Magnetic Fields and Forces

The Lorentz Force

Part 1 of 7 — $\vec{F} = q\vec{v} \times \vec{B}$

The Magnetic Force on a Moving Charge

A charged particle moving through a magnetic field experiences a force:

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

Key properties:

• The force is perpendicular to both $\vec{v}$ and $\vec{B}$
• The force does no work ($\vec{F} \perp \vec{v} \implies \vec{F} \cdot \vec{v} = 0$)
• The force changes the direction of motion but not the speed

Magnitude

$|\vec{F}| = |q|vB\sin\theta$

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

$\theta$	Force
$0°$ or $180°$	$F = 0$ (parallel to $\vec{B}$)
$90°$	$F = qvB$ (maximum)
General	$F = qvB\sin\theta$

Computing the Cross Product

For $\vec{v} = v_x\hat{x} + v_y\hat{y} + v_z\hat{z}$ and $\vec{B} = B_x\hat{x} + B_y\hat{y} + B_z\hat{z}$:

$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ v_x & v_y & v_z \\ B_x & B_y & B_z \end{vmatrix}$

$= (v_yB_z - v_zB_y)\hat{x} + (v_zB_x - v_xB_z)\hat{y} + (v_xB_y - v_yB_x)\hat{z}$

Example

$\vec{v} = 2\hat{x} + 3\hat{y}$ m/s, $\vec{B} = 4\hat{z}$ T, $q = -e$

$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 2 & 3 & 0 \\ 0 & 0 & 4 \end{vmatrix} = (12 - 0)\hat{x} + (0 - 8)\hat{y} + (0)\hat{z} = 12\hat{x} - 8\hat{y}$

$\vec{F} = q(\vec{v} \times \vec{B}) = (-e)(12\hat{x} - 8\hat{y}) = -12e\hat{x} + 8e\hat{y}$

The Full Lorentz Force

When both electric and magnetic fields are present:

$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$

Velocity Selector

If $\vec{E} = E\hat{y}$ and $\vec{B} = B\hat{z}$, a positive charge moving in the $\hat{x}$ direction:

• Electric force: $\vec{F}_E = qE\hat{y}$
• Magnetic force: $\vec{F}_B = q(v\hat{x} \times B\hat{z}) = -qvB\hat{y}$

For the particle to pass undeflected:

$qE = qvB \implies v = \frac{E}{B}$

Only particles with speed $v = E/B$ pass through, regardless of charge or mass.

Summary — Part 1

Concept	Formula
Magnetic force	$\vec{F} = q\vec{v} \times \vec{B}$
Magnitude	$F =
Full Lorentz	$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
Velocity selector	$v = E/B$
Work done	$W = 0$ (always)

Next up: Circular motion of charged particles in magnetic fields — Part 2.

Circular Motion in Magnetic Fields

Part 2 of 7 — Cyclotron Motion

Uniform Circular Motion

When $\vec{v} \perp \vec{B}$, the magnetic force provides the centripetal acceleration:

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

Solving for the cyclotron radius:

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

Period and Frequency

The cyclotron period (time for one revolution):

$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$

The cyclotron frequency:

$f = \frac{1}{T} = \frac{qB}{2\pi m} \qquad \omega = \frac{qB}{m}$

Critical insight: The period and frequency are independent of velocity. Faster particles trace larger circles but at the same frequency.

Helical Motion

If $\vec{v}$ has components both parallel and perpendicular to $\vec{B}$:

$\vec{v} = v_\parallel \hat{b} + v_\perp \hat{n}$

• $v_\parallel$ is unchanged (no force along $\vec{B}$)
• $v_\perp$ produces circular motion with $r = mv_\perp/(qB)$

The result is a helix with:

$\text{pitch} = v_\parallel T = \frac{2\pi m v_\parallel}{qB}$

Decomposition

If the initial velocity makes angle $\alpha$ with $\vec{B}$:

$v_\parallel = v\cos\alpha, \qquad v_\perp = v\sin\alpha$

$r = \frac{mv\sin\alpha}{qB}, \qquad \text{pitch} = \frac{2\pi m v\cos\alpha}{qB}$

The Cyclotron

A cyclotron accelerates charged particles using:

1. A magnetic field to bend particles in semicircles
2. An oscillating electric field across a gap to accelerate them

Since $T = 2\pi m/(qB)$ is independent of speed, the AC frequency stays constant (non-relativistic limit).

