AP Physics C: E&M Unit 4 Notes — Understanding Magnetic Forces

Magnetic Force on Moving Charges

What the magnetic force is (and what makes it different)

A magnetic force is the force a magnetic field exerts on a moving electric charge. The key word is moving: a stationary charge can create an electric field, and it can feel an electric force, but it does not feel a magnetic force from a magnetic field unless it has velocity.

This matters because it explains a huge range of phenomena: why particle beams curve in accelerators, how mass spectrometers separate ions, and why electric currents experience forces in motors. In AP Physics C: E&M, you use the magnetic force law constantly as the “bridge” between magnetic fields (what exists in space) and observable motion/deflection (what you can measure).

The Lorentz force idea

In general, a charge can experience both electric and magnetic effects. The combined force is called the Lorentz force:

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

In this section, the focus is the magnetic part:

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

Where:

  • q is the charge (positive or negative)
  • \vec{v} is the particle’s velocity
  • \vec{B} is the magnetic field
  • \times indicates a **cross product**, meaning the force is perpendicular to both \vec{v} and \vec{B}.

Magnitude and angle dependence

From the cross product, the magnitude is:

F_B = |q|vB\sin\theta

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

Interpretation you should internalize:

  • If \vec{v} is parallel or antiparallel to \vec{B}, then \theta = 0 or \pi and \sin\theta = 0, so F_B = 0. The particle doesn’t get deflected.
  • The force is largest when \vec{v} is perpendicular to \vec{B}, giving F_B = |q|vB.

This angle dependence is a common source of errors—students often plug in |q|vB automatically even when motion is not perpendicular.

Direction: right-hand rule, then fix the sign of the charge

Because \vec{F}_B = q(\vec{v} \times \vec{B}), direction comes from a cross product.

A reliable method:

  1. Use your right hand for \vec{v} \times \vec{B} assuming a positive charge.
  2. If the actual charge is negative (like an electron), reverse the direction.

A quick direction reminder (for a positive charge):

  • Point fingers along \vec{v}.
  • Curl toward \vec{B}.
  • Thumb points along \vec{F}_B.

A crucial property: magnetic forces do no work

Magnetic force is always perpendicular to velocity (when it’s nonzero). That means it can change the direction of motion but not the speed.

Work is:

W = \int \vec{F}\cdot d\vec{r}

If \vec{F}_B is always perpendicular to the displacement direction, then \vec{F}_B\cdot d\vec{r} = 0 and:

W_B = 0

So the kinetic energy stays constant:

\Delta K = 0

This becomes essential when you analyze circular motion in a magnetic field: the field “bends” the path without speeding the particle up.

Units and what a tesla really means

The SI unit of magnetic field is the tesla (T). From F_B = |q|vB\sin\theta, you can view:

1\ \text{T} = \frac{1\ \text{N}}{1\ \text{C}\cdot 1\ \text{m/s}}

This is a useful sanity check for unit consistency in free-response work.

Example 1: force magnitude and direction on a moving charge

A proton with speed v = 3.0\times 10^6\ \text{m/s} enters a uniform magnetic field B = 0.20\ \text{T}. Its velocity is perpendicular to the field.

Step 1: choose the correct formula
Perpendicular motion means \sin\theta = 1:

F_B = |q|vB

Step 2: substitute values
Use |q| = 1.60\times 10^{-19}\ \text{C}:

F_B = (1.60\times 10^{-19})(3.0\times 10^6)(0.20)

F_B = 9.6\times 10^{-14}\ \text{N}

Step 3: direction
Direction is given by \vec{v} \times \vec{B} for a positive charge (proton). If it were an electron, the force would be opposite.

Exam Focus
  • Typical question patterns
    • Given q, v, B, and an angle, find force magnitude and/or direction.
    • Use “no work” to argue speed is constant while direction changes.
    • Combine electric and magnetic forces in a region (often leading to a zero-deflection condition).
  • Common mistakes
    • Forgetting the \sin\theta factor (assuming perpendicular when not stated).
    • Getting the right-hand rule correct for \vec{v} \times \vec{B} but forgetting to reverse for a negative charge.
    • Thinking the magnetic field changes the particle’s speed (it changes direction only, unless an electric field is also present).

Magnetic Force on Current-Carrying Wires

Why currents feel magnetic forces

A current is many charges moving together. Since a moving charge experiences q\vec{v} \times \vec{B}, it makes sense that a wire carrying current in a magnetic field experiences a force too. This is the core physics behind electric motors: current in coils plus magnetic fields produces a torque and motion.

At the macroscopic level, you don’t want to track individual electrons. Instead, you use a compact, experimentally verified result for a straight wire segment.

