Electromagnetism · Chapter 15

Magnetic Fields

A moving charge creates a magnetic field, and a magnetic field exerts a force on moving charges. These two facts, combined with Faraday's law, unify electricity and magnetism into a single theory.

Prerequisites

Electric fields (Ch. 13) \cdot DC circuits (Ch. 14) \cdot Vectors and cross products (Math Ch

Learning Goals

Use the Lorentz force law to predict the magnitude and direction of magnetic forces.
Explain why magnetic forces change particle direction but do no work.
Derive cyclotron radius and period for motion perpendicular to a uniform magnetic field.
Compute magnetic fields from common current geometries.
Apply Ampère's law when symmetry makes the line integral simple.

15.1 The Magnetic Force

A charged particle moving with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$ experiences the Lorentz force:

\mathbf{F} = q\mathbf{v}\times\mathbf{B}

(15.1)

The cross product means the force is perpendicular to both the velocity and the field. This has three immediate consequences: (1) a stationary charge feels no magnetic force; (2) a charge moving parallel to $\mathbf{B}$ feels no force; (3) the force does no work — it can change direction but not speed.

The magnitude is $F = |q|vB\sin\theta$ , where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{B}$ . The direction is given by the right-hand rule: point fingers in the direction of $\mathbf{v}$ , curl toward $\mathbf{B}$ , and the thumb points in the direction of $\mathbf{F}$ (for positive $q$ ; reverse for negative $q$ ).

Definition 15.1 — Cyclotron Motion

A charged particle moving perpendicular to a uniform magnetic field follows a circular path. The magnetic force provides the centripetal acceleration:

|q|vB = \frac{mv^2}{r}

→

r = \frac{mv}{|q|B}

This radius

r

is the cyclotron radius (or Larmor radius). The period

T = 2\pi m/(|q|B)

is independent of velocity — the basis of the cyclotron particle accelerator.

Figure 15.1. Magnetic field simulator. Toggle among a straight wire, solenoid, and moving charge: reverse the current to see field directions flip, increase current to see field strength rise, and change charge sign to reverse the Lorentz-force direction in cyclotron motion.

15.2 Magnetic Fields from Currents

Just as a charge creates an electric field, a moving charge (current) creates a magnetic field. The fundamental law for this is the Biot–Savart law: each current element $I\,d\mathbf{l}$ contributes a field $d\mathbf{B}$ at position $\mathbf{r}$ :

d\mathbf{B} = \frac{\mu_0}{4\pi}\frac{I\,d\mathbf{l}\times\hat{\mathbf{r}}}{r^2}

(15.2)

where $\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m/A}$ is the permeability of free space. For practical geometries, we integrate this law to find closed-form results.

Theorem 15.1 — Magnetic Field of Common Current Configurations

Infinite straight wire at distance

r

B = \mu_0I/(2\pi r)

, circles the wire by right-hand ruleCircular loop of radius

R

at center:

B = \mu_0I/(2R)

, along the axisSolenoid (

n

turns/meter, inside):

B = \mu_0nI

, uniform and axial

15.3 Ampère's Law

Ampère's law is the magnetic analogue of Gauss's law. For any closed loop (Amperian loop), the line integral of $\mathbf{B}$ around the loop equals $\mu_0$ times the current threading the loop:

\oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_\mathrm{enc}

(15.3)

Like Gauss's law, Ampère's law is always true but only useful for deriving fields when the geometry is highly symmetric (infinite wire, solenoid, toroid). For the infinite wire, choose a circular Amperian loop of radius $r$ : $B(2\pi r) = \mu_0I$ , immediately giving $B = \mu_0I/(2\pi r)$ .

Example 15.1 — Force Between Two Parallel Wires

Two parallel wires $0.5\,\mathrm{m}$ apart carry currents $I_1 = 10\,\mathrm{A}$ and $I_2 = 20\,\mathrm{A}$ in the same direction. Find the force per unit length between them.

