Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

4.2.9 Recap

Recap

Magnetic medium: the aligned atomic current loops are described by the magnetic polarisation $M$ [T], with bound currents $J_{m} = \nabla \times M / μ_{0}$ in the volume and $J_{m s} = M \times a_{n} / μ_{0}$ on the surface. $H$ has only the free current as its source, $\nabla \times H = J$ in every medium, and $B = μ_{0} H + M$ , Eq. (4.2.4).
Simple media: $M = μ_{0} χ_{m} H$ , $μ_{r} = 1 + χ_{m}$ , $B = μ H$ with $μ = μ_{0} μ_{r}$ , Eq. (4.2.8). Diamagnetic ( $μ_{r} ≲ 1$ ), paramagnetic ( $μ_{r} ≳ 1$ ) and ferromagnetic ( $μ_{r} ≫ 1$ , nonlinear, saturating at $B_{s} \approx 2 T$ for iron, with hysteresis).
Fundamental equations, Eq. (4.2.9): $\nabla \times H = J$ and $\nabla \cdot B = 0$ in any medium, integral forms $\oint H \cdot d l = \int J \cdot d S$ and $\oint B \cdot d S = 0$ .
Boundary conditions, Eqs. (4.2.11), (4.2.12): $H_{1 τ} - H_{2 τ} = J_{n S}$ (continuous without surface current) and $B_{1 n} = B_{2 n}$ , hence $tan θ_{1} / tan θ_{2} = μ_{1} / μ_{2}$ : flux leaves iron almost normally.
Inductance: $Φ_{12} = \int_{S_{2}} B_{1} \cdot d S$ , flux linkage $Λ_{12} = N_{2} Φ_{12}$ , mutual inductance $L_{12} = Λ_{12} / I_{1} = L_{21}$ , self inductance $L_{11} = N_{1} Φ_{11} / I_{1}$ . The current cancels: $L$ depends on geometry and $μ$ only, $L \propto N^{2}$ . Solenoid $μ N^{2} S / l$ , toroid $μ N^{2} S / (2 π R)$ , gapped core $μ_{0} N^{2} S / (l_{g} + l_{c} / μ_{r})$ .
Energy: $W_{m} = \frac{1}{2} \int H \cdot B d V = \frac{1}{2} L I^{2}$ , density $\frac{1}{2} H \cdot B$ , two circuits $\frac{1}{2} L_{11} I_{1}^{2} + L_{12} I_{1} I_{2} + \frac{1}{2} L_{22} I_{2}^{2} \geq 0$ , so $k = L_{12} / L_{11} L_{22} \leq 1$ . The energy sits where $μ$ is small: in the air gap.
Conduction: Ohm's law $J = σ E$ , $J = J_{v} + J_{c}$ , $R = l / (σ A)$ . Joule phenomenon: $P = \int J_{c} \cdot E d V = σ \int E^{2} d V$ appears as heat, the magnetic force does no work.
Dipole and torque: a loop is a dipole $m = I S a_{n}$ with $B = \frac{μ m}{4 π r ^{3}} (2 cos θ a_{r} + sin θ a_{θ})$ , and in a uniform field it feels $T = m \times B$ (also $T = r \times F$ ). Hence the DC motor, the meter and the loudspeaker ( $F = B l I$ ).