Sources of Magnetic Fields

🧲 Magnetic Force on Moving Charges

Part 1 of 7 — The Lorentz Force

Magnetic Force

$\vec{F} = q\vec{v} \times \vec{B}$

Magnitude: $F = qvB\sin\theta$

Circular Motion in B Field

$qvB = \frac{mv^2}{r}$

$r = \frac{mv}{qB}$

$\omega = \frac{qB}{m}$ (cyclotron frequency)

🔑 A charged particle in a uniform $\vec{B}$ moves in a circle (or helix). The magnetic force provides centripetal acceleration.

Helical Motion and Crossed Fields

Why a helix? Split the velocity into components parallel and perpendicular to $\vec{B}$. The perpendicular part $v_\perp$ feels the force and circles with radius $r = \frac{mv_\perp}{qB}$; the parallel part $v_\parallel$ feels no force ($\sin\theta = 0$) and drifts at constant speed. Together they trace a helix whose pitch (advance per turn) is $p = v_\parallel T = v_\parallel\frac{2\pi m}{qB}$.

The velocity selector. Cross an electric field with a magnetic field so their forces oppose. A charge goes straight only when they cancel:

$qE = qvB \quad\Longrightarrow\quad v = \frac{E}{B}.$

This selects a single speed regardless of charge or mass — the front end of a mass spectrometer.

Speed never changes. Because $\vec{F}\perp\vec{v}$, the magnetic force does zero work, so $|\vec{v}|$ is constant; only the direction turns. Kinetic energy is conserved in any purely magnetic field.

Worked Example — Radius and Period of Circular Motion

A proton ($m = 1.67\times10^{-27}\text{ kg}$, $q = 1.6\times10^{-19}\text{ C}$) enters a uniform field $B = 0.50\text{ T}$ perpendicular to its velocity $v = 2.0\times10^{6}\text{ m/s}$. Find (a) the radius of its path and (b) the period of revolution.

Step 1 — Force provides centripetal acceleration. The magnetic force is the only force, so $qvB = \frac{mv^2}{r}$.

Step 2 — Solve for radius. Cancel one $v$: $r = \frac{mv}{qB} = \frac{(1.67\times10^{-27})(2.0\times10^{6})}{(1.6\times10^{-19})(0.50)}$. Numerator $= 3.34\times10^{-21}$; denominator $= 8.0\times10^{-20}$. So $r = 4.2\times10^{-2}\text{ m} = 4.2\text{ cm}$.

Step 3 — Period. The cyclotron period is $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$. Note that $v$ cancels:

$T = \frac{2\pi (1.67\times10^{-27})}{(1.6\times10^{-19})(0.50)} = 1.3\times10^{-7}\text{ s}.$

Key insight: the cyclotron period (and frequency $\omega = qB/m$) is independent of speed and radius — faster particles trace bigger circles in exactly the same time. This is the principle that makes the cyclotron work.

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🔌 Force on Current-Carrying Wires

Part 2 of 7 — Wires in Magnetic Fields

Force on a Wire

$\vec{F} = I\vec{L} \times \vec{B}$

Magnitude: $F = BIL\sin\theta$

where $L$ is the length of wire in the field.

Torque on a Current Loop

$\vec{\tau} = \vec{\mu} \times \vec{B}$

where $\vec{\mu} = NIA\hat{n}$ is the magnetic dipole moment.

$|\tau| = NIAB\sin\theta$

🔑 This is the principle behind electric motors — a current loop in a magnetic field experiences a torque.

Force on a Bent Wire, and the Motor Principle

For a wire of arbitrary shape, sum the force on each element: $\vec{F} = I\int d\vec{l}\times\vec{B}$. In a uniform field, $\vec{B}$ factors out and the integral reduces to the straight-line displacement between endpoints:

$\vec{F} = I\,\vec{L}_{\text{net}}\times\vec{B},$

where $\vec{L}_{\text{net}}$ points from start to finish. A semicircle of radius $R$ therefore feels the same force as a straight wire of length $2R$ joining its ends.

Closed loop ⇒ zero net force. For any closed loop $\oint d\vec{l} = 0$, so the net force vanishes in a uniform field — yet the loop still feels a torque $\tau = \mu B\sin\theta$ that twists it to align $\vec{\mu}$ with $\vec{B}$.

The motor. A DC motor uses a commutator to flip the current direction every half-turn, so the torque always pushes the loop the same way around. The magnetic potential energy $U = -\vec{\mu}\cdot\vec{B} = -\mu B\cos\theta$ is lowest when aligned — the loop "wants" to rotate toward that configuration.

Worked Example — Torque on a Loop and Integrating Force

A rectangular coil of $N = 50$ turns, width $w = 0.10\text{ m}$ and height $h = 0.20\text{ m}$, carries $I = 2.0\text{ A}$ in a uniform field $B = 0.30\text{ T}$. The coil's normal makes $\theta = 30^\circ$ with $\vec{B}$. Find the torque.