Force on a straight current segment

For a straight wire segment of length L in a uniform magnetic field,

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

Here:

  • I is the current
  • \vec{L} is a vector of magnitude L pointing in the direction of conventional current
  • \vec{B} is the magnetic field

Magnitude form:

F = ILB\sin\theta

where \theta is the angle between the current direction and \vec{B}.

Conceptual meaning: the wire is pushed sideways when current is not parallel to the field. If the current is parallel to \vec{B}, there’s no force.

Connecting to the moving-charge picture (micro-to-macro)

You can understand where I\vec{L} \times \vec{B} comes from by thinking about charge carriers in the wire:

  • Charges drift with some average drift velocity.
  • Each carrier feels q\vec{v} \times \vec{B}.
  • Summed over all carriers in the wire segment, the total force becomes proportional to the current I and the segment length.

You typically won’t re-derive it in detail on the exam, but this connection helps prevent a common misconception: the force is not on the “wire material” magically—it’s on the moving charges, transmitted to the lattice.

Direction: another right-hand rule

The direction comes from \vec{L} \times \vec{B}:

  • Point fingers along conventional current direction (along \vec{L}).
  • Curl toward \vec{B}.
  • Thumb gives the force direction on the wire.

This is analogous to \vec{v} \times \vec{B} for a positive charge.

Forces between parallel current-carrying wires (high-yield application)

Two long parallel wires carrying currents exert forces on each other because each wire creates a magnetic field that acts on the other wire.

For two long, straight, parallel wires separated by distance r, the magnitude of the force per unit length is:

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

Where:

  • I_1 and I_2 are the currents
  • r is the separation
  • \mu_0 is the permeability of free space

Direction rule (you should memorize the qualitative result):

  • Currents in the same direction attract.
  • Currents in opposite directions repel.

This often shows up as a conceptual question plus a short calculation.

Example 2: force on a wire segment in a uniform field

A straight wire segment of length L = 0.50\ \text{m} carries current I = 4.0\ \text{A}. It lies in a uniform magnetic field B = 0.30\ \text{T} such that the wire is perpendicular to the field.

Step 1: choose magnitude formula
Perpendicular implies \sin\theta = 1:

F = ILB

Step 2: compute

F = (4.0)(0.50)(0.30)

F = 0.60\ \text{N}

Step 3: direction
Use \vec{L} \times \vec{B} with the right-hand rule. The direction depends on the diagram (field into/out of page is common).

Example 3: force per length between two parallel wires

Two long parallel wires are r = 0.040\ \text{m} apart. They carry currents I_1 = 8.0\ \text{A} and I_2 = 3.0\ \text{A} in the same direction. Find the magnitude of force per unit length.

Step 1: use the formula

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

Step 2: substitute (using \mu_0 = 4\pi\times 10^{-7}\ \text{N/A}^2)

\frac{F}{L} = \frac{(4\pi\times 10^{-7})(8.0)(3.0)}{2\pi(0.040)}

Step 3: simplify
Cancel \pi and compute:

\frac{F}{L} = \frac{(4\times 10^{-7})(24)}{2(0.040)}

\frac{F}{L} = \frac{9.6\times 10^{-6}}{0.080}

\frac{F}{L} = 1.2\times 10^{-4}\ \text{N/m}

Because the currents are in the same direction, the force is attractive.

Exam Focus
  • Typical question patterns
    • Compute magnitude/direction of force on a wire using F = ILB\sin\theta.
    • Determine whether two wires attract or repel, then compute F/L.
    • Combine magnetic force on a wire with equilibrium (set net force to zero, often involving weight or tension).
  • Common mistakes
    • Using electron flow direction instead of conventional current direction in the right-hand rule.
    • Treating L as “total wire length” when only the portion in the uniform field matters.
    • Mixing up which angle goes into \sin\theta (it’s between the wire/current direction and \vec{B}).

Motion of Charged Particles in Magnetic Fields

Why magnetic fields cause circular and helical motion

Because \vec{F}_B is perpendicular to \vec{v}, the magnetic force naturally acts like a centripetal force: it changes the direction of velocity without changing its magnitude. That’s exactly what happens in uniform circular motion.

The big idea is to decompose the velocity into components relative to the field:

  • \vec{v}_\parallel: component parallel to \vec{B} (feels no magnetic force)
  • \vec{v}_\perp: component perpendicular to \vec{B} (feels maximum deflecting effect)

Only \vec{v}_\perp contributes to the magnetic force:

F_B = |q|v_\perp B

So:

  • If v_\parallel = 0 and v_\perp \neq 0, the motion is a circle.
  • If both components exist, the motion is a helix (spiral) around the field lines.

Uniform circular motion case (velocity perpendicular to field)

Set magnetic force equal to centripetal force:

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

Solve for the radius (often called the gyroradius):

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

This equation is extremely high-yield: if a particle curves gently, r is large; stronger fields or larger charge magnitude give tighter circles.

