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 around the field direction.

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 and $\vec{L}$ points along the conventional current.

Force Between Parallel Wires

Wire 1 makes a field $B_1 = \mu_0 I_1/(2\pi d)$ at wire 2, so the force per unit length on wire 2 is

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

🔑 Currents in the same direction attract; currents in opposite directions repel.

Force on a Bent Wire, and a Loop Near a Long Wire

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 in a uniform field ⇒ zero net force. For any closed loop $\oint d\vec{l} = 0$, so the net force vanishes.

Closed loop in a NONuniform field ⇒ net force. Put a rectangular loop beside a long straight wire, in the same plane. The field $B = \mu_0 I_1/(2\pi r)$ is stronger at the near side than at the far side. The two sides perpendicular to the wire feel equal and opposite forces that cancel, but the two sides parallel to the wire do not:

$F_{\text{net}} = \frac{\mu_0 I_1 I_2 L}{2\pi}\left(\frac{1}{a} - \frac{1}{a + w}\right)$

where $L$ is the length of the parallel sides, $a$ the distance to the near side, and $w$ the loop's width. The near side wins, so the loop is pulled toward the wire when its near-side current is parallel to $I_1$.

Worked Example — Net Force on a Loop Beside a Long Wire

A long straight wire carries $I_1 = 20\text{ A}$. A rectangular loop in the same plane carries $I_2 = 5.0\text{ A}$; its sides parallel to the wire have length $L = 0.30\text{ m}$, the near side is $a = 0.050\text{ m}$ from the wire, and the far side is $0.15\text{ m}$ from it. The near-side current runs parallel to $I_1$. Find the net force on the loop.

Step 1 — Near side. $F_{\text{near}} = \frac{\mu_0 I_1 I_2 L}{2\pi a} = \frac{(2\times10^{-7})(20)(5.0)(0.30)}{0.050} = \frac{6.0\times10^{-6}}{0.050} = 1.2\times10^{-4}\text{ N}$, toward the wire (parallel currents attract).

Step 2 — Far side. $F_{\text{far}} = \frac{6.0\times10^{-6}}{0.15} = 4.0\times10^{-5}\text{ N}$, away from the wire (its current is antiparallel to $I_1$).

Step 3 — The perpendicular sides. Each element of the top side has a partner on the bottom side at the same distance carrying current the opposite way, so their forces cancel.

Step 4 — Net force. $F_{\text{net}} = 1.2\times10^{-4} - 4.0\times10^{-5} = 8.0\times10^{-5}\text{ N}$, toward the wire.

Takeaway. "Closed loop ⇒ zero force" holds only in a uniform field. In a field that varies with position, add the force on each side separately.

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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 (perpendicular bisector). For a segment of length $L$ and a point a distance $d$ from its midpoint, each end is seen at angle $\theta$ from the perpendicular, with $\sin\theta = \frac{L/2}{\sqrt{d^2 + L^2/4}}$. Integrating gives

$B = \frac{\mu_0 I}{4\pi d}\left(2\sin\theta\right) = \frac{\mu_0 I}{2\pi d}\sin\theta.$

For a very long wire $\theta \to 90^\circ$, so $\sin\theta \to 1$ and $B \to \frac{\mu_0 I}{2\pi d}$ — recovering the familiar result.

Use Biot-Savart for the cases Ampère's law cannot handle: a straight segment (on its perpendicular bisector), an arc (at its center), and a loop (on its axis). 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 (long wire, solenoid, slab, thick wire with current density) → Ampère's law.
• Low symmetry (finite segment on its perpendicular bisector, arc at its center, loop on its axis) → Biot-Savart integral.

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

• Point charge → $\vec{F} = q\vec{v}\times\vec{B}$.
• Current-carrying wire → $\vec{F} = I\vec{L}\times\vec{B}$.
• Closed loop → add the forces on its sides (zero net force in a uniform field).

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 Gauss's law for electric fields and Faraday's law from the induction unit, these make up Maxwell's four equations (Ampère's law gains Maxwell's addition, a changing electric field, which you need only qualitatively).

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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