Magnetic Fields from Currents

Branch note: This page develops the three current-produced field patterns required by the 9749 syllabus. It starts with direction, then adds magnitude and formula limits.

Overview

This branch explains how conventional current produces magnetic fields around a straight wire, a circular coil, and a long solenoid.

Core Ideas

• Reversing current reverses $\vec{B}$ while preserving the geometric pattern.
• The right-hand grip rule must be matched to the geometry: straight wire versus coil or solenoid.
• Each formula has a specific observation point or region and an air/free-space model.
• Multiple magnetic fields add vectorially.

Exam Relevance

Use this page for sketching current-produced fields, selecting the correct field expression, identifying direction with the right-hand grip rule, and making proportional comparisons.

1. Current creates a magnetic field

A steady conventional current produces a steady magnetic field. The field is a vector, so a complete answer may require both its magnitude and direction.

Changing the current magnitude changes $B$. Reversing the current reverses $\vec{B}$ while leaving the geometric pattern unchanged.

2. Direction rules and viewpoints

Straight conductor

Right thumb $=$ conventional current; curled fingers $=$ circular field direction.

Figure: For a straight wire perpendicular to the page, the magnetic field circles the wire. Current out of the page gives anticlockwise field lines; reversing the current reverses the field direction.

For a wire perpendicular to the page:

Current	Field as seen by the reader
$\odot$ out of page	anticlockwise
$\otimes$ into page	clockwise

Coil or solenoid

Curl the fingers in the conventional-current direction around the turns. The thumb points along the field through the coil or inside the solenoid and towards the north pole.

This is the same physical right-hand grip idea but a different assignment from the straight-wire version.

3. Long straight wire

Field lines are concentric circles centred on the wire. At every point, $\vec{B}$ is tangent to the appropriate circle.

For an external point at shortest perpendicular distance $d$ from a long straight wire in air or free space,

$$B=\frac{{\mu }_{0}I}{2\pi d}.$$

The wire must be long relative to the distances considered, and the point should be far from the ends. The expression is not the field formula inside a conductor of finite radius.

At fixed $d$, $B\propto I$. At fixed $I$, $B\propto 1/d$.

Worked example

For $I=5.0\mathrm{A}$ and $d=0.020\mathrm{m}$,

$$B=\frac{(4\pi \times 10^{-7})(5.0)}{2\pi (0.020)}=5.0\times 10^{-5}\mathrm{T}.$$

The formula alone does not supply clockwise or anticlockwise direction; use the grip rule.

4. Flat circular coil

A circular loop can be viewed as many short current elements. At the centre, their field contributions point in the same axial direction and reinforce.

For $N$ closely coincident turns of common radius $r$ in air or free space,

$$B_{\text{centre}}=\frac{{\mu }_{0}NI}{2r}.$$

Figure: At the centre of a flat circular coil, field contributions from the turns reinforce along the coil axis. The equation $B={\mu }_{0}NI/(2r)$ applies at the centre of closely coincident turns, not everywhere around the coil.

At fixed $I$ and $r$, doubling $N$ doubles the centre field. At fixed $N$ and $I$, doubling $r$ halves it.

Warning

The formula applies at the centre. It does not say that the field is uniform throughout the coil or globally greatest at the centre.

Worked example

For $N=40$, $I=0.30\mathrm{A}$ and $r=0.060\mathrm{m}$,

$$B_{\text{centre}}=\frac{(4\pi \times 10^{-7})(40)(0.30)}{2(0.060)}=1.26\times 10^{-4}\mathrm{T}.$$

5. Long solenoid

A solenoid is a helical coil with many turns. In the central region of a long solenoid, fields from neighbouring turns reinforce inside. Outside a finite solenoid, contributions largely—but not exactly—cancel.

For a long, closely wound air-core solenoid,

$$B={\mu }_{0}nI,n=\frac{N}{L}.$$

The field is approximately uniform in the central region and the external pattern resembles that of a bar magnet.

Figure: The solenoid field is approximately uniform only in the central region of a long solenoid. The external field is weaker but not exactly zero, and the whole pattern forms closed loops.

Worked example

A $0.50\mathrm{m}$ solenoid has 1000 turns and carries $0.40\mathrm{A}$.

$$n=\frac{1000}{0.50}=2000{\mathrm{m}}^{-1},$$ $$B=(4\pi \times 10^{-7})(2000)(0.40)=1.01\times 10^{-3}\mathrm{T}.$$

Do not substitute a ferrous core into this air-core formula without an additional material model.

6. Formula selection table

Geometry	Required location	Magnitude in air/free space	Direction
long straight wire	external point at perpendicular distance $d$	$B=\frac{{\mu }_{0}I}{2\pi d}$	thumb=current, fingers=$\vec{B}$
flat $N$-turn circular coil	centre of common-radius turns	$B=\frac{{\mu }_{0}NI}{2r}$	fingers=current, thumb=$\vec{B}$
long air-core solenoid	central region, away from ends	$B={\mu }_{0}nI$, $n=N/L$	fingers=current, thumb=$\vec{B}$ and N

Here

$${\mu }_0=4\pi \times 10^{-7}\mathrm{N}{\mathrm{A}}^{-2}.$$

Figure: Choose the formula only after identifying the source geometry and observation point. The symbols $d$, $r$, $N$, $L$, and $n=N/L$ refer to different physical quantities and should not be interchanged.

7. Application: field superposition

Magnetic fields obey vector superposition:

$${\vec{B}}_{\text{resultant}}={\vec{B}}_1+{\vec{B}}_2+\cdots .$$

Figure: Magnetic fields from multiple wires add as vectors. At the midpoint between equal currents out of the page, the two field contributions oppose; cancellation depends on equal magnitudes and opposite directions, not on current values alone.

At a chosen point:

1. find the direction of each contribution using the grip rule;
2. calculate each magnitude with the correct distance;
3. choose a positive direction and add signed components.

Two fields of equal magnitude cancel only if they point in exactly opposite directions. Equal currents alone do not guarantee cancellation; geometry and distance matter.

Enrichment

Quantitative multiple-wire superposition is a useful application rather than an explicit outcome (a)–(d). It prepares you for the force between parallel currents in Topic 17.

8. Fast reasoning checks

• A larger current strengthens the field but does not reverse it.
• Reversing current reverses the field but does not change its shape.
• For a coil, use total turns $N$; for a solenoid, use turn density $n=N/L$.
• State what is held constant in every proportional comparison.
• A formula gives magnitude; a hand rule gives direction.

Links

• Magnetic Fields
• Solenoids and Electromagnets
• Magnetic Fields Common Exam Traps
• Magnetic Force
• Current Electricity Fundamentals
• Vectors