Step 1 — Magnetic moment. $\mu = NIA = NI(wh) = (50)(2.0)(0.10\times0.20) = (50)(2.0)(0.020) = 2.0\ \text{A}\cdot\text{m}^2$.

Step 2 — Apply the torque law. $\tau = \mu B\sin\theta = (2.0)(0.30)\sin30^\circ = (0.60)(0.50) = 0.30\ \text{N}\cdot\text{m}$.

Why the cross product? (calculus view). For a curved or angled wire, the total force is the line integral $\vec{F} = I\int d\vec{l}\times\vec{B}$. In a uniform field $\vec{B}$ can come outside the integral: $\vec{F} = I\left(\int d\vec{l}\right)\times\vec{B}$. For a closed loop, $\int d\vec{l} = 0$, so the net force is zero — but the torque is generally nonzero, which is exactly what spins a motor. The two horizontal sides of our coil carry opposite-direction currents, producing forces that form a couple of magnitude $\tau = \mu B\sin\theta$.

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🔬 Biot-Savart Law

Part 3 of 7 — Magnetic Field from Current

The Biot-Savart Law

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

Common Results

$\mu_0 = 4\pi \times 10^{-7}$ T·m/A

🔑 Right-hand rule: curl fingers in direction of current → thumb points in direction of $\vec{B}$.

How to Attack a Biot-Savart Integral

The law $d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$ is the magnetic analog of Coulomb's law for fields. A reliable recipe:

1. Pick the field point and draw $\hat{r}$ from a typical current element to it.
2. Evaluate the cross product $d\vec{l}\times\hat{r}$ — its magnitude is $dl\sin\theta$, and its direction (right-hand rule) tells you which way $d\vec{B}$ points.
3. Exploit symmetry to cancel components before integrating. (On a loop's axis, only the axial component survives.)
4. Integrate what is left, pulling constants outside.

Finite straight wire. Integrating for a straight segment subtending angles $\theta_1$ and $\theta_2$ at perpendicular distance $d$ gives

$B = \frac{\mu_0 I}{4\pi d}\left(\sin\theta_1 + \sin\theta_2\right).$

For an infinite wire both angles approach $90^\circ$, so $\sin\theta_1 + \sin\theta_2 \to 2$ and $B \to \frac{\mu_0 I}{2\pi d}$ — recovering the familiar result.

Use Biot-Savart when symmetry is too low for Ampère's law (finite segments, arcs, off-axis points). Use Ampère when symmetry is high.

Worked Example — Integrating Biot-Savart for a Loop's Axis

Find the on-axis field of a circular loop of radius $R$ carrying current $I$, at distance $x$ from the center, by integrating the Biot-Savart law. Then evaluate the center.

Step 1 — Set up $dB$. Each element $d\vec{l}$ is perpendicular to $\hat{r}$ (the vector to the axial point), so $|d\vec{l}\times\hat{r}| = dl$ and $dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{r^2}$, where $r = \sqrt{R^2 + x^2}$.

Step 2 — Symmetry kills the perpendicular components. Components transverse to the axis cancel in pairs; only the axial part survives, scaled by $\cos\alpha = \frac{R}{\sqrt{R^2+x^2}}$:

$dB_x = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2+x^2}\cdot\frac{R}{\sqrt{R^2+x^2}}.$

Step 3 — Integrate around the loop. Everything except $dl$ is constant, and $\oint dl = 2\pi R$:

$B_x = \frac{\mu_0 I R}{4\pi (R^2+x^2)^{3/2}}(2\pi R) = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}.$

Step 4 — At the center ($x = 0$). $B = \frac{\mu_0 I R^2}{2R^3} = \frac{\mu_0 I}{2R}$, recovering the standard center-of-loop result.

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🔁 Ampere’s Law

Part 4 of 7 — Symmetry and Magnetic Fields

Ampere’s Law

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$

When to Use Ampere’s Law

Use when there is sufficient symmetry to simplify $\oint \vec{B} \cdot d\vec{l}$:

🔑 Ampere's law is the magnetic analog of Gauss's law — use symmetry!

Choosing the Amperian Loop

Ampère's law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ is always true, but it only solves for $B$ when you can pull $B$ out of the integral. That requires picking a loop along which $\vec{B}$ is either constant-and-parallel or perpendicular-to-$d\vec{l}$ (contributing nothing):

The full field of a thick wire (radius $a$, uniform current) has two regimes:

• Inside ($r < a$): $B = \frac{\mu_0 I r}{2\pi a^2}$ — grows linearly with $r$.
• Outside ($r > a$): $B = \frac{\mu_0 I}{2\pi r}$ — falls as $1/r$.