Energy Gained

After $N$ full turns (each crossing the gap twice):

$K = 2NqV_{\text{gap}}$

The maximum kinetic energy (at radius $R$ of the cyclotron):

$K_{\max} = \frac{q^2B^2R^2}{2m}$

Derivation: $r = mv/(qB)$ at the edge gives $v = qBR/m$, so:

$K = \frac{1}{2}mv^2 = \frac{q^2B^2R^2}{2m}$

Summary — Part 2

Quantity	Formula
Cyclotron radius	$r = mv/(qB)$
Period	$T = 2\pi m/(qB)$
Cyclotron frequency	$\omega = qB/m$
Helical pitch	$v_\parallel \cdot T$
Max cyclotron KE	$K = q^2B^2R^2/(2m)$

Next up: The mass spectrometer — Part 3.

Mass Spectrometer

Part 3 of 7 — Separating Ions by Mass

How It Works

A mass spectrometer combines three stages:

1. Ion source — atoms are ionized (given charge $q$)
2. Velocity selector — crossed $\vec{E}$ and $\vec{B}_1$ select $v = E/B_1$
3. Deflection chamber — uniform $\vec{B}_2$ bends ions in semicircles

In the deflection chamber:

$r = \frac{mv}{qB_2}$

After a semicircle, the ion hits a detector at distance:

$d = 2r = \frac{2mv}{qB_2}$

Since $v = E/B_1$ is fixed:

$d = \frac{2mE}{qB_1B_2}$

Different masses land at different positions, allowing identification.

Alternative: Accelerating Potential

Instead of a velocity selector, ions may be accelerated through a potential difference $V_0$:

$qV_0 = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV_0}{m}}$

In the deflection region:

$r = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV_0}{m}} = \frac{1}{B}\sqrt{\frac{2mV_0}{q}}$

$\boxed{r = \frac{1}{B}\sqrt{\frac{2mV_0}{q}}}$

So $r \propto \sqrt{m}$ when accelerated through a fixed potential.

Resolving Power

Two masses $m$ and $m + \Delta m$ are resolved when their semicircular paths are separated by more than the detector resolution $\delta$:

$\Delta d = 2\Delta r = \frac{2\Delta m \cdot E}{q B_1 B_2} \quad (\text{velocity selector})$

$\Delta d = \frac{\Delta m}{B}\sqrt{\frac{2V_0}{mq}} \quad (\text{accelerating potential})$

Applications of Mass Spectrometry

Application	What's Measured
Isotope identification	Mass-to-charge ratio $m/q$
Chemical analysis	Molecular ion masses
Carbon dating	$^{14}$C/$^{12}$C ratio
Doping detection	Trace element concentrations

The Thomson Experiment ($e/m$ of Electron)

J.J. Thomson used crossed $\vec{E}$ and $\vec{B}$ fields:

1. With both fields: $v = E/B$ (velocity selector)
2. With only $\vec{E}$: deflection gives $qE = ma$, so $a = qE/m$
3. Time in field: $t = L/v$
4. Deflection: $y = \frac{1}{2}at^2 = \frac{qEL^2}{2mv^2}$

$\frac{q}{m} = \frac{2yv^2}{EL^2} = \frac{2yE}{B^2L^2}$

Summary — Part 3

Configuration	Radius Formula
Velocity selector ($v$ fixed)	$r = mv/(qB)$, $r \propto m$
Accelerating potential ($V_0$ fixed)	$r = \sqrt{2mV_0/q}/B$, $r \propto \sqrt{m}$
Thomson experiment	$q/m = 2yE/(B^2L^2)$

Next up: Force on current-carrying wires — Part 4.

Force on Current-Carrying Wires

Part 4 of 7 — $\vec{F} = I\vec{L} \times \vec{B}$

From Charges to Currents

A current is moving charges. For $N$ charges each with charge $q$ and drift velocity $\vec{v}_d$ in a wire of length $L$ and cross-section $A$:

$\vec{F} = Nq\vec{v}_d \times \vec{B} = (nAL)(q\vec{v}_d \times \vec{B})$

Since $I = nqv_d A$:

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

where $\vec{L}$ points in the direction of conventional current.

Magnitude

$F = BIL\sin\theta$

where $\theta$ is the angle between the wire and $\vec{B}$.

The Differential Form

For a curved wire or non-uniform field, use the differential element:

$d\vec{F} = I\,d\vec{\ell} \times \vec{B}$

The total force is:

$\vec{F} = I \int d\vec{\ell} \times \vec{B}$

Important Theorem

For a uniform field, the force on any curved wire depends only on the endpoints:

$\vec{F} = I\left(\int d\vec{\ell}\right) \times \vec{B} = I\vec{L}_{\text{net}} \times \vec{B}$

where $\vec{L}_{\text{net}}$ is the displacement vector from start to end.

