)) × A l. The first bracket is built only from the field inside, B = μ₀nI; the second is the volume that field fills. So U = (B²/2μ₀) × volume, and the energy per unit volume is u_B = B²/(2μ₀) — a statement about the field alone, with no turns, no current and no coil left in it. The electric analogue is u_E = ½ε₀E², and the magnetic one is far more generous: at B = 0.50 T it is 9.9 × 10⁴ J m⁻³, while an electric field at the breakdown strength of air, 3.0 × 10⁶ V m⁻¹, manages only 40 J m⁻³.

Once the energy belongs to the field, geometry with no obvious coil can be handled too. Take a coaxial cable of inner radius a and shield radius b. The run looks the same at every point along its length and the same under any rotation about its axis, so B can only be tangential to circles about that axis with one and the same magnitude all the way round — which is precisely what lets B come outside ∮B⋅dl. Ampère's law on such a circle of radius r between the conductors encloses only the core current, so B(2πr) = μ₀I and B = μ₀I/(2πr); beyond the shield the enclosed current is I − I = 0, and so is B. The density therefore varies as 1/r²: u_B = μ₀I²/(8π²r²). Slice the gap into thin cylindrical shells of volume dV = 2πr l dr, so that dU = (μ₀I²l/4π)(dr/r), and integrate from a to b: U = (μ₀I²l/4π) ln(b/a). The 1/r² falloff and the 2πr growth of each shell conspire into a logarithm, which is why making the cable fatter barely raises its stored energy. Reading L = 2U/I² off that result gives L/l = (μ₀/2π) ln(b/a) — an inductance obtained without counting a single turn.

The loop equation for a battery, resistor and inductor in series is ε = IR + L dI/dt. Multiply every term by the current: εI = I²R + L I (dI/dt). Each term is now a power. εI is what the source delivers, I²R is what the resistor turns into heat, and L I (dI/dt) is what the field banks, so nothing is left unaccounted for. Take ε = 12 V, R = 6.0 Ω and L = 0.50 H. The instant the switch closes I = 0, the source delivers nothing, and the whole 12 V sits across the coil. When the current has climbed to 1.0 A the source supplies 12 W, the resistor takes 6.0 W and 6.0 W goes into the field. At the final 2.0 A the storage term vanishes because dI/dt does: the source supplies 24 W, the resistor takes all 24 W, and the field simply holds its fixed ½ × 0.50 H × (2.0 A)² = 1.0 J.

Short the coil and resistor onto themselves and the field, not the battery, drives the circuit: I = I₀ e^(−Rt/L). The resistor dissipates I²R at every instant, and the total is ∫ I₀²R e^(−2Rt/L) dt over all time, which equals I₀²R × L/(2R) = ½LI₀². The books balance exactly: the field surrenders precisely what was put in. Because the power falls as e^(−2t/τ) with τ = L/R, half the store is gone by t = 0.35τ and 86% of it by one full time constant. The same accounting explains switch arcs. Open the circuit and the resistance jumps towards infinity, so τ collapses and the stored joule has to leave in microseconds; the coil will raise whatever emf that takes, which is why a flyback diode is fitted across a relay coil.

Because u_B goes as B², field energy climbs steeply and then meets engineering walls. The 1.5 T bore of an MRI magnet holds 9.0 × 10⁵ J m⁻³, so a cubic metre of it stores 0.90 MJ — enough to lift a 1.0 tonne car 91 m, and still only about a fortieth of the chemical energy in a litre of petrol. Push to 8.0 T and the density reaches 2.5 × 10⁷ J m⁻³, but the winding must then be superconducting, and the same u_B appears as an outward magnetic pressure the structure has to survive. Iron cores cheat in a different way: they raise B for a given current, but as the core saturates L stops being a constant, and ½LI² has to be replaced by ∫ I dλ — on a measured plot of flux linkage λ against current, the area between the curve and the λ axis, which is no longer half a rectangle.