D.3 Motion in electromagnetic fields

IB Physics HL· D.3 Motion in electromagnetic fields· 20 min read

1. Lorentz Force on Moving Charges★★☆☆☆⏱ 5 min

📘 Definition

Lorentz Force

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

The total electromagnetic force on a charged particle moving through electric and magnetic fields. For pure magnetic fields, $F = q v B sin θ$ , where $θ$ is the angle between velocity $v$ and magnetic flux density $B$ .

Example:

A proton moving perpendicular to a 0.5 T field at $2 \times 1 0^{6}$ m/s experiences a magnetic force of $1.6 \times 1 0^{- 13}$ N.

The direction of the magnetic force is always perpendicular to both the velocity of the charge and the magnetic field, found using the right-hand rule for positive charges. For negative charges, the force direction is reversed.

📐 Worked Example

An electron moving at $3.0 \times 1 0^{6}$ m/s enters a uniform 0.2 T magnetic field at 90° to the field lines. Calculate the magnitude of the magnetic force acting on the electron.

1
Recall the force formula for perpendicular motion:
2
$F = qvB$
3
Substitute the known values $q = 1.6 \times 1 0^{- 19}$ C:
4
$F = (1.6 \times 10^{-19})(3.0 \times 10^6)(0.2)$
5
Calculate the final magnitude:
6
$F = 9.6 \times 10^{-14} \text{ N}$

2. Circular Motion in Uniform Magnetic Fields★★☆☆☆⏱ 6 min

When a charged particle enters a uniform magnetic field with velocity perpendicular to the field, the magnetic force is always perpendicular to velocity. This acts as a centripetal force, causing uniform circular motion with constant speed, because no work is done by the magnetic force.

🔬 Derivation

Goal:

Derive the expression for the radius of the circular path

Starting from:

Equate magnetic force to centripetal force for perpendicular motion

1
Magnetic force provides centripetal force:
2
$qvB = \frac{mv^2}{r}$
3
Cancel velocity $v$ from both sides:
4
$qB = \frac{mv}{r}$
5
Rearrange to solve for radius $r$ :
6
$r = \frac{mv}{qB}$

Result:

Radius is directly proportional to particle momentum $m v$ , inversely proportional to charge $q$ and magnetic flux density $B$ .

📐 Worked Example

A proton ( $m = 1.67 \times 1 0^{- 27}$ kg, $q = + 1.6 \times 1 0^{- 19}$ C) moves in a circular path of radius 0.2 m in a uniform 0.15 T magnetic field. Calculate its speed.

1
Start with the radius formula:
2
$r = \frac{mv}{qB}$
3
Rearrange to isolate $v$ :
4
$v = \frac{rqB}{m}$
5
Substitute values:
6
$v = \frac{(0.2)(1.6 \times 10^{-19})(0.15)}{1.67 \times 10^{-27}} \approx 2.9 \times 10^6 \, \text{m/s}$

3. Combined Electric and Magnetic Fields★★★☆☆⏱ 7 min

When both electric and magnetic fields are present, the total Lorentz force is the vector sum of the electric force $q E$ and magnetic force $q v \times B$ . A common IB exam application is the velocity selector, which uses crossed (perpendicular) electric and magnetic fields to select particles of a specific velocity.

📘 Definition

Velocity Selector

A device that allows only particles with a specific velocity to pass through undeflected. For undeflected motion, electric and magnetic forces cancel each other out.

📐 Worked Example

A velocity selector has a 200 V/m electric field perpendicular to a 0.05 T magnetic field. What velocity allows any charged particle to pass through undeflected?

1
For undeflected motion, force magnitudes are equal:
2
$qE = qvB$
3
Charge $q$ cancels out from both sides:
4
$E = vB$
5
Rearrange for $v$ and substitute values:
6
$v = \frac{E}{B} = \frac{200}{0.05} = 4000 \, \text{m/s}$

✓ Quick check

Test your understanding of velocity selectors:

A positive ion moving faster than the selected velocity will experience:
- Net force in direction of electric force
- Net force opposite to electric force
- No net force
- Net force along direction of motion
Reveal answer
1 —
For $v > \frac{E}{B}$ , magnetic force $q v B > q E$ , so net force points along the magnetic force direction, which is opposite to the electric force for crossed fields.

