Magnetic Fields Due to Currents

A2 · 16 min

Every current is surrounded by a magnetic field. The shape of that field depends on the shape of the conductor: circles around a straight wire, a bar-magnet pattern around a solenoid. This note covers the three field patterns the syllabus asks you to sketch (long straight wire, flat circular coil, long solenoid), the effect of an iron core, and the forces between two current-carrying conductors. Sketching these patterns accurately, with arrows, is a routine Paper 4 question, and the forces between parallel wires test whether you can combine a field pattern with Fleming's left-hand rule.

The right-hand grip rule

In 1820 Oersted noticed that a compass needle near a wire swung round when a current was switched on. The current had created a magnetic field. For every current-carrying conductor, the direction of the field is given by one rule.

Key result

Right-hand grip rule. Grip the conductor with your right hand, thumb pointing in the direction of the conventional current. Your fingers curl in the direction of the magnetic field lines.

The same hand can be used the other way round for coils and solenoids: curl the fingers in the direction of the current around the coil, and the thumb points along the field inside the coil, towards the coil's north pole.

Long straight wire

The field lines around a long straight wire are concentric circles centred on the wire, in planes perpendicular to it.

The circles get further apart with increasing distance: the field gets weaker as you move away from the wire.
The direction is given by the right-hand grip rule. Seen from the end, a current out of the page gives anticlockwise field lines; a current into the page gives clockwise lines.
Reversing the current reverses the direction of every field line; increasing the current makes the field stronger everywhere.

Field of a long straight wire seen end-on. Left: current out of the page, field anticlockwise. Right: current into the page, field clockwise. The spacing of the circles increases with distance because the field weakens.

When sketching, draw at least three circles with the gaps between them visibly increasing, and put an arrow on each one.

Tip

For a long straight wire, the flux density at distance $r$ is $B = \dfrac{\mu_0 I}{2\pi r}$ , where $\mu_0 = 4\pi \times 10^{-7}\ \text{H m}^{-1}$ is the permeability of free space. You do not need to recall this equation for the 9702 syllabus; if a question needs it, it will be given. It does show why the field lines spread out: $B \propto 1/r$ .

Flat circular coil

Bend the wire into a flat circular loop (or a coil of a few turns). Each short section of the coil is surrounded by its own small circles of field. Near the wire on each side, you see these circles. Inside the loop, the fields from all parts of the coil point the same way and add up.

The resulting pattern:

Near the wire, the field lines are small loops circling the wire, like those around a straight wire.
Through the centre of the coil, the field lines are nearly straight and perpendicular to the plane of the coil, and the field is strongest there.
Further out, the lines spread out and curve round, joining up on the other side. The pattern looks like the field of a very short bar magnet, with one face of the coil acting as a north pole and the other as a south pole.

A quick check of the poles: look at a face of the coil. If the current flows anticlockwise as you look at it, that face is a north pole; if it flows clockwise, that face is a south pole. (Draw an N and an S with arrows on their ends: the arrows on the letter N run anticlockwise, those on S clockwise.)

Long solenoid

A solenoid is a long coil of many turns. Its field is the sum of the fields of many flat coils side by side.

Inside the solenoid (away from the ends) the field lines are parallel, straight and equally spaced: the field is uniform and strong.
Outside, the field is like that of a bar magnet: lines emerge from one end (the north pole), loop round the outside and re-enter at the other end (the south pole). The field outside is much weaker than inside.
At the ends the lines begin to spread out; the flux density at the very end is about half its value at the centre.

A long solenoid cut along its axis. The field inside is uniform and points to the right, so the right-hand end is a north pole. Outside, the field loops back from N to S like the field of a bar magnet.

Check the direction with the grip rule: curl the right hand's fingers along the current in the turns (out of the page at the top, into the page at the bottom) and the thumb points to the right along the axis, towards the north pole.

What makes a solenoid's field stronger

The field inside a solenoid increases with:

the current $I$ (the field is proportional to it),
the number of turns per unit length (packing the turns more closely),
a ferrous core inside the coil.

The length and diameter of the solenoid hardly affect the uniform field at its centre, provided it is long compared with its diameter.

Tip

For a long air-cored solenoid, $B = \mu_0 n I$ , where $n$ is the number of turns per unit length. This is beyond the 9702 syllabus but explains the factors above.