The two pieces match at the surface ($r = a$), where the field peaks at $\frac{\mu_0 I}{2\pi a}$. Plotting $B$ vs. $r$ gives a triangle-then-tail shape — a classic free-response figure.

Worked Example — Field Inside a Thick Wire

A long cylindrical wire of radius $a = 2.0\text{ mm}$ carries a total current $I = 8.0\text{ A}$ distributed uniformly over its cross-section. Find $B$ at $r = 1.0\text{ mm}$ (inside the wire).

Step 1 — Choose an Amperian loop. By cylindrical symmetry, $\vec{B}$ is tangential with constant magnitude on a circle of radius $r$, so $\oint \vec{B}\cdot d\vec{l} = B(2\pi r)$.

Step 2 — Enclosed current. The current density is $J = \frac{I}{\pi a^2}$. The loop of radius $r$ encloses

$I_{\text{enc}} = J(\pi r^2) = I\frac{r^2}{a^2}.$

Step 3 — Apply Ampère's law. $B(2\pi r) = \mu_0 I\frac{r^2}{a^2}$, giving

$B = \frac{\mu_0 I r}{2\pi a^2}.$

Inside the wire $B$ grows linearly with $r$ (unlike the $1/r$ falloff outside).

Step 4 — Plug in. $B = \frac{(4\pi\times10^{-7})(8.0)(1.0\times10^{-3})}{2\pi(2.0\times10^{-3})^2} = \frac{(2\times10^{-7})(8.0)(1.0\times10^{-3})}{(2.0\times10^{-3})^2} = \frac{1.6\times10^{-9}}{4.0\times10^{-6}} = 4.0\times10^{-4}\text{ T}.$

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🌀 Magnetic Flux

Part 5 of 7 — Flux Through Surfaces

Magnetic Flux

$\Phi_B = \int \vec{B} \cdot d\vec{A}$

For uniform $B$ through flat surface: $\Phi_B = BA\cos\theta$

Units: Weber (Wb) = $T\cdot m^{2}$

Gauss’s Law for Magnetism

$\oint \vec{B} \cdot d\vec{A} = 0$

🔑 No magnetic monopoles — field lines have no beginning or end. Total flux through any closed surface is zero.

Uniform vs. Non-Uniform Flux

When $\vec{B}$ is uniform over a flat surface, the integral collapses: $\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the surface normal $\hat{n}$. Flux is maximal when the normal lies along $\vec{B}$ and zero when the surface is edge-on to the field.

When $\vec{B}$ varies across the surface (for example, near a current-carrying wire where $B\propto 1/r$), you must actually integrate: slice the area into strips on which $B$ is nearly constant, then $\Phi_B = \int B\,dA$. Integrating a $1/r$ field across a strip produces the logarithm $\ln\frac{a+b}{a}$ you will see in the worked example.

Why This Matters for Induction

Flux is the bridge to Faraday's law: an EMF appears only when $\Phi_B$ changes in time. Gauss's law for magnetism, $\oint\vec{B}\cdot d\vec{A}=0$, guarantees the field lines you count entering a closed surface exactly equal those leaving — there are no magnetic charges to act as sources or sinks. This is one of the four Maxwell equations.

Worked Example — Flux from a Wire Through a Loop (Integration)

A long straight wire carries current $I$. A rectangular loop of height $\ell$ lies in the same plane, with its near side a distance $a$ from the wire and its far side at $a + b$. Find the total flux through the loop.

Step 1 — The field varies across the loop. At distance $r$ from the wire, $B(r) = \frac{\mu_0 I}{2\pi r}$, pointing perpendicular to the loop's plane. Because $B$ depends on $r$, we must integrate rather than use $BA$.

Step 2 — Set up the strip. Take a thin strip of width $dr$ at distance $r$, with area $dA = \ell\,dr$. The flux through it is

$d\Phi_B = B(r)\,dA = \frac{\mu_0 I}{2\pi r}\,\ell\,dr.$

Step 3 — Integrate from $a$ to $a+b$.

$\Phi_B = \frac{\mu_0 I \ell}{2\pi}\int_a^{a+b}\frac{dr}{r} = \frac{\mu_0 I \ell}{2\pi}\ln\!\left(\frac{a+b}{a}\right).$

Numbers. With $I = 20\text{ A}$, $\ell = 0.10\text{ m}$, $a = 0.01\text{ m}$, $b = 0.04\text{ m}$: $\Phi_B = \frac{(4\pi\times10^{-7})(20)(0.10)}{2\pi}\ln(5) = (4.0\times10^{-7})\ln 5 = 6.4\times10^{-7}\text{ Wb}$. The logarithm is the signature of a $1/r$ field integrated over distance.