Consequence: A closed loop in a uniform field feels zero net force (but generally nonzero torque).

Example: Semicircular Wire

A semicircular wire of radius $R$ in the $xy$-plane carries current $I$ in a uniform field $\vec{B} = B\hat{y}$.

$\vec{L}_{\text{net}} = 2R\hat{x} \quad \Rightarrow \quad \vec{F} = I(2R\hat{x}) \times (B\hat{y}) = 2IRB\hat{z}$

Force Between Parallel Wires

Two long parallel wires separated by distance $d$, carrying currents $I_1$ and $I_2$:

Wire 1 creates field $B_1 = \mu_0 I_1/(2\pi d)$ at wire 2.

Force per unit length on wire 2:

$\frac{F}{L} = I_2 B_1 = \frac{\mu_0 I_1 I_2}{2\pi d}$

Currents	Force
Same direction	Attractive
Opposite direction	Repulsive

This defines the ampere: two parallel wires 1 m apart, each carrying 1 A, attract with $F/L = 2 \times 10^{-7}$ N/m.

Summary — Part 4

Formula	Application
$\vec{F} = I\vec{L} \times \vec{B}$	Straight wire, uniform $\vec{B}$
$d\vec{F} = Id\vec{\ell} \times \vec{B}$	Curved wire or non-uniform $\vec{B}$
$F/L = \mu_0 I_1 I_2/(2\pi d)$	Parallel wires
Closed loop, uniform $\vec{B}$	$\vec{F}_{\text{net}} = 0$

Next up: Torque on current loops and magnetic dipoles — Part 5.

Torque on Current Loops

Part 5 of 7 — Magnetic Dipole Moment

Torque on a Rectangular Loop

Consider a rectangular loop (sides $a$ and $b$) carrying current $I$ in a uniform field $\vec{B}$.

Forces on the sides of length $a$ (perpendicular to $\vec{B}$): $F = BIa$

These forces form a couple with moment arm $b\sin\theta$:

$\tau = BIa \cdot b\sin\theta = BIA\sin\theta$

where $A = ab$ is the area of the loop and $\theta$ is the angle between $\vec{B}$ and the normal to the loop.

The Magnetic Dipole Moment

$\vec{\mu} = NIA\hat{n}$

where $N$ is the number of turns and $\hat{n}$ is the unit normal (right-hand rule with current direction).

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

$\tau = \mu B\sin\theta = NIAB\sin\theta$

Potential Energy of a Dipole

The potential energy of a magnetic dipole in a field is:

$U = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta$

Orientation	$\theta$	$U$	Stability
$\vec{\mu} \parallel \vec{B}$	$0$	$-\mu B$	Stable equilibrium
$\vec{\mu} \perp \vec{B}$	$90°$	$0$	—
$\vec{\mu}$ antiparallel $\vec{B}$	$180°$	$+\mu B$	Unstable equilibrium

Work to Rotate

Work done by external agent to rotate from $\theta_1$ to $\theta_2$:

$W = \Delta U = -\mu B(\cos\theta_2 - \cos\theta_1) = \mu B(\cos\theta_1 - \cos\theta_2)$

Connection to Torque

$\tau = -\frac{dU}{d\theta} = -\frac{d}{d\theta}(-\mu B\cos\theta) = -\mu B\sin\theta$

The negative sign confirms torque rotates toward $\theta = 0$ (stable equilibrium).

General Loop Shape

The torque formula $\vec{\tau} = \vec{\mu} \times \vec{B}$ works for any flat loop shape (not just rectangular):

$\vec{\mu} = NIA\hat{n}$

For a non-planar loop, the dipole moment is:

$\vec{\mu} = \frac{I}{2}\oint \vec{r} \times d\vec{\ell}$

Force on a Dipole in a Non-Uniform Field

In a non-uniform field, there is a net force:

$\vec{F} = \nabla(\vec{\mu} \cdot \vec{B})$

For a dipole aligned along $z$ in a field with gradient $\partial B/\partial z$:

$F_z = \mu \frac{\partial B}{\partial z}$

This is how the Stern-Gerlach experiment separates magnetic moments.