The effect of a ferrous core

A ferrous material is one that contains iron (soft iron, many steels). Placing a soft-iron core inside a solenoid increases the flux density enormously: by a factor of hundreds or even thousands.

Why? Iron is ferromagnetic. It contains tiny regions (domains) in which the atoms' own magnetic fields are all aligned. Normally the domains point in random directions and their fields cancel. The solenoid's field lines up the domains along the axis, so their fields add to the field of the current. The total field is the solenoid's field plus the much larger field of the magnetised iron.

Soft iron is used for the cores of electromagnets because it is magnetised easily when the current is switched on and loses its magnetism almost completely when the current is switched off. That is what makes electromagnets useful in relays, door locks, scrap-metal cranes and loudspeakers.

Forces between current-carrying conductors

Two parallel wires carrying currents push or pull on each other. No new physics is needed to explain this: each wire sits in the magnetic field of the other, so each feels a force $F = BIL$ .

Take two long parallel wires X and Y, with currents in the same direction (both out of the page):

The field of X at the position of Y: circles around X, anticlockwise (current out of page). At Y, which is to the right of X, the anticlockwise field points up the page. It is perpendicular to Y.
Y carries current out of the page in an upward field. Fleming's left-hand rule (first finger up, second finger out of the page) gives a force on Y to the left, towards X.
By the same argument with the roles swapped, X feels a force to the right, towards Y.

Key result

Parallel currents in the same direction attract.
Parallel currents in opposite directions repel.
The forces on the two wires are equal in size and opposite in direction, even when the currents are different. They are a Newton's third law pair.

The third-law point is often tested. If X carries $10\ \text{A}$ and Y carries $2\ \text{A}$ , the field of X at Y is five times the field of Y at X, but Y's current is five times smaller. $F = BIL$ comes out the same for both.

Direction of the force between two conductors

Choose one wire as the source. Use the right-hand grip rule to find the direction of its field at the position of the other wire.
Apply Fleming's left-hand rule to the other wire, using that field and the other wire's current.
The force on the source wire is equal and opposite.
Check: like currents attract, unlike currents repel.

Tip

Before 2019 the ampere was defined using this force: the current which, in two infinitely long parallel wires $1\ \text{m}$ apart in a vacuum, produces a force of $2 \times 10^{-7}\ \text{N}$ per metre of length. The ampere is now defined by fixing the value of the elementary charge $e$ , but the force between wires is still how currents are compared in precise work.

Worked examples

Compass needles around a wire

A long vertical wire passes through a horizontal card. The current in the wire flows upwards. Four plotting compasses are placed on the card at points north, east, south and west of the wire. Ignoring the Earth's field, state the direction each compass points.

Solution

Looking down from above, the current flows towards you (out of the card), so by the right-hand grip rule the field lines are anticlockwise circles when viewed from above.

Going anticlockwise round the wire as seen from above (north, west, south, east):

North of the wire: the field points west.
West of the wire: the field points south.
South of the wire: the field points east.
East of the wire: the field points north.

Each compass needle lines up tangent to the circle through its position.

Force between parallel wires

Two long parallel wires X and Y carry currents of $5.0\ \text{A}$ and $3.0\ \text{A}$ in opposite directions. The magnetic flux density at Y due to the current in X is $2.0 \times 10^{-5}\ \text{T}$ .

(a) Calculate the force on a $0.80\ \text{m}$ length of Y and state whether the wires attract or repel. (b) State the force on the same length of X. (c) Calculate the flux density at X due to Y.

Solution

(a) The field of X is perpendicular to Y, so

F = BIL = 2.0 \times 10^{-5} \times 3.0 \times 0.80 = 4.8 \times 10^{-5}\ \text{N}

The currents are in opposite directions, so the wires repel.

(b) By Newton's third law, X feels a force of $4.8 \times 10^{-5}\ \text{N}$ in the opposite direction (away from Y).

B_Y = \frac{F}{I_X L} = \frac{4.8 \times 10^{-5}}{5.0 \times 0.80} = 1.2 \times 10^{-5}\ \text{T}

The fields at each wire are different, but the forces are equal.

Comparing a coil and a solenoid

A student uses a Hall probe to compare the field at the centre of a flat circular coil with the field at the centre of a long solenoid. Describe how the field varies near the centre of each, and explain the effect on each of placing a soft-iron rod along its axis.