Field from wire 1:

B_1 = \frac{\mu_0I_1}{2\pi d} = \frac{(4\pi\times10^{-7})(10)}{2\pi(0.5)} = 4\times10^{-6}\,\mathrm{T}

Force on wire 2:

F/L = I_2B_1 = 20(4\times10^{-6}) = 8\times10^{-5}\,\mathrm{N/m}

Direction:Same-direction currents attract (opposite-direction repel).

Note:

This experiment defines the ampere: 2\times10^{-7} N/m force per meter between wires 1 m apart car

rying 1 A each.

Example 15.2 — Cyclotron Radius of a Proton

A proton ( $m = 1.67 \times 10^{-27}\,\mathrm{kg}$ , $q = 1.6 \times 10^{-19}\,\mathrm{C}$ ) moves at $2 \times 10^6\,\mathrm{m/s}$ perpendicular to a $0.1\,\mathrm{T}$ magnetic field. Find the radius of its circular orbit.

Formula:

r = \frac{mv}{qB}

Substitute:

r = \frac{(1.67\times10^{-27})(2\times10^6)}{(1.6\times10^{-19})(0.1)}

Calculate:

r = \frac{3.34\times10^{-21}}{1.6\times10^{-20}} = 0.209\,\mathrm{m} \approx 21\,\mathrm{cm}

15.4 The Magnetic Force on a Current

A current-carrying wire in a magnetic field experiences a force — since each mobile charge experiences $\mathbf{F} = q\mathbf{v}\times\mathbf{B}$ , the wire as a whole feels a net force. For a straight segment of length $L$ carrying current $I$ in field $\mathbf{B}$ :

\mathbf{F} = I\mathbf{L}\times\mathbf{B} \qquad \text{(magnitude: }F = BIL\sin\theta\text{)}

(15.4)

This is the operating principle of every electric motor: a current loop in a magnetic field experiences a torque $\tau = NIAB\sin\theta$ ( $N$ turns, area $A$ , tilt angle $\theta$ ), which causes rotation. The torque is maximized when the loop is parallel to the field ( $\theta = 90^\circ$ ) and zero when it is perpendicular (aligned with $\mathbf{B}$ ) — requiring a commutator to maintain continuous rotation.

Definition 15.2 — Common Traps

Magnetic force needs motion: a stationary charge feels no magnetic force.
Perpendicular matters: only the velocity component perpendicular to B curves the path.
The force does no work: it changes direction, not speed, for an isolated charged particle.
Right-hand rules depend on sign: reverse the direction for negative charges.
Ampère's law needs symmetry: the law is general, but the shortcut works only when B is constant along the chosen loop.

Exercises — 15.1–15.4 Magnetic Fields

A proton moves at 2\times10^{6} m/s perpendicular to a 0.1 T field. Find the cyclotron radius

Straightforward

Two parallel wires are 0.5 m apart carrying I_{1} = 10 A and I_{2} = 20 A. Find the force per me

ter.

N/m

Straightforward

Find the magnetic field 0.5 m from an infinite straight wire carrying 10 A.

Intermediate

4.Explain why the period of circular cyclotron motion is independent of the particle's speed. How does this make the cyclotron particle accelerator possible?

Intermediate

5.Describe how a

mass spectrometer separates ions by mass-to-charge ratio. Derive the detector separation \Delta

of the same charge differing in mass by \Deltam

Challenging

Key Takeaways

$Lorentz force: F = q(v \times B) — perpendicular to both velocity and field, does no work$ .
$Cyclotron radius r = mv/|q|B; period T = 2\pim/|q|B is speed-independent$ .
Biot–Savart law gives field from current elements; Ampère's law gives field from symmetric configurations.
$Infinite wire: B = \mu_{0}I/2\pir; solenoid: B = \mu_{0}nI (uniform, inside$ .
Parallel wires attract if currents are in the same direction, repel if opposite.
$Motor torque: \tau = NIAB sin \theta — the rotating coil is maximally torqued when parallel to B$ .