Summary — Part 5

Quantity	Formula
Magnetic moment	$\vec{\mu} = NIA\hat{n}$
Torque	$\vec{\tau} = \vec{\mu} \times \vec{B}$
Potential energy	$U = -\vec{\mu} \cdot \vec{B}$
Force (non-uniform $\vec{B}$)	$\vec{F} = \nabla(\vec{\mu} \cdot \vec{B})$

Next up: Problem-solving workshop — Part 6.

Problem-Solving Workshop

Part 6 of 7 — AP Physics C: E&M Style Problems

Strategy for Magnetic Force Problems

1. Identify the charged object: single particle, wire, or loop.
2. Write the appropriate force law:
   • Particle: $\vec{F} = q\vec{v} \times \vec{B}$
   • Wire: $\vec{F} = I\vec{L} \times \vec{B}$ or $d\vec{F} = Id\vec{\ell} \times \vec{B}$
   • Dipole: $\vec{\tau} = \vec{\mu} \times \vec{B}$
3. Evaluate cross products carefully using the determinant method.
4. Apply Newton's second law or energy methods as needed.

Worked Example: Hall Effect

A conducting slab (width $w$, thickness $t$) carries current $I$ in the $+x$ direction through a field $\vec{B} = B\hat{z}$.

Step 1: Current carriers (electrons) drift with $\vec{v}_d = -v_d\hat{x}$.

Step 2: Magnetic force on electrons: $\vec{F} = (-e)(-v_d\hat{x}) \times (B\hat{z}) = ev_dB(\hat{x} \times \hat{z}) = -ev_dB\hat{y}$

Electrons accumulate on the $-y$ face, creating a transverse electric field.

Step 3: Equilibrium: $eE_H = ev_dB$, so $E_H = v_dB$.

Step 4: Hall voltage: $V_H = E_H w = v_d B w$

Since $I = nev_d(wt)$, we get $v_d = I/(newt)$:

$V_H = \frac{IB}{net} = \frac{R_H IB}{t}$

where $R_H = 1/(ne)$ is the Hall coefficient.

Workshop Summary

Problem Type	Key Approach
Same KE comparison	$r \propto \sqrt{m}/q$
Hall effect	$V_H = IB/(net)$
Angular momentum	$L = qBR^2$
Curved wire, uniform $\vec{B}$	Use $\vec{L}_{\text{net}}$

Next up: Review and applications — Part 7.

Review & Applications

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Concept	Formula
Lorentz force	$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$
Cyclotron radius	$r = mv/(qB)$
Cyclotron frequency	$\omega = qB/m$
Force on wire	$\vec{F} = I\vec{L} \times \vec{B}$
Differential force	$d\vec{F} = Id\vec{\ell} \times \vec{B}$
Parallel wires	$F/L = \mu_0 I_1 I_2/(2\pi d)$
Magnetic moment	$\vec{\mu} = NIA\hat{n}$
Torque	$\vec{\tau} = \vec{\mu} \times \vec{B}$
Dipole energy	$U = -\vec{\mu} \cdot \vec{B}$
Hall voltage	$V_H = IB/(net)$

Applications in Modern Physics

1. MRI (Magnetic Resonance Imaging)

Protons in hydrogen atoms precess at the Larmor frequency: $f = \frac{\gamma B}{2\pi}$ where $\gamma = qB/(2m)$ for a classical magnetic moment. Varying $B$ with gradient coils selects spatial slices.

2. Particle Accelerators

Charged particles are bent by magnetic fields and accelerated by electric fields. The magnetic rigidity is: $B\rho = \frac{p}{q} = \frac{mv}{q} \quad (\text{or } \frac{\gamma mv}{q} \text{ relativistically})$

3. Aurora Borealis

Charged particles from the solar wind spiral along Earth's magnetic field lines ($v_\parallel$ carries them toward the poles; $v_\perp$ gives helical motion). They excite atmospheric molecules, producing light.

4. Electric Motors

A current loop in a magnetic field experiences torque $\vec{\tau} = \vec{\mu} \times \vec{B}$. A commutator reverses current each half-turn to maintain continuous rotation.

🎉 Topic Complete!

You've mastered Magnetic Forces for AP Physics C: E&M:

Part	Topic	Status
1	Lorentz force	✅
2	Circular motion in B fields	✅
3	Mass spectrometer	✅
4	Force on current-carrying wire	✅
5	Torque on current loop	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key takeaway: The cross product is central to all magnetic force problems. Master the determinant method, understand that magnetic forces do no work, and remember that closed loops in uniform fields have zero net force but generally nonzero torque.