Solution

At the centre of the flat coil, the field lines are perpendicular to the plane of the coil and the field is strongest there; moving along the axis away from the coil, the field falls off quickly, and the lines spread out.

At the centre of the long solenoid, the field is uniform: moving the probe along the axis or sideways (inside the coil) gives the same reading until the probe nears an end, where the reading falls to about half.

A soft-iron core greatly increases the flux density in both cases. The field of the current magnetises the iron (it aligns the magnetic domains), and the field of the magnetised iron adds to that of the current.

Two wires and a neutral point

Two long parallel wires P and Q are $6.0\ \text{cm}$ apart. P carries $8.0\ \text{A}$ and Q carries $4.0\ \text{A}$ , in the same direction. The flux density at distance $r$ from a long wire carrying current $I$ is $B = kI/r$ , where $k = 2.0 \times 10^{-7}\ \text{T m A}^{-1}$ .

(a) Calculate the force per unit length on Q and state its direction. (b) Find the point on the line joining the wires where the resultant flux density is zero.

Solution

(a) Field at Q due to P:

B_P = \frac{kI_P}{r} = \frac{2.0 \times 10^{-7} \times 8.0}{0.060} = 2.67 \times 10^{-5}\ \text{T}

This is perpendicular to Q, so the force per unit length is

\frac{F}{L} = B_P I_Q = 2.67 \times 10^{-5} \times 4.0 = 1.1 \times 10^{-4}\ \text{N m}^{-1}

Like currents attract, so the force on Q is towards P.

(b) Between the wires, the two fields are in opposite directions (for example, with both currents out of the page, P's field points up at a point to its right, and Q's field points down at a point to its left). The resultant is zero where the magnitudes are equal. Let the point be distance $x$ from P:

\frac{k \times 8.0}{x} = \frac{k \times 4.0}{0.060 - x} \quad\Rightarrow\quad 8.0(0.060 - x) = 4.0x \quad\Rightarrow\quad x = 0.040\ \text{m}

The neutral point is $4.0\ \text{cm}$ from P ( $2.0\ \text{cm}$ from Q), closer to the wire with the smaller current.

Watch out

Using the left hand for the field. The right-hand grip rule gives the field around a current. Fleming's left-hand rule gives the force on a current in a field. Questions on forces between wires need both, in that order.

Watch out

Drawing equally spaced circles. The field around a straight wire gets weaker with distance, so the circles must be drawn further apart further out. Equally spaced circles lose the sketch mark.

Watch out

Thinking the larger current feels the larger force. The two forces between parallel wires are a Newton's third law pair: always equal and opposite, whatever the two currents.

Exam tip

"Sketch the magnetic field pattern" (2 to 3 marks): correct shape; correct spacing (closer where stronger; uniform inside a solenoid); arrows showing direction. Draw at least three lines and keep them smooth; field lines must not touch or cross.
For a solenoid, show the parallel, equally spaced lines inside, lines spreading at the ends and closing outside, and label N and S if asked.
"Explain why the two wires attract" (3 marks): each wire is in the magnetic field of the other; the field is perpendicular to the current; a force $BIL$ acts on each wire (direction from Fleming's left-hand rule), towards the other wire.
"Explain why the forces are equal": Newton's third law. Mention it by name.
"State the effect of a ferrous core": the flux density is (greatly) increased. One mark is for "increased"; a second may need the reason (the core is magnetised by the field of the current).

Summary

Currents produce magnetic fields. Direction from the right-hand grip rule: thumb along current, fingers curl with the field.
Long straight wire: concentric circles, further apart with distance.
Flat circular coil: loops around each side of the wire; nearly straight lines perpendicular to the coil through its centre; like a short bar magnet. Anticlockwise current seen from a face makes that face N.
Long solenoid: uniform field inside (parallel, equally spaced lines); bar-magnet pattern outside; field at the ends is about half that at the centre.
The solenoid's field increases with current and turns per unit length, and is greatly increased by a ferrous (soft-iron) core.
Parallel wires: each lies in the other's field, so each feels $BIL$ . Like currents attract, unlike repel; the forces are equal and opposite (Newton's third law).

