# D.4 Magnetic effects of electric currents

> IB Physics HL · Theme D: Fields
> Source: https://www.owlsprep.com/study/ib-physics-hl-u4-d-4-magnetic-effects-of/

This sub-topic explores magnetic forces generated by moving charges and electric currents. You will learn to calculate force on moving charges, current-carrying wires, and interacting parallel currents, and apply right-hand rules for direction.

**Prerequisites:** [Magnetic field fundamentals](https://www.owlsprep.com/study/ib-physics-hl-u4-d-1-introduction-to-fields/); Uniform circular motion and centripetal force

## Learning objectives

- Calculate magnetic force on moving charges and current-carrying conductors
- Apply right-hand rules to find force and field direction
- Determine force between two parallel current-carrying wires
- Solve problems for charged particles moving in magnetic fields

## Force on a Moving Charged Particle

**Magnetic Lorentz Force** — The magnetic force on a point charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$. Magnitude is $F = qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

*Notation:* \vec{F} = q \vec{v} \times \vec{B}

*Example:* A 1×10⁶ m/s proton in a 2 T perpendicular field experiences ~3.2×10⁻¹⁹ N force.

> **Right-hand rule for direction**
>
> Point fingers of your right hand along $\vec{v}$, curl fingers toward $\vec{B}$, thumb points to force direction for positive charges. Reverse direction for negative charges.

**Worked example:** An electron moves at $2.0 \times 10^6$ m/s horizontally into a vertical 0.50 T magnetic field pointing downwards. Find the magnitude and direction of the force on the electron.

1. Extract given values: $q = -1.6 \times 10^{-19}$ C, $v = 2.0 \times 10^6$ m/s, $B = 0.50$ T, $\theta = 90^\circ$ so $\sin\theta = 1$
2. Calculate magnitude of force:
3. $$F = |q|vB = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50) = 1.6 \times 10^{-19}\ \text{N}$$
4. Find direction: Right-hand rule for positive charge gives force out of the page. Electron is negative, so force points into the page.

## Force on a Current-Carrying Conductor

**Force on a straight current-carrying wire** — A current is a flow of moving charges, so a wire placed in a magnetic field experiences a net magnetic force. $I$ is current, $L$ is length of wire in the field, $\theta$ is the angle between current direction and $\vec{B}$.

*Notation:* F = BIL\sin\theta

> **tip**
>
> If the wire is parallel to the magnetic field, $\theta = 0$ so $F = 0$: no net force acts on the wire.

**Worked example:** A 50 cm straight wire carrying 2.0 A current is placed at $30^\circ$ to a uniform 0.3 T magnetic field. Calculate the magnitude of the force on the wire.

1. Convert length to SI units: $L = 50$ cm $= 0.50$ m
2. Substitute values into the force formula:
3. $$F = BIL\sin\theta = (0.3)(2.0)(0.50)\sin(30^\circ)$$
4. $\sin(30^\circ) = 0.5$, so:
5. $$F = 0.15\ \text{N}$$

## Force Between Parallel Current-Carrying Wires

**Derivation:** Derive force per unit length between two parallel wires

*Starting from:* Magnetic field due to long straight wire, force on a current-carrying wire

1. Magnetic flux density at distance $r$ from wire 1 (current $I_1$): $B_1 = \frac{\mu_0 I_1}{2\pi r}$
2. This field acts on wire 2 (length $L$, current $I_2$), placed parallel to wire 1, current is perpendicular to $B_1$
3. Force on wire 2: $F = B_1 I_2 L = \frac{\mu_0 I_1 I_2 L}{2\pi r}$
4. Divide by $L$ to get force per unit length: $f = \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$

*Conclusion:* Currents in same direction attract each other; opposite directions repel.

**Worked example:** Two parallel wires 10 cm apart carry 10 A and 15 A currents in opposite directions. Find the force per unit length, and state if it is attractive or repulsive.

1. Identify values: $I_1 = 10$ A, $I_2 = 15$ A, $r = 0.10$ m, $\mu_0 = 4\pi \times 10^{-7}$ T m A⁻¹
2. Substitute into formula:
3. $$f = \frac{\mu_0 I_1 I_2}{2\pi r} = \frac{4\pi \times 10^{-7} \times 10 \times 15}{2\pi \times 0.10} = 3 \times 10^{-4}\ \text{N m}^{-1}$$
4. Opposite currents repel, so the force is repulsive.

