Magnetic fields due to currents (solenoid, flat coil, force between conductors)

Physics· 12 min read

1. Magnetic Field Inside a Long Straight Solenoid★★★☆☆⏱ 3 min

An ideal infinitely long solenoid produces a perfectly uniform magnetic field inside its core, with near-zero magnetic field outside the coil. The direction of the field along the solenoid axis is found using the right-hand grip rule: curl your fingers in the direction of current flow, and your extended thumb points to the north pole end of the solenoid.

📘 Definition

Solenoid Magnetic Flux Density

$B = μ_{0} n I$

For an ideal long solenoid, the uniform internal magnetic flux density depends only on the permeability of free space, number of turns per unit length $n$ , and current $I$ , and is independent of the solenoid cross-sectional radius.

📐 Worked Example

A solenoid of length 0.4 m has 1200 turns, and carries a current of 2.5 A. Calculate the magnetic flux density at its centre, assuming it behaves as an ideal long solenoid. Take $μ_{0} = 4 π \times 1 0^{- 7} H m^{- 1}$ .

1
Step 1: Calculate number of turns per unit length $n$
2
$n = \frac{N}{L} = \frac{1200}{0.4} = 3000 \, \text{turns m}^{-1}$
3
Step 2: Substitute values into the solenoid B formula
4
$B = 4\pi \times 10^{-7} \times 3000 \times 2.5$
5
Step 3: Compute final result
6
$B \approx 9.42 \times 10^{-3} \, \text{T}$

Exam tip:

CIE examiners regularly penalise answers that incorrectly include the solenoid radius in B calculations. The ideal solenoid B formula has no area or radius term, so do not waste time calculating cross-sectional area for these problems.

2. Magnetic Field at the Centre of a Flat Circular Coil★★★☆☆⏱ 3 min

Every small segment of current in a flat circular coil contributes a magnetic field component at the centre of the coil. All perpendicular components cancel out, leaving a net field that points entirely along the coil's central axis, aligned to the right-hand grip rule for circular current.

📘 Definition

Flat Circular Coil Central Flux Density

$B = \frac{μ _{0} N I}{2 r}$

At the exact centre of a flat circular coil of $N$ turns, radius $r$ carrying current $I$ , the magnetic flux density points along the coil's central axis.

📐 Worked Example

A flat circular coil with 50 turns of radius 0.08 m carries a current of 1.2 A. Calculate the magnetic flux density at its centre.

1
Step 1: Confirm all values are in SI units: $N = 50$ , $I = 1.2 A$ , $r = 0.08 m$
2
$B = \frac{4\pi \times 10^{-7} \times 50 \times 1.2}{2 \times 0.08}$
3
Step 2: Simplify and compute final value
4
$B \approx 4.71 \times 10^{-4} \, \text{T}$

✓ Quick check

Test your understanding of the flat coil B formula

If you double the number of turns and halve the radius of the coil, by what factor does the central B change?
- A) 2x
- B) 4x
- C) 0.5x
- D) No change
Reveal answer
B —
B is proportional to $N / r$ , so $2 N / (0.5 r) = 4 * (N / r)$ , so B quadruples.

3. Force Between Two Parallel Current-Carrying Conductors★★★★☆⏱ 3 min

Each long straight current-carrying wire generates a circular magnetic field around itself, which exerts a force on any other nearby parallel current-carrying wire. If the currents flow in the same direction, the force is attractive; if currents flow opposite directions, the force is repulsive.

🔬 Derivation

Goal:

Derive force per unit length between two parallel conductors

Starting from:

Start from $F = B I L$ for a current-carrying wire in an external magnetic field

1
1. Wire 1 carrying $I_{1}$ produces magnetic flux density at distance $d$ : $B = \frac{μ _{0} I _{1}}{2 π d}$
2
1. This B field acts on Wire 2 of length $L$ carrying $I_{2}$ , so total force $F = B I_{2} L$
3
1. Substitute B and rearrange to get force per unit length

Result:

Force per unit length: $\frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$

📐 Worked Example

Two long parallel wires are separated by 0.02 m, carrying currents of 10 A and 15 A in the same direction. Calculate the force per unit length acting between them.

1
$\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 10 \times 15}{2\pi \times 0.02}$
2
Cancel $π$ terms and simplify
3
$\frac{F}{L} = 1.5 \times 10^{-3} \, \text{N m}^{-1}$
4
Final note: Force is attractive as currents flow in the same direction

4. SI Definition of the Ampere★★☆☆☆⏱ 2 min

The ampere is one of the 7 SI base units, and it is formally defined using the measurable force between two parallel current-carrying conductors, rather than using the derived quantity of charge flow.

📘 Definition

Ampere (SI Base Unit)

The constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, placed 1 metre apart in vacuum, would produce a force between these conductors equal to $2 \times 1 0^{- 7}$ newtons per metre of length.

5. Common Pitfalls

Wrong move:

Using the solenoid B formula for a flat circular coil

Why:

Confusing the two similar $μ_{0}$ formulas, leading to incorrect denominator values

Correct move:

Label formulas clearly, confirm if the problem refers to a solenoid or flat coil before substituting values.

Wrong move:

Forgetting to convert coil radius from cm to m before calculation

Why:

Units of $μ_{0}$ are in SI units, so all lengths must be in metres for consistent results

Correct move:

Convert all given lengths to SI units before substituting into any magnetic field formula.

Wrong move:

Stating that the force between two parallel wires is always repulsive

Why:

Mixing up direction rules for parallel and anti-parallel currents

Correct move:

Use the 'Same Stick, Opposite Off' mnemonic to confirm force direction quickly.

Wrong move:

Including the solenoid cross-sectional area in B calculation

Why:

Incorrectly assuming B depends on area like magnetic flux $Φ = B A$

Correct move:

Recall that ideal solenoid B depends only on n and I, no area term exists in the formula.

Wrong move:

Defining the ampere as 1 coulomb per second

Why:

Using the derived informal definition instead of the formal SI definition required for CIE A-Level

Correct move:

Memorise the full parallel conductor definition to score full marks on definition questions.

6. Quick Reference Cheatsheet

Scenario	Formula	Key Notes
Ideal long solenoid	$B = μ_{0} n I$	Uniform internal field, zero outside, no radius dependence
Centre of flat circular coil	$B = \frac{μ _{0} N I}{2 r}$	Field points along central axis
Parallel conductors force per unit length	$\frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π d}$	Attractive for same current direction

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.

2024 · Paper 4
Solenoid flux density calculation
2023 · Paper 2
Force between parallel conductors
2022 · Paper 4
Flat coil B derivation and definition of ampere

What's Next

Mastering these current-generated magnetic field rules is critical for upcoming topics on electromagnetic induction, alternating current transformers, and charged particle motion in combined electric and magnetic fields. You will frequently combine these B formulas with $F = B I L$ and $F = B Q v$ in 6+ mark structured questions, so ensure you can recall all three formulas without reference. This content is also a core prerequisite for the extended response questions on electrical generators that appear in almost every Paper 4 exam.