Part 3 · Chapter 13

Magnetic Forces, Materials, and Devices

A field is only as interesting as what it pushes. Here magnetism does work: it deflects charges, drives motors, draws wires together, and turns ordinary iron into something that traps a field. From the Lorentz force to the hysteresis loop, this is where the theory meets the machine.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • The Lorentz force on a moving charge and why magnetic force does no work.
  • The force on a current-carrying conductor and between two parallel wires.
  • Torque on a loop and the magnetic dipole moment \(\vec{m}\).
  • Magnetization \(\vec{M}\), bound currents, and the role of permeability.
  • The three material classes and the B–H hysteresis loop of ferromagnets.
  • Magnetic boundary conditions and the magnetic-circuit analogy.
Section 13-1

The Lorentz Force

A charge \(Q\) moving with velocity \(\vec{u}\) through fields \(\vec{E}\) and \(\vec{B}\) feels the Lorentz force — the union of the electric push and the magnetic deflection:

Lorentz force
\[ \vec{F} = Q\vec{E} + Q\vec{u}\times\vec{B} \]

The magnetic part is always perpendicular to the velocity (it is a cross product), so it can change a charge's direction but never its speed. A purely magnetic force therefore does no work — it bends a charged particle into a circle without speeding it up, the principle behind cyclotrons, mass spectrometers, and the beam steering in many pulsed-power and accelerator systems.

🔑
Magnetic force turns, never accelerates
\[ \vec{F}_m = Q\vec{u}\times\vec{B} \perp \vec{u} \quad\Rightarrow\quad \frac{d}{dt}\!\left(\tfrac12 m u^2\right)=0 \]

Because the force is always sideways, the kinetic energy is constant. In a uniform \(\vec{B}\) a charge spirals at the cyclotron radius \(r = mu/QB\) — faster particles and weaker fields make wider circles.

Section 13-2

Force on a Conductor

A current is moving charge, so a current-carrying wire in a field feels a force — the sum of the Lorentz forces on its drifting charges. For a straight segment \(\vec{L}\) carrying current \(I\) in a uniform field:

Force on a current element and segment
\[ d\vec{F} = I\,d\vec{l}\times\vec{B}, \qquad \vec{F} = I\vec{L}\times\vec{B} \]
B (out) I F = IL×B
F = IL×B is perpendicular to both the current and the field
Section 13-3

Force Between Currents

Each wire sits in the field of the other, so two parallel currents exert forces on each other. Using the wire field \(B=\mu_0 I/2\pi\rho\) from Chapter 11, the force per unit length between wires a distance \(d\) apart is:

Force per length between parallel wires
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]

Currents in the same direction attract; opposite currents repel — the reverse of the electrostatic rule for charges. This force once defined the ampere itself, and it is the mechanical stress that high-current busbars and pulsed-power conductors must be braced against.

Section 13-4

Torque & Dipole Moment

A current loop in a uniform field feels no net force — but it can feel a torque that tries to rotate it. Defining the loop's magnetic dipole moment \(\vec{m}=I\vec{S}\) (current times the vector area, normal by the right-hand rule), the torque is a clean cross product:

Torque on a magnetic dipole
\[ \vec{m} = I\vec{S}, \qquad \vec{T} = \vec{m}\times\vec{B} \]

The torque is greatest when the loop lies parallel to the field and zero when its moment aligns with \(\vec{B}\) — the loop wants to swing until \(\vec{m}\parallel\vec{B}\). This is the operating principle of every electric motor and moving-coil meter. Note the exact parallel with the electric dipole of Chapter 6, where \(\vec{T}=\vec{p}\times\vec{E}\).

Section 13-5

Magnetization & Materials

Inside matter, the orbiting and spinning electrons act as countless atomic current loops, each a tiny dipole. Their alignment per unit volume is the magnetization \(\vec{M}\) (A/m) — the magnetic twin of polarization \(\vec{P}\). It modifies the relation between \(\vec{B}\) and \(\vec{H}\), and collapses (for linear materials) into a single permeability:

B, H and magnetization
\[ \vec{B} = \mu_0(\vec{H} + \vec{M}), \qquad \vec{B} = \mu\vec{H}, \qquad \mu = \mu_0\mu_r \]