## Circular Motion of Charged Particles

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force is always perpendicular to velocity. This means the force does no work (it does not change speed), only changes direction of motion, resulting in uniform circular motion.

**Derivation:** Derive radius of the circular path

1. Magnetic force provides centripetal force: $qvB = \frac{mv^2}{r}$
2. Cancel $v$ from both sides: $qB = \frac{mv}{r}$
3. Rearrange for $r$: $r = \frac{mv}{qB} = \frac{p}{qB}$ where $p$ is momentum

*Conclusion:* Radius is proportional to the momentum of the particle, inversely proportional to charge and magnetic field strength.

**Worked example:** A proton ($m = 1.67 \times 10^{-27}$ kg, $q = 1.6 \times 10^{-19}$ C) moves in a 0.20 m radius circular path in a 0.15 T uniform magnetic field. Calculate the speed of the proton.

1. Rearrange the radius formula to solve for $v$:
2. $$v = \frac{qBr}{m}$$
3. Substitute values:
4. $$v = \frac{(1.6 \times 10^{-19})(0.15)(0.20)}{1.67 \times 10^{-27}} \approx 2.9 \times 10^6\ \text{m s}^{-1}$$

> **info**
>
> This relationship is the core principle behind mass spectrometry, used to identify unknown particles by their deflection in a magnetic field.

## Common pitfalls

- **Wrong:** Forgetting to reverse force direction for negative charges when using right-hand rules
  - Why it fails: All standard right-hand rules are defined for positive charges
  - Correct: Always check the sign of the charge, reverse the force direction for any negative charge
- **Wrong:** Measuring $\theta$ between the force and magnetic field instead of between velocity/current and magnetic field
  - Why it fails: The $\sin\theta$ term depends on the angle between the moving charge and the field, not the force
  - Correct: Always measure $\theta$ between velocity (for a point charge) or current direction (for a wire) and the magnetic field vector
- **Wrong:** Claiming magnetic force does work on a moving charged particle
  - Why it fails: Magnetic force is always perpendicular to the displacement of the charge
  - Correct: Recognize that magnetic force only changes direction of motion, not speed or kinetic energy, so it does no work
- **Wrong:** Using centimetres instead of meters for length/distance in force calculations
  - Why it fails: All constants like $\mu_0$ are defined in SI units, so inconsistent units give wrong answers
  - Correct: Always convert all lengths to meters before substituting into formulas
- **Wrong:** Assuming parallel currents always attract regardless of direction
  - Why it fails: Confusion over the direction of the magnetic field produced by the first wire
  - Correct: Memorize: same direction = attraction, opposite direction = repulsion

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| Force on moving charge | $F = qvB\sin\theta$ | $\theta$ = angle between $\vec{v}$ and $\vec{B}$ |
| Force on current wire | $F = BIL\sin\theta$ | $\theta$ = angle between $I$ and $\vec{B}$ |
| Force per unit length (parallel wires) | $f = \frac{\mu_0 I_1 I_2}{2\pi r}$ | Same direction: attract, opposite: repel |
| Radius of circular path | $r = \frac{mv}{qB} = \frac{p}{qB}$ | Valid when $\vec{v} \perp \vec{B}$ |
| Right-hand rule direction |  | Reverse force direction for negative charges |

## What's next

Understanding magnetic effects of currents is the foundation for explaining how electric motors, generators, and mass spectrometers work, and it leads directly to electromagnetic induction, the next core topic in the IB Physics fields theme. This sub-topic also underpins the study of electromagnetic waves and Lorentz force interactions used in particle physics, with practical applications in particle accelerators and medical imaging technologies like MRI. Mastering these concepts will prepare you for both Paper 1 and Paper 2 exam questions on fields, which frequently combine electric and magnetic force problems.

- [D.5 Electromagnetic induction](https://www.owlsprep.com/study/ib-physics-hl-u4-d-5-electromagnetic-induction/)

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