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

> Physics · CIE A-Level 9702
> Source: https://www.owlsprep.com/study/cie-9702-u23-magnetic-fields-due-to-currents/

This module covers magnetic flux density formulas for ideal solenoids, flat circular coils, the force between parallel current-carrying wires, and the formal SI definition of the ampere for CIE A-Level 9702.

**Prerequisites:** [Recall of magnetic field lines, Fleming's Left Hand Rule, and $F=BIL$ for current in a uniform magnetic field](https://www.owlsprep.com/study/cie-9702-u23-intro-to-magnetic-fields/); [Basic vector addition for perpendicular magnetic field components](https://www.owlsprep.com/study/cie-9702-u01-vector-addition/)

## Learning objectives

- Recall and apply the formula for uniform magnetic flux density inside a long straight ideal solenoid
- Calculate magnetic flux density at the centre of a flat circular current-carrying coil
- Derive and use the expression for force per unit length between two parallel current-carrying conductors
- State the formal SI definition of the ampere aligned to CIE assessment requirements
- Solve multi-step problems combining all three current-generated magnetic field scenarios

## Magnetic Field Inside a Long Straight Solenoid

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.

**Solenoid Magnetic Flux Density** — 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.

*Notation:* $B = \mu_0 n I$

**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 $\mu_0 = 4\pi \times 10^{-7} \, \text{H m}^{-1}$.

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.

## Magnetic Field at the Centre of a Flat Circular Coil

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.

**Flat Circular Coil Central Flux Density** — 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.

*Notation:* $B = \frac{\mu_0 N I}{2r}$

**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: $N=50$, $I=1.2 \, \text{A}$, $r=0.08 \, \text{m}$
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. $$B \approx 4.71 \times 10^{-4} \, \text{T}$$

**Check your understanding**

Test your understanding of the flat coil B formula

1. 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

   *Why:* B is proportional to $N/r$, so $2N/(0.5r) = 4*(N/r)$, so B quadruples.

## Force Between Two Parallel Current-Carrying Conductors

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:** Derive force per unit length between two parallel conductors

*Starting from:* Start from $F=BIL$ for a current-carrying wire in an external magnetic field

1. 1. Wire 1 carrying $I_1$ produces magnetic flux density at distance $d$: $B = \frac{\mu_0 I_1}{2\pi d}$
2. 2. This B field acts on Wire 2 of length $L$ carrying $I_2$, so total force $F = B I_2 L$
3. 3. Substitute B and rearrange to get force per unit length

*Conclusion:* Force per unit length: $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

**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 $\pi$ 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

> **mnemonic**
>
> Same direction = Stick (Attractive), Opposite = Off (Repulsive) to remember force direction instantly in exams.

## SI Definition of the Ampere

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.

**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 $2 \times 10^{-7}$ newtons per metre of length.

**Exam command terms**

CIE uses strict command term requirements for this topic, you must meet all criteria to score full marks:

- **Define the ampere** — You must reference the exact parallel conductor scenario, no partial marks for vague answers referencing 1 coulomb per second.

- **Explain why the force between two wires is attractive** — You must link the B field from one wire, then apply Fleming's Left Hand Rule to the second wire to derive force direction.

## Common pitfalls

- **Wrong:** Using the solenoid B formula for a flat circular coil
  - Why it fails: Confusing the two similar $\mu_0$ formulas, leading to incorrect denominator values
  - Correct: Label formulas clearly, confirm if the problem refers to a solenoid or flat coil before substituting values.
- **Wrong:** Forgetting to convert coil radius from cm to m before calculation
  - Why it fails: Units of $\mu_0$ are in SI units, so all lengths must be in metres for consistent results
  - Correct: Convert all given lengths to SI units before substituting into any magnetic field formula.
- **Wrong:** Stating that the force between two parallel wires is always repulsive
  - Why it fails: Mixing up direction rules for parallel and anti-parallel currents
  - Correct: Use the 'Same Stick, Opposite Off' mnemonic to confirm force direction quickly.
- **Wrong:** Including the solenoid cross-sectional area in B calculation
  - Why it fails: Incorrectly assuming B depends on area like magnetic flux $\Phi = BA$
  - Correct: Recall that ideal solenoid B depends only on n and I, no area term exists in the formula.
- **Wrong:** Defining the ampere as 1 coulomb per second
  - Why it fails: Using the derived informal definition instead of the formal SI definition required for CIE A-Level
  - Correct: Memorise the full parallel conductor definition to score full marks on definition questions.

## Cheatsheet

| Scenario | Formula | Key Notes |
| --- | --- | --- |
| Ideal long solenoid | $B = \mu_0 n I$ | Uniform internal field, zero outside, no radius dependence |
| Centre of flat circular coil | $B = \frac{\mu_0 N I}{2r}$ | Field points along central axis |
| Parallel conductors force per unit length | $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$ | Attractive for same current direction |

## 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 $F=BIL$ and $F=BQv$ 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.

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