The relative permeability \(\mu_r=1+\chi_m\) sorts materials into three classes — and ferromagnets break the linear picture entirely, retaining magnetization after the field is removed:

Class\(\mu_r\)BehaviourExample
Diamagneticslightly < 1Weakly repelled by a fieldCopper, water
Paramagneticslightly > 1Weakly attractedAluminium
Ferromagnetic\(\gg 1\) (≈10²–10⁵)Strongly attracted; retains magnetismIron, steel
H B remanence coercivity
Ferromagnetic hysteresis: B lags H; loop area is energy lost per cycle

In a ferromagnet \(\vec{B}\) depends on the material's history, tracing a hysteresis loop as \(H\) cycles. The loop's width gives the remanence (field kept at \(H=0\)) and coercivity (reverse field needed to erase it); its enclosed area is the energy dissipated as heat each cycle — the core loss that matters in every transformer and inductor.

Section 13-6

Boundaries & Circuits

Like the electric field, the magnetic field bends at a material interface. The rules mirror Chapter 7: the normal \(\vec{B}\) is continuous (from \(\nabla\cdot\vec{B}=0\)), and the tangential \(\vec{H}\) is continuous when no free surface current flows:

Magnetic boundary conditions
\[ B_{1n} = B_{2n}, \qquad H_{1t} - H_{2t} = K \]

When a field is guided around a core of high \(\mu\) — a transformer or inductor — the flux behaves like current in a wire, and we analyse it as a magnetic circuit. The magnetomotive force \(\mathcal{F}=NI\) drives flux \(\Psi_m\) against reluctance \(\mathcal{R}\), in a direct echo of Ohm's law:

Magnetic Ohm's law
\[ \mathcal{F} = NI = \Psi_m\,\mathcal{R}, \qquad \mathcal{R} = \frac{l}{\mu A} \]
One analogy, two circuits. Voltage drives current through resistance; magnetomotive force drives flux through reluctance. Series and parallel reluctances combine exactly like resistors, so an entire magnetic core — including its air gap — reduces to a simple circuit. This is how engineers design the magnetics inside transformers, relays, and the energy-storage inductors of pulsed-power stages.
Section 13-7

Worked Examples

1 Force on a moving charge

Problem. A \(+2\ \text{mC}\) charge moves at \(\vec{u}=10^5\,\hat{a}_x\ \text{m/s}\) through \(\vec{B}=0.5\,\hat{a}_z\ \text{T}\). Find the force.

Solution. Apply \(\vec{F}=Q\vec{u}\times\vec{B}\):

Working
\[ \vec{F} = (2\times10^{-3})(10^5)(0.5)(\hat{a}_x\times\hat{a}_z) = -100\,\hat{a}_y\ \text{N} \]
2 Force on a wire

Problem. A \(0.5\ \text{m}\) wire carries \(8\ \text{A}\) perpendicular to \(\vec{B}=0.3\ \text{T}\). Find the force.

Solution. Use \(F=BIL\) (perpendicular case):

Working
\[ F = (0.3)(8)(0.5) = 1.2\ \text{N} \]
3 Force between wires

Problem. Two wires \(2\ \text{cm}\) apart each carry \(20\ \text{A}\) in the same direction. Find the force per metre.

Solution. Apply \(F/L=\mu_0 I_1 I_2/2\pi d\):

Working
\[ \frac{F}{L} = \frac{(4\pi\times10^{-7})(20)(20)}{2\pi(0.02)} = 4\ \text{mN/m (attractive)} \]
4 Torque on a coil

Problem. A \(50\)-turn coil of area \(4\ \text{cm}^2\) carries \(2\ \text{A}\) in \(\vec{B}=0.6\ \text{T}\), plane parallel to the field. Find the torque.

Solution. Use \(T=NBIS\) (maximum, plane parallel to \(\vec{B}\)):

Working
\[ T = (50)(0.6)(2)(4\times10^{-4}) = 0.024\ \text{N·m} \]
5 Cyclotron radius

Problem. An electron (\(m=9.11\times10^{-31}\ \text{kg}\)) moves at \(10^6\ \text{m/s}\) perpendicular to \(\vec{B}=0.01\ \text{T}\). Find the orbit radius.

