Magnetic Fields and Magnetic Forces

AP Physics C: E&M· AP Physics C: E&M CED — Magnetic Fields· 14 min read

1. Core Concepts of Magnetic Fields★★☆☆☆⏱ 3 min

A magnetic field ( $B$ ) is a vector field that only exerts force on moving electric charge. Unlike electric fields, magnetic forces never do work on charge because the force is always perpendicular to the charge's displacement. The SI unit for magnetic field is the tesla (T), where $1 T = 1 N / (A \cdotp m)$ .

Magnetic forces are tested in three key contexts on the AP exam: force on a single moving point charge, force on a bulk current-carrying wire, and torque on a current-carrying loop. This topic is also the foundation for mass spectrometry problems involving charged particle trajectories.

2. The Lorentz Force Law★★☆☆☆⏱ 3 min

📘 Definition

Lorentz Force Law

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

Describes the total force exerted on a charged particle moving through combined electric and magnetic fields, the starting point for all magnetic force calculations.

If no electric field is present, the law simplifies to $F = q v \times B$ . The magnitude of the magnetic force is $F = q v B sin θ$ , where $θ$ is the angle between $v$ and $B$ .

Direction follows the right-hand rule for cross products: for positive $q$ , point the fingers of your right hand along $v$ , curl them toward $B$ , and your thumb points to the force direction. For negative $q$ , the force is exactly opposite this direction. A key property: magnetic force is always perpendicular to velocity, so it changes only direction of motion, not speed, so it never does work.

📐 Worked Example

A proton with charge $q = 1.6 \times 1 0^{- 19}$ C moves at $v = 2.0 \times 1 0^{6}$ m/s along the +x axis, through a uniform magnetic field $B = 0.50$ T $\hat{k}$ (out of the page). What is the magnitude and direction of the magnetic force on the proton?

1
Identify given vectors:
$\vec{v} = 2.0 \times 10^6 \hat{i} \text{ m/s}, \vec{B} = 0.50 \hat{k} \text{ T}, q>0$
2
Calculate the cross product:
$\vec{v} \times \vec{B} = (2.0 \times 10^6 \hat{i}) \times (0.50 \hat{k}) = 1.0 \times 10^6 (\hat{i} \times \hat{k}) = -1.0 \times 10^6 \hat{j} \text{ m·T/s}$
3
Multiply by charge to get force:
$\vec{F} = (1.6 \times 10^{-19})(-1.0 \times 10^6 \hat{j}) = -1.6 \times 10^{-13} \hat{j} \text{ N}$
4
Final result: magnitude $1.6 \times 1 0^{- 13}$ N, direction along the negative y-axis.

Exam tip:

Always explicitly check the sign of the charge before writing your final direction. AP MCQs almost always include an option that matches the magnitude but has the wrong direction for negative charges to catch this common mistake.

3. Force on Current-Carrying Wires★★★☆☆⏱ 3 min

A current in a wire is a continuous stream of moving charges, so the force on a wire can be derived directly from the Lorentz force law. For an infinitesimal segment of wire of length $d l$ , where $d l$ points in the direction of conventional current, the force on the segment is $d F = I d l \times B$ , where $I$ is the current.

For a uniform magnetic field, integrating over the full length of the wire gives:

\vec{F} = I \left(\int d\vec{l}\right) \times \vec{B}

For a straight wire of total length $L$ , this simplifies to $F = I L \times B$ , with magnitude $F = I L B sin θ$ , where $θ$ is the angle between $L$ (current direction) and $B$ . A key result: the net force on any closed current loop in a uniform magnetic field is zero, because the integral of $d l$ around a closed loop equals zero.

📐 Worked Example

A 30 cm long straight wire carries 2.0 A of current along the +y axis, in a uniform magnetic field $B = 1.2$ T $\hat{i} + 0.8$ T $\hat{k}$ . Find the magnitude of the net magnetic force on the wire.

