D.4 Magnetic effects of electric currents

IB Physics HL· D.4· 10 min read

1. Force on a Moving Charged Particle★★☆☆☆⏱ 3 min

📘 Definition

Magnetic Lorentz Force

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

The magnetic force on a point charge $q$ moving with velocity $v$ in a magnetic field $B$ . Magnitude is $F = q v B sin θ$ , where $θ$ is the angle between $v$ and $B$ .

Example:

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

📐 Worked Example

An electron moves at $2.0 \times 1 0^{6}$ 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: $q = - 1.6 \times 1 0^{- 19}$ C, $v = 2.0 \times 1 0^{6}$ m/s, $B = 0.50$ T, $θ = 9 0^{\circ}$ so $sin θ = 1$
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.

2. Force on a Current-Carrying Conductor★★☆☆☆⏱ 2 min

📘 Definition

Force on a straight current-carrying wire

F = BIL\sin\theta

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, $θ$ is the angle between current direction and $B$ .

📐 Worked Example

A 50 cm straight wire carrying 2.0 A current is placed at $3 0^{\circ}$ 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
$sin (3 0^{\circ}) = 0.5$ , so:
5
$F = 0.15\ \text{N}$

3. Force Between Parallel Current-Carrying Wires★★★☆☆⏱ 4 min

🔬 Derivation

Goal:

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 $I_{1}$ ): $B_{1} = \frac{μ _{0} I _{1}}{2 π r}$
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: $F = B_{1} I_{2} L = \frac{μ _{0} I _{1} I _{2} L}{2 π r}$
4
Divide by $L$ to get force per unit length: $f = \frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π r}$

Result:

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, $μ_{0} = 4 π \times 1 0^{- 7}$ 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.

4. Circular Motion of Charged Particles★★★☆☆⏱ 3 min

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

Goal:

Derive radius of the circular path

1
Magnetic force provides centripetal force: $q v B = \frac{m v ^{2}}{r}$
2
Cancel $v$ from both sides: $q B = \frac{m v}{r}$
3
Rearrange for $r$ : $r = \frac{m v}{q B} = \frac{p}{q B}$ where $p$ is momentum

Result:

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

📐 Worked Example

A proton ( $m = 1.67 \times 1 0^{- 27}$ kg, $q = 1.6 \times 1 0^{- 19}$ 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
$v = \frac{(1.6 \times 10^{-19})(0.15)(0.20)}{1.67 \times 10^{-27}} \approx 2.9 \times 10^6\ \text{m s}^{-1}$

5. Common Pitfalls

Wrong move:

Forgetting to reverse force direction for negative charges when using right-hand rules

Why:

All standard right-hand rules are defined for positive charges

Correct move:

Always check the sign of the charge, reverse the force direction for any negative charge

Wrong move:

Measuring $θ$ between the force and magnetic field instead of between velocity/current and magnetic field

Why:

The $sin θ$ term depends on the angle between the moving charge and the field, not the force

Correct move:

Always measure $θ$ between velocity (for a point charge) or current direction (for a wire) and the magnetic field vector

Wrong move:

Claiming magnetic force does work on a moving charged particle

Why:

Magnetic force is always perpendicular to the displacement of the charge

Correct move:

Recognize that magnetic force only changes direction of motion, not speed or kinetic energy, so it does no work

Wrong move:

Using centimetres instead of meters for length/distance in force calculations

Why:

All constants like $μ_{0}$ are defined in SI units, so inconsistent units give wrong answers

Correct move:

Always convert all lengths to meters before substituting into formulas

Wrong move:

Assuming parallel currents always attract regardless of direction

Why:

Confusion over the direction of the magnetic field produced by the first wire

Correct move:

Memorize: same direction = attraction, opposite direction = repulsion

6. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Force on moving charge	$F = q v B sin θ$	$θ$ = angle between $v$ and $B$
Force on current wire	$F = B I L sin θ$	$θ$ = angle between $I$ and $B$
Force per unit length (parallel wires)	$f = \frac{μ _{0} I _{1} I _{2}}{2 π r}$	Same direction: attract, opposite: repel
Radius of circular path	$r = \frac{m v}{q B} = \frac{p}{q B}$	Valid when $v ⊥ B$
Right-hand rule direction		Reverse force direction for negative charges

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 · P1
Force between parallel wires
2024 · P2
Circular motion of proton
2023 · P1
Force on moving electron

Going deeper

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