# D.3 Motion in electromagnetic fields

> IB Physics HL · Theme D: Fields
> Source: https://www.owlsprep.com/study/ib-physics-hl-u4-d-3-motion-in-electromagnetic/

This sub-topic explores how charged particles move in magnetic and combined electromagnetic fields. We cover force rules, circular and helical motion, and common applications assessed regularly in IB exams.

**Prerequisites:** [Magnetic fields and magnetic force](https://www.owlsprep.com/study/ib-physics-hl-u4-d-2-magnetic-fields/); [Uniform circular motion and centripetal force](https://www.owlsprep.com/study/ib-physics-global-mechanics-circular-motion/)

## Learning objectives

- Calculate the magnetic force on moving charged particles in uniform fields
- Derive the formula for cyclotron radius of charged particles in magnetic fields
- Analyze motion of particles in combined electric and magnetic fields
- Apply concepts to common applications like velocity selectors

## Lorentz Force on Moving Charges

**Lorentz Force** — The total electromagnetic force on a charged particle moving through electric and magnetic fields. For pure magnetic fields, $F = qvB\sin\theta$, where $\theta$ is the angle between velocity $v$ and magnetic flux density $B$.

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

*Example:* A proton moving perpendicular to a 0.5 T field at $2 \times 10^6$ m/s experiences a magnetic force of $1.6 \times 10^{-13}$ 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.

> **mnemonic**
>
> Right-hand rule: Fingers point along velocity, curl towards magnetic field, thumb points to force direction. Flip thumb for negative charges.

**Worked example:** An electron moving at $3.0 \times 10^6$ 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 $q = 1.6 \times 10^{-19}$ C:
4. $$F = (1.6 \times 10^{-19})(3.0 \times 10^6)(0.2)$$
5. Calculate the final magnitude:
6. $$F = 9.6 \times 10^{-14} \text{ N}$$

## Circular Motion in Uniform Magnetic Fields

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:** 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. $$r = \frac{mv}{qB}$$

*Conclusion:* Radius is directly proportional to particle momentum $mv$, inversely proportional to charge $q$ and magnetic flux density $B$.

**Worked example:** A proton ($m = 1.67 \times 10^{-27}$ kg, $q = +1.6 \times 10^{-19}$ 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. $$v = \frac{(0.2)(1.6 \times 10^{-19})(0.15)}{1.67 \times 10^{-27}} \approx 2.9 \times 10^6 \, \text{m/s}$$

## Combined Electric and Magnetic Fields

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

**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. $$v = \frac{E}{B} = \frac{200}{0.05} = 4000 \, \text{m/s}$$

**Check your understanding**

Test your understanding of velocity selectors:

1. 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

   *Answer:* Net force opposite to electric force

   *Why:* For $v > \frac{E}{B}$, magnetic force $qvB > qE$, so net force points along the magnetic force direction, which is opposite to the electric force for crossed fields.

## Helical Motion at an Angle to the Field

When velocity has a component parallel to the magnetic field, the parallel component experiences zero magnetic force because $v_{\parallel} \times B = 0$. 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 $2 \times 10^6$ 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 $v_\perp$:
4. $$r = \frac{m_e v_\perp}{q_e B}$$
5. Substitute $m_e = 9.11 \times 10^{-31}$ kg, $q_e = 1.6 \times 10^{-19}$ C:
6. $$r = \frac{(9.11 \times 10^{-31})(1 \times 10^6)}{(1.6 \times 10^{-19})(0.1)} \approx 5.7 \times 10^{-5} \, \text{m}$$

## Common pitfalls

- **Wrong:** Getting force direction wrong for negative charges by using the right-hand rule directly
  - Why it fails: Right-hand rule gives force direction for positive charges only
  - Correct: Always reverse the force direction predicted by the right-hand rule for electrons and negative ions
- **Wrong:** Using total velocity instead of perpendicular velocity for helical motion radius
  - Why it fails: Only the perpendicular component of velocity contributes to the magnetic force and circular motion
  - Correct: Always calculate $v_\perp = v\sin\theta$ when velocity is at an angle to the magnetic field
- **Wrong:** Claiming magnetic force does work on charged particles, changing their speed
  - Why it fails: Magnetic force is always perpendicular to velocity, so work done is zero
  - Correct: Speed remains constant in pure magnetic fields; only the direction of motion changes
- **Wrong:** Thinking selected velocity in a velocity selector depends on particle charge or mass
  - Why it fails: Charge cancels out when equating electric and magnetic forces
  - Correct: Remember $v = \frac{E}{B}$, which is independent of all particle properties, only depends on $E$ and $B$
- **Wrong:** Writing the radius formula as $r = \frac{qB}{mv}$ instead of $r = \frac{mv}{qB}$
  - Why it fails: Higher momentum particles produce larger radius paths, not smaller
  - Correct: Quickly re-derive the formula during the exam by equating $qvB = \frac{mv^2}{r}$ to confirm

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Magnetic force | $F = qvB\sin\theta$ | $\theta$ = angle between $v$ and $B$, $F \perp v$ and $F \perp B$ |
| Circular path radius | $r = \frac{mv}{qB}$ | Applies when $v \perp B$, proportional to momentum |
| Total Lorentz force | $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ | Vector sum of electric and magnetic force |
| Undeflected velocity | $v = \frac{E}{B}$ | Crossed fields, independent of particle mass and charge |
| Helical path radius | $r = \frac{mv\sin\theta}{qB}$ | Parallel velocity component remains constant |

## 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.

- [D.4 Magnetic effects of electric currents](https://www.owlsprep.com/study/ib-physics-hl-u4-d-4-magnetic-effects-of/)
- [D.5 Electromagnetic induction (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u4-d-5-electromagnetic-induction/)

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