1
Write the length vector aligned with current:
$\vec{L} = 0.30 \hat{j} \text{ m}$
2
Calculate the cross product:
$\vec{L} \times \vec{B} = 0.30 \hat{j} \times (1.2 \hat{i} + 0.8 \hat{k}) = 0.30(-1.2 \hat{k} + 0.8 \hat{i}) = 0.24 \hat{i} - 0.36 \hat{k} \text{ m·T}$
3
Multiply by current to get force vector:
$\vec{F} = 2.0(0.24 \hat{i} - 0.36 \hat{k}) = 0.48 \hat{i} - 0.72 \hat{k} \text{ N}$
4
Calculate the magnitude of the force:
$|\vec{F}| = \sqrt{(0.48)^2 + (0.72)^2} = \sqrt{0.7488} \approx 0.87 \text{ N}$

4. Motion of Charged Particles in Uniform Magnetic Fields★★★☆☆⏱ 3 min

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force that maintains uniform circular motion. Equating the Lorentz force magnitude to centripetal force gives:

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

Rearranging gives the radius of the circular path:

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

where $p = m v$ is the particle's momentum. The cyclotron angular frequency is $ω = \frac{q B}{m}$ , which is independent of the particle's speed or radius, the property that makes cyclotron particle accelerators work. If the particle has a velocity component parallel to $B$ , that component experiences no force, so the particle follows a helical path. This behavior is the basis of mass spectrometry, where particles of different mass are separated by their radius of curvature.

📐 Worked Example

A mass spectrometer uses a 0.25 T uniform magnetic field to separate singly ionized ( $q = + e$ ) carbon isotopes. Carbon-12 has mass $1.99 \times 1 0^{- 26}$ kg, and enters the B field perpendicular to $B$ with speed $1.2 \times 1 0^{5}$ m/s. What is the diameter of its circular path?

1
Recall the radius formula for perpendicular motion: $r = \frac{m v}{q B}$
2
Substitute given values:
$r = \frac{(1.99 \times 10^{-26} \text{ kg})(1.2 \times 10^5 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C})(0.25 \text{ T})} \approx 0.060 \text{ m} = 6.0 \text{ cm}$
3
The question asks for diameter, not radius: multiply the radius by 2
4
Final result: diameter $d = 2 (0.060 m) = 0.12$ m = 12 cm

Exam tip:

AP problems intentionally ask for diameter (the quantity measured experimentally in mass spectrometers) much more often than radius. Always confirm what quantity you are asked for before writing your final answer.

5. Torque on Current-Carrying Loops★★★★☆⏱ 2 min

Even though the net force on a closed current loop in a uniform magnetic field is zero, there is a net torque that rotates the loop to align its magnetic dipole moment with the magnetic field. The torque formula is:

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

where $μ$ , the magnetic dipole moment of the loop, has magnitude $μ = I A$ , with $I$ the current and $A$ the area of the loop. The direction of $μ$ is along the normal to the plane of the loop, found via right-hand rule: curl your fingers along the direction of current, and your thumb points to $μ$ . The magnitude of torque is $τ = I A B sin θ$ , where $θ$ is the angle between $μ$ and $B$ . The potential energy of the dipole in the B field is $U = - μ \cdot B$ , so the lowest energy (equilibrium) state is when $μ$ is aligned with $B$ . This relationship is the operating principle of electric motors and galvanometers.

📐 Worked Example

A rectangular 2.0 cm × 3.0 cm current loop carries 5.0 A of current, and is placed in a 0.4 T uniform magnetic field. The angle between the plane of the loop and the magnetic field is 60 degrees. What is the magnitude of the torque on the loop?

1
Calculate the area of the loop:
$A = (0.02 \text{ m})(0.03 \text{ m}) = 6.0 \times 10^{-4} \text{ m}^2$
2
Find $θ$ , the angle between the normal (dipole moment) and B: $θ = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$
3
Calculate torque using the magnitude formula:
$\tau = IAB\sin\theta = (5.0 A)(6.0 \times 10^{-4} \text{ m}^2)(0.4 T)(\sin 30^\circ) = 6.0 \times 10^{-4} \text{ N·m}$

Exam tip:

If the problem gives the angle between the plane of the loop and B, always subtract from 90 degrees to get the correct $θ$ for the torque formula.