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🛠️ Magnetic Fields Workshop

Part 6 of 7 — Strategies

Choosing the Right Law

Field vs. Force — Don't Mix Them Up

Magnetic problems split into two families. Identify which you are in first:

Family A — Find the FIELD a current creates.

• High symmetry (infinite wire, solenoid, toroid, thick wire) → Ampère's law.
• Low symmetry (finite segment, arc, off-axis point) → Biot-Savart integral.

Family B — Find the FORCE/TORQUE on something in a known field.

• Point charge → $\vec{F} = q\vec{v}\times\vec{B}$.
• Current-carrying wire → $\vec{F} = I\vec{L}\times\vec{B}$.
• Current loop → torque $\vec{\tau} = \vec{\mu}\times\vec{B}$, with $\vec{\mu} = NI A\hat{n}$.

Two-step problems (like parallel wires) chain them: use Family A to get one wire's field, then Family B to get the force on the other. The force per length between long parallel wires, $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$, is the canonical example — same direction currents attract, opposite repel.

Worked Example — Choosing and Combining Tools

A long straight wire carries $I_1 = 6.0\text{ A}$. A second long parallel wire, a distance $d = 0.030\text{ m}$ away, carries $I_2 = 4.0\text{ A}$ in the same direction over a length $L = 1.5\text{ m}$. Find the force between them and its direction.

Step 1 — Field of wire 1 at wire 2 (Ampère / standard result). $B_1 = \frac{\mu_0 I_1}{2\pi d} = \frac{(4\pi\times10^{-7})(6.0)}{2\pi(0.030)} = \frac{(2\times10^{-7})(6.0)}{0.030} = 4.0\times10^{-5}\text{ T}$.

Step 2 — Force on wire 2 (force-on-current law). $F = B_1 I_2 L = (4.0\times10^{-5})(4.0)(1.5) = 2.4\times10^{-4}\text{ N}$.

Step 3 — Direction (right-hand rules). $\vec{B}_1$ at wire 2 points into the page (say), and $\vec{F} = I_2\vec{L}\times\vec{B}_1$ points toward wire 1: parallel currents attract.

Takeaway — the decision tree: use Ampère's law (or a memorized symmetric result) to get the field, then $\vec{F} = I\vec{L}\times\vec{B}$ for the force. Reserve the Biot-Savart integral for fields of finite or curved segments lacking symmetry.

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📋 Magnetic Fields Review

Part 7 of 7 — Summary

Key Formulas

The Big Picture of Magnetostatics

Currents make fields; fields push currents. Those are the two halves of the unit.

Making fields. All field results trace back to Biot-Savart, $d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$. When symmetry is high, Ampère's law $\oint\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ shortcuts the integral. Standard answers to memorize: long wire $\frac{\mu_0 I}{2\pi r}$, loop center $\frac{\mu_0 I}{2R}$, solenoid $\mu_0 nI$.

Feeling forces. A field exerts $\vec{F} = q\vec{v}\times\vec{B}$ on a charge and $\vec{F} = I\vec{L}\times\vec{B}$ on a wire. Because these are cross products, the force is perpendicular to the motion, so magnetic forces do no work — they bend paths into circles and helices but never change speed.

Two field laws you can quote.

• $\oint\vec{B}\cdot d\vec{A} = 0$ (Gauss for magnetism — no monopoles).
• $\oint\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ (Ampère, magnetostatic form).

Together with the two electric Maxwell equations and Faraday's law from the induction unit, these complete the electromagnetic picture.

Worked Example — Velocity Selector and Mass Spectrometer

A velocity selector has perpendicular fields $E = 3.0\times10^{4}\text{ V/m}$ and $B_1 = 0.10\text{ T}$. Ions then enter a region of field $B_2 = 0.20\text{ T}$. (a) Find the selected speed. (b) A singly-charged ion ($q = 1.6\times10^{-19}\text{ C}$) then bends in a circle of radius $r = 0.050\text{ m}$; find its mass.

Part (a) — Balance electric and magnetic forces. An ion passes straight through only if $qE = qvB_1$, so the speed is independent of charge:

$v = \frac{E}{B_1} = \frac{3.0\times10^{4}}{0.10} = 3.0\times10^{5}\text{ m/s}.$

Part (b) — Circular motion in $B_2$. From $r = \frac{mv}{qB_2}$, solve for mass:

$m = \frac{qB_2 r}{v} = \frac{(1.6\times10^{-19})(0.20)(0.050)}{3.0\times10^{5}} = \frac{1.6\times10^{-21}}{3.0\times10^{5}} = 5.3\times10^{-27}\text{ kg}.$

This pair of ideas — a velocity selector feeding a momentum analyzer — is exactly how a mass spectrometer separates isotopes: equal-speed ions of different mass trace circles of different radius via $r = mv/(qB)$.

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