Solution. Use \(r=mu/QB\):

Working
\[ r = \frac{(9.11\times10^{-31})(10^6)}{(1.6\times10^{-19})(0.01)} = 5.7\times10^{-4}\ \text{m} \]
6 Magnetic circuit

Problem. A toroidal core (\(\mu_r=2000\), mean length \(0.4\ \text{m}\), area \(2\ \text{cm}^2\)) is wound with \(200\) turns carrying \(1\ \text{A}\). Find the flux.

Solution. Find reluctance, then \(\Psi_m=NI/\mathcal{R}\):

Working
\[ \mathcal{R} = \frac{0.4}{(2000\mu_0)(2\times10^{-4})} = 7.96\times10^{5}, \quad \Psi_m = \frac{200}{7.96\times10^{5}} = 0.25\ \text{mWb} \]
Review

Chapter Summary

Lorentz force

\(\vec{F}=Q\vec{E}+Q\vec{u}\times\vec{B}\); magnetic part \(\perp\vec{u}\), does no work; orbit radius \(mu/QB\).

On a conductor

\(\vec{F}=I\vec{L}\times\vec{B}\); perpendicular to both current and field.

Between wires

\(F/L=\mu_0 I_1I_2/2\pi d\); same-direction currents attract.

Torque

\(\vec{m}=I\vec{S}\), \(\vec{T}=\vec{m}\times\vec{B}\); the basis of the motor.

Materials

\(\vec{B}=\mu_0(\vec{H}+\vec{M})=\mu\vec{H}\); dia/para/ferro; hysteresis loss in ferromagnets.

Circuits

\(B_{1n}=B_{2n}\), \(H_{1t}-H_{2t}=K\); \(\mathcal{F}=NI=\Psi_m\mathcal{R}\), \(\mathcal{R}=l/\mu A\).

Practice

Problems

For each item, identify whether it is a force, torque, material, or circuit problem, then apply the matching relation. Difficulty rises down the list.

  1. A proton moves at \(2\times10^6\,\hat{a}_y\ \text{m/s}\) in \(\vec{B}=0.2\,\hat{a}_z\ \text{T}\). Find the force.
  2. A \(1\ \text{m}\) wire carries \(15\ \text{A}\) at \(30^\circ\) to a \(0.4\ \text{T}\) field. Find the force.
  3. Two wires \(5\ \text{cm}\) apart carry \(10\ \text{A}\) and \(25\ \text{A}\) in opposite directions. Find the force per metre and its sense.
  4. A \(100\)-turn coil of area \(6\ \text{cm}^2\) carries \(0.5\ \text{A}\) in \(\vec{B}=0.8\ \text{T}\). Find the maximum torque.
  5. Find the cyclotron radius of a \(5\ \text{MeV}\) proton in a \(1.5\ \text{T}\) field.
  6. A material has \(\chi_m=4\times10^{-4}\). Classify it and find \(\mu_r\).
  7. An iron core (\(\mu_r=4000\)) has \(B=1.2\ \text{T}\). Find \(H\) and \(M\).
  8. A magnetic circuit has a core (\(\mu_r=1500\), length \(0.5\ \text{m}\)) and a \(1\ \text{mm}\) air gap, area \(4\ \text{cm}^2\). Find the total reluctance and the flux for \(NI=600\ \text{A·t}\).
  9. At an iron–air interface, \(\vec{B}\) in iron makes \(80^\circ\) with the normal. Find the angle in air (\(\mu_{r,\text{iron}}=2000\)).
  10. Estimate the hysteresis loss in a core if one loop encloses \(500\ \text{J/m}^3\) and the material cycles at \(50\ \text{Hz}\), volume \(10^{-3}\ \text{m}^3\).
  11. Show that the magnetic force on a charge cannot change its kinetic energy.
  12. Explain why the air gap usually dominates the reluctance of a magnetic circuit even though it is far shorter than the iron path.
Tip: two cross products run this whole chapter — \(\vec{F}=I\vec{L}\times\vec{B}\) for force and \(\vec{T}=\vec{m}\times\vec{B}\) for torque — and both are just the Lorentz force in disguise. For materials, lean on the polarization analogy from Chapter 7: \(\vec{M}\) is to magnetism what \(\vec{P}\) was to dielectrics, and \(\mu_r\) plays the role of \(\varepsilon_r\). And whenever you see a wound core with an air gap, stop computing fields and draw the reluctance circuit instead — the gap almost always dominates.