6. Concept Check

✓ Quick check

Test your understanding with this AP-style multiple choice question:

An electron moving with velocity $v = 2.0 \times 1 0^{5} \hat{i}$ m/s enters a velocity selector region with uniform crossed fields $E = 100 \hat{j}$ V/m and $B = B \hat{k}$ T. The net force on the electron is zero. What is the magnitude of B?
- $2.0 \times 1 0^{- 7}$ T
- $5.0 \times 1 0^{- 4}$ T
- $2.0 \times 1 0^{- 3}$ T
- $5.0 \times 1 0^{6}$ T
Reveal answer
1 —
For zero net force, the magnitude of electric force equals the magnitude of magnetic force: $∣ q E ∣ = ∣ q v B ∣$ . The charge cancels out regardless of sign, so $B = E / v = 100/ (2.0 \times 1 0^{5}) = 5.0 \times 1 0^{- 4}$ T.

7. Common Pitfalls

Wrong move:

Forgetting to flip the force direction for negative charges, giving the same direction as a positive charge with identical velocity.

Why:

Students memorize the right-hand rule for positive charge and do not account for the sign of $q$ in the cross product.

Correct move:

Always explicitly note the sign of $q$ after finding the right-hand direction, and flip direction if $q$ is negative.

Wrong move:

Stopping after calculating radius when the question asks for the diameter of a particle’s circular path, getting an answer half the correct value.

Why:

Students default to outputting radius from the common formula, and do not read the question’s request carefully.

Correct move:

Circle the quantity the question asks for (radius, diameter, momentum, frequency) before starting calculations, and double-check at the end.

Wrong move:

Calculating non-zero net work done by magnetic force, using $W = F d$ .

Why:

Students confuse magnetic force with other constant forces and forget its key direction property.

Correct move:

If asked for work done by magnetic force, immediately state it is zero, because force is always perpendicular to displacement.

Wrong move:

Using the angle between the plane of the current loop and B as $θ$ in the torque formula.

Why:

Problems often describe orientation relative to the plane, so students mix up plane angle and normal angle.

Correct move:

Recall $θ$ is the angle between B and the dipole moment (normal), so $θ = 9 0^{\circ} - ϕ$ , where $ϕ$ is the angle between the plane of the loop and B.

Wrong move:

Claiming net force on a closed current loop is zero in a non-uniform magnetic field.

Why:

Students memorize the zero net force rule and forget it only applies to uniform B.

Correct move:

Only use the zero net force rule for uniform B; integrate $d F = I d l \times B$ for non-uniform B to find net force.

8. Quick Reference Cheatsheet

Category	Formula	Notes
Lorentz Force	$F = q (E + v \times B)$	Direction flipped for negative $q$ ; does no work because $F ⊥ v$
Force on straight wire	$F = I L \times B$	$L$ points in direction of current; applies only to uniform $B$
Radius of charged particle circle	$r = \frac{m v}{q B} = \frac{p}{q B}$	Only applies when velocity is perpendicular to uniform $B$
Cyclotron angular frequency	$ω = \frac{q B}{m}$	Independent of particle speed/radius for non-relativistic speeds
Magnetic dipole moment	$μ = I A \overset{n}{^}$	$\overset{n}{^}$ is normal to loop, found via right-hand rule along current
Torque on current loop	$τ = μ \times B$	Net force on loop is zero only for uniform $B$
Potential energy of dipole	$U = - μ \cdot B$	Minimum energy when $μ$ aligned with $B$
Force on curved wire (uniform B)	$F = I L_{e n d} \times B$	$L_{e n d}$ is straight vector from start to end of curved wire

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Lorentz force direction problem
2022 · FRQ
Mass spectrometry radius problem

What's Next

This sub-topic is the foundational prerequisite for all remaining topics in Unit 4 (Magnetic Fields), starting with the calculation of magnetic fields produced by currents via the Biot-Savart law and Ampere's law. To find the force between two parallel current-carrying wires, you need the force on a current-carrying wire from this chapter, combined with the B-field produced by the second wire that you will derive next. Without mastering the direction rules and force formulas from this chapter, you cannot correctly solve problems involving solenoids, motors, or electromagnetic induction in later units. This topic is also the basis for all real-world electromagnetic device problems that commonly appear on AP exam FRQs.