4. Helical Motion at an Angle to the Field★★★☆☆⏱ 6 min

When velocity has a component parallel to the magnetic field, the parallel component experiences zero magnetic force because $v_{∥} \times B = 0$ . Only the perpendicular component contributes to circular motion, resulting in a helical path along the magnetic field lines.

📐 Worked Example

An electron enters a 0.1 T magnetic field at $2 \times 1 0^{6}$ m/s at 30° to the field lines. Calculate the radius of the helical path.

1
Calculate the perpendicular component of velocity:
2
$v_\perp = v\sin\theta = (2 \times 10^6)\sin(30^\circ) = 1 \times 10^6 \, \text{m/s}$
3
Use the radius formula with $v_{⊥}$ :
4
$r = \frac{m_e v_\perp}{q_e B}$
5
Substitute $m_{e} = 9.11 \times 1 0^{- 31}$ kg, $q_{e} = 1.6 \times 1 0^{- 19}$ C:
6
$r = \frac{(9.11 \times 10^{-31})(1 \times 10^6)}{(1.6 \times 10^{-19})(0.1)} \approx 5.7 \times 10^{-5} \, \text{m}$

5. Common Pitfalls

Wrong move:

Getting force direction wrong for negative charges by using the right-hand rule directly

Why:

Right-hand rule gives force direction for positive charges only

Correct move:

Always reverse the force direction predicted by the right-hand rule for electrons and negative ions

Wrong move:

Using total velocity instead of perpendicular velocity for helical motion radius

Why:

Only the perpendicular component of velocity contributes to the magnetic force and circular motion

Correct move:

Always calculate $v_{⊥} = v sin θ$ when velocity is at an angle to the magnetic field

Wrong move:

Claiming magnetic force does work on charged particles, changing their speed

Why:

Magnetic force is always perpendicular to velocity, so work done is zero

Correct move:

Speed remains constant in pure magnetic fields; only the direction of motion changes

Wrong move:

Thinking selected velocity in a velocity selector depends on particle charge or mass

Why:

Charge cancels out when equating electric and magnetic forces

Correct move:

Remember $v = \frac{E}{B}$ , which is independent of all particle properties, only depends on $E$ and $B$

Wrong move:

Writing the radius formula as $r = \frac{q B}{m v}$ instead of $r = \frac{m v}{q B}$

Why:

Higher momentum particles produce larger radius paths, not smaller

Correct move:

Quickly re-derive the formula during the exam by equating $q v B = \frac{m v ^{2}}{r}$ to confirm

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Magnetic force	$F = q v B sin θ$	$θ$ = angle between $v$ and $B$ , $F ⊥ v$ and $F ⊥ B$
Circular path radius	$r = \frac{m v}{q B}$	Applies when $v ⊥ B$ , proportional to momentum
Total Lorentz force	$F = q (E + v \times B)$	Vector sum of electric and magnetic force
Undeflected velocity	$v = \frac{E}{B}$	Crossed fields, independent of particle mass and charge
Helical path radius	$r = \frac{m v s i n θ}{q B}$	Parallel velocity component remains constant

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.

2025 · 2
Circular motion in magnetic field question
2024 · 1
Velocity selector multiple choice
2023 · 2
Helical motion calculation

Going deeper

official documentIB Physics HL Data BookletContains all core formulas for this topic

What's Next

Motion in electromagnetic fields is a core assessed topic for IB Physics HL, appearing regularly in both multiple choice and extended response questions. This sub-topic builds on your understanding of magnetic force and circular motion, and forms the foundation for understanding particle detection and acceleration in modern physics. Next, you will explore electromagnetic induction, which describes how changing magnetic fields generate electric currents, a concept with widespread technological applications.