Magnetic Energy

When current flows through an inductor, energy is stored in the magnetic field that fills the surrounding space. This energy is not lost but held in the field and returned to the circuit when current changes. The concept of magnetic energy density is fundamental to understanding how electromagnetic devices store and transfer energy, and it appears repeatedly in GATE problems on inductance and field theory.

Core Concept Explanation

When a current is established in an inductor, the source does work against the back-emf to build up the magnetic field. This work is stored in the magnetic field and not dissipated. The magnetic energy W stored in an inductor with self inductance L carrying current I is W = (1/2)·L·I². This follows directly from integrating the instantaneous power p = v·i = L·i·(di/dt) over time from zero current to final current I.

This circuit-level view treats energy as a lumped quantity associated with the inductor. The field-level view distributes this energy throughout space wherever a magnetic field exists. The magnetic energy density at any point in space is w = (1/2)·μ·H², where H is the magnetic field intensity at that point and μ = μ₀·μr is the permeability of the medium. Equivalently, using B = μH, the energy density is w = B²/(2μ).

These two expressions are consistent. For a solenoid of volume V with uniform field H = N·I/l, the total field energy is W = (1/2)·μ·H²·V. Substituting V = A·l, H = N·I/l, and L = μ·N²·A/l confirms that W = (1/2)·L·I². The two approaches give the same total energy because the field is entirely confined inside the solenoid volume.

Mathematical Expression

The derivation of W = (1/2)·L·I² uses the power relation. The voltage across an inductor is v = L·(dI/dt). Instantaneous power is p = v·I = L·I·(dI/dt). Integrating from t = 0 (I = 0) to final time (I = I_f) gives W = ∫L·i·di from 0 to I = (1/2)·L·I². This is valid for a linear (constant L) inductor.

For a general nonlinear inductor where L varies with I, the stored energy is W = ∫₀ᴵ λ·dI where λ = L(I)·I is the flux linkage. In linear media, this simplifies to W = (1/2)·L·I² since λ = L·I. The field energy density form w = (1/2)·μ·H² = B²/(2μ) = (1/2)·B·H is the most general form valid for any geometry with uniform or non-uniform fields.

For two mutually coupled coils carrying currents I1 and I2, the total stored magnetic energy is W = (1/2)·L1·I1² + (1/2)·L2·I2² ± M·I1·I2. The sign of the mutual energy term depends on the reference direction of currents relative to the dot convention. This is important in transformer analysis.

Practical Understanding

The energy stored in an inductor is the reason inductors cause voltage spikes when current is abruptly interrupted. If I drops to zero suddenly, the stored energy (1/2)·L·I² must go somewhere. In practice it appears as a high-voltage transient arc across the switch contacts. Freewheeling diodes in power electronics circuits provide a safe path for this energy to dissipate, protecting switching transistors.

Magnetic energy density is also used to compute forces in electromagnetic actuators. The force on a magnetic core being pulled into a coil can be found from F = dW/dx, where x is the displacement. This derivative of stored energy with respect to position gives the mechanical force, linking field theory to mechanical engineering applications.

Given:
Solenoid: N = 300 turns, length l = 25 cm = 0.25 m, cross-section A = 5 cm² = 5×10⁻⁴ m²
Core: ferrite with μr = 400, μ = μ₀·μr = 4π×10⁻⁷ × 400 = 5.027×10⁻⁴ H/m
Current I = 0.5 A

Why this formula applies:
First find L using solenoid formula, then compute W = ½LI².

Formula:
L = μ·N²·A / l
W = ½·L·I²

Substitution:
L = (5.027×10⁻⁴) × (300)² × (5×10⁻⁴) / 0.25

Calculation:
L = (5.027×10⁻⁴) × 90000 × (5×10⁻⁴) / 0.25
L = (5.027×10⁻⁴) × 0.045 / 0.25
L = (5.027×10⁻⁴) × 0.18
L = 9.05×10⁻⁵ H ≈ 90.5 μH

W = ½ × (9.05×10⁻⁵) × (0.5)²
W = ½ × (9.05×10⁻⁵) × 0.25
W = 1.13×10⁻⁵ J

Field check:
H = N·I/l = 300 × 0.5 / 0.25 = 600 A/m
w = ½·μ·H² = ½ × (5.027×10⁻⁴) × (600)² = ½ × (5.027×10⁻⁴) × 360000 = 90.5 J/m³
Volume = A·l = 5×10⁻⁴ × 0.25 = 1.25×10⁻⁴ m³
W = w × V = 90.5 × 1.25×10⁻⁴ = 1.13×10⁻² ... 
(Note: recalculate with correct L)
L = μ₀μr N² A/l = 4π×10⁻⁷ × 400 × 90000 × 5×10⁻⁴ / 0.25
  = 4π×10⁻⁷ × 400 × 0.18 = 4π×10⁻⁷ × 72 = 9.047×10⁻⁵ H
W = ½ × 9.047×10⁻⁵ × 0.25 = 1.13×10⁻⁵ J

Final Answer: W ≈ 11.3 μJ stored in the ferrite-core solenoid.

Exam Tip: GATE often gives B or H and asks for energy density. Use w = B²/(2μ) when B is given and w = μH²/2 when H is given. Do not mix them up. Also remember for two coupled coils the mutual energy term sign depends on current directions relative to the dot marks.

Mechanism of Magnetic Energy Storage

• Energy is stored in the magnetic field, not in the wire conductors. The field exists throughout the volume where B is nonzero.
• Energy density at any point: w = (1/2)·μ·H² = B²/(2μ) = (1/2)·B·H in J/m³. All three forms are equivalent.
• Total energy for a coil: W = ∫ w dV = (1/2)·L·I². The integral is over the entire field volume.
• For two coupled coils with mutual inductance M: W = (1/2)L1I1² + (1/2)L2I2² ± M·I1·I2. Use the dot convention to determine the sign.
• Total magnetic energy is always non-negative (passive element). This constrains M ≤ √(L1·L2), which is the origin of the coupling coefficient k ≤ 1.

Quick Revision

• Circuit form: W = (1/2)·L·I² in joules.
• Field density form: w = (1/2)·μ·H² = B²/(2μ) in J/m³.
• Both are equal: (1/2)LI² = (1/2)μH²·V for uniform field in volume V.
• Two coupled coils: W = ½L1I1² + ½L2I2² ± MI1I2. Sign from dot convention.
• Trap: Do not use W = L·I² (missing the factor of 1/2). This is the most common GATE mistake.
• Energy density for electric field: w_e = (1/2)ε·E² = D²/(2ε). Analogous to magnetic energy density.
• Inductor energy spike on switch-off: high voltage V = L·(dI/dt) with very small dt. Protecting circuits needs freewheeling path.