Cyclotron (angular) frequency and period

From circular motion, the angular speed is:

\omega = \frac{v}{r}

Substitute r = mv/(|q|B):

\omega = \frac{|q|B}{m}

The period of revolution is:

T = \frac{2\pi}{\omega}

So:

T = \frac{2\pi m}{|q|B}

Two important insights:

  • The period does not depend on the speed (in the nonrelativistic regime). That surprises many students.
  • Heavier particles have longer periods; stronger fields shorten the period.

Helical motion (general velocity)

If the particle has a parallel component v_\parallel, it continues moving along the field at constant speed (no force parallel to \vec{B}), while simultaneously executing circular motion from v_\perp.

Key quantities:

  • Radius:

r = \frac{mv_\perp}{|q|B}

  • Same angular frequency as before:

\omega = \frac{|q|B}{m}

  • Pitch of the helix (distance advanced along the field per revolution):

\text{pitch} = v_\parallel T

This decomposition is often the cleanest way to solve multi-step particle trajectory problems.

Direction of curvature: sign matters

The sense (clockwise vs counterclockwise) of the circular motion depends on the sign of q. The force is q(\vec{v} \times \vec{B}), so a negative charge curves opposite the direction a positive charge would.

A common mistake is to compute the radius correctly (it uses |q|) but then draw the curvature direction as if the charge were positive.

Crossed fields and the velocity selector (application of force balance)

A classic setup has perpendicular electric and magnetic fields, with a particle moving through them. If the particle is to go straight (no deflection), the net transverse force must be zero:

q\vec{E} + q\vec{v} \times \vec{B} = \vec{0}

In magnitude terms (for perpendicular directions arranged to oppose):

qE = qvB

So the selected speed is:

v = \frac{E}{B}

This is called a velocity selector because only particles with that speed pass through undeflected. Note that the charge cancels, so the selected speed is independent of sign and magnitude of charge (though direction of deflection for non-selected speeds does depend on sign).

Example 4: radius of circular motion in a magnetic field

An electron enters a region of uniform magnetic field B = 0.050\ \text{T} with speed v = 2.0\times 10^7\ \text{m/s} perpendicular to the field. Find the radius of its path.

Step 1: use the circular motion radius formula

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

Step 2: substitute constants
Use m_e = 9.11\times 10^{-31}\ \text{kg} and |q| = 1.60\times 10^{-19}\ \text{C}:

r = \frac{(9.11\times 10^{-31})(2.0\times 10^7)}{(1.60\times 10^{-19})(0.050)}

Step 3: compute numerator and denominator

\text{numerator} = 1.822\times 10^{-23}

\text{denominator} = 8.0\times 10^{-21}

r = 2.28\times 10^{-3}\ \text{m}

So the electron moves in a circle of radius about 2.3\ \text{mm}.

Example 5: helical motion pitch

A proton moves in a uniform magnetic field B = 0.20\ \text{T} with speed v = 1.0\times 10^6\ \text{m/s} at an angle such that v_\parallel = 6.0\times 10^5\ \text{m/s}. Find the pitch of the helix.

Step 1: find the period

T = \frac{2\pi m}{|q|B}

Using m_p = 1.67\times 10^{-27}\ \text{kg} and |q| = 1.60\times 10^{-19}\ \text{C}:

T = \frac{2\pi(1.67\times 10^{-27})}{(1.60\times 10^{-19})(0.20)}

Compute the denominator:

(1.60\times 10^{-19})(0.20) = 3.2\times 10^{-20}

So:

T = 2\pi\left(\frac{1.67\times 10^{-27}}{3.2\times 10^{-20}}\right)

T = 2\pi(5.22\times 10^{-8})

T = 3.28\times 10^{-7}\ \text{s}

Step 2: pitch is distance traveled along the field in one period

\text{pitch} = v_\parallel T

\text{pitch} = (6.0\times 10^5)(3.28\times 10^{-7})

\text{pitch} = 1.97\times 10^{-1}\ \text{m}

So the helix advances about 0.20\ \text{m} along the field per turn.

Exam Focus
  • Typical question patterns
    • A charged particle enters a uniform \vec{B} field: determine whether the path is straight, circular, or helical, then find r or T.
    • Use centripetal reasoning: set |q|v_\perp B = mv_\perp^2/r.
    • Crossed-field “no deflection” problems leading to v = E/B.
  • Common mistakes
    • Using total speed v instead of perpendicular component v_\perp in radius and force calculations.
    • Forgetting that magnetic force does no work and incorrectly changing the speed during motion in a purely magnetic field.
    • Getting the curvature direction wrong for negative charges (radius uses |q|, but direction uses the sign of q).