Practice questions

Question

State the rule used to find the direction of the magnetic field around a current-carrying wire, and describe the field pattern of a long straight wire.
Describe the magnetic field inside and outside a long solenoid carrying a steady current.
Explain why the field of a solenoid is increased when a soft-iron core is inserted, and why soft iron rather than steel is used in an electromagnet.
Two long parallel wires carry currents in the same direction. Explain, using the field pattern of one wire and Fleming's left-hand rule, why the wires attract.
Two long parallel wires $5.0\ \text{cm}$ apart carry currents of $10\ \text{A}$ and $15\ \text{A}$ in opposite directions. Using $B = kI/r$ with $k = 2.0 \times 10^{-7}\ \text{T m A}^{-1}$ , calculate the force per unit length on each wire and state its direction.
A flat circular coil lies in the plane of the page. Looking at the page, the current flows clockwise. State the direction of the field at the centre of the coil, and which face of the coil acts as a north pole.
Two long parallel wires A and B, $10\ \text{cm}$ apart, carry currents of $6.0\ \text{A}$ and $2.0\ \text{A}$ in opposite directions. Using $B = kI/r$ , find the position where the resultant flux density is zero.
A long wire carrying $9.0\ \text{A}$ runs parallel to a second wire carrying $3.0\ \text{A}$ . The force on a $2.0\ \text{m}$ length of the second wire is $2.7 \times 10^{-4}\ \text{N}$ . (a) Calculate the flux density at the second wire due to the first. (b) Calculate the flux density at the first wire due to the second. (c) Using $B = kI/r$ with $k = 2.0 \times 10^{-7}\ \text{T m A}^{-1}$ , find the separation of the wires.

Answers

The right-hand grip rule: grip the wire with the right thumb along the conventional current; the fingers curl in the direction of the field. The field lines are concentric circles centred on the wire, in planes perpendicular to it, becoming further apart with increasing distance.
Inside (away from the ends): strong, uniform field, with parallel, straight, equally spaced lines along the axis. Outside: a weaker field shaped like that of a bar magnet, with lines leaving one end (N), looping round and entering the other end (S). The field spreads out at the ends.
The field of the current magnetises the iron, aligning its magnetic domains; the field of the magnetised iron adds to the field of the current, increasing the flux density greatly. Soft iron is magnetised easily and loses its magnetism almost completely when the current is switched off, so the electromagnet can be switched off; steel stays magnetised.
Wire 1 produces circular field lines around it. At the position of wire 2, this field is perpendicular to wire 2. With both currents out of the page and wire 2 to the right of wire 1, wire 1's (anticlockwise) field at wire 2 points up the page. Fleming's left-hand rule (field up, current out of the page) gives a force on wire 2 to the left, towards wire 1. Similarly the force on wire 1 is towards wire 2, so they attract.
Field at the second wire due to the first: $B = 2.0 \times 10^{-7} \times 10/0.050 = 4.0 \times 10^{-5}\ \text{T}$ . $F/L = BI = 4.0 \times 10^{-5} \times 15 = 6.0 \times 10^{-4}\ \text{N m}^{-1}$ . The same size acts on the first wire (Newton's third law). Opposite currents repel, so each force is directed away from the other wire.
Clockwise current as seen from the front: the field at the centre points into the page, perpendicular to the plane of the coil (grip rule). The face you are looking at is a south pole; the back face (the side the field emerges from) is the north pole.
With opposite currents, the fields reinforce between the wires, so the zero point lies outside, on the side of the smaller current (B). Let it be $x$ beyond B: $6.0/(x + 0.10) = 2.0/x$ , so $6.0x = 2.0x + 0.20$ and $x = 0.050\ \text{m}$ . The neutral point is $5.0\ \text{cm}$ beyond B, on the side away from A ( $15\ \text{cm}$ from A).
(a) $B_1 = F/(I_2 L) = 2.7 \times 10^{-4}/(3.0 \times 2.0) = 4.5 \times 10^{-5}\ \text{T}$ . (b) The force on the first wire is also $2.7 \times 10^{-4}\ \text{N}$ , so $B_2 = F/(I_1L) = 2.7 \times 10^{-4}/(9.0 \times 2.0) = 1.5 \times 10^{-5}\ \text{T}$ . (c) $r = kI_1/B_1 = 2.0 \times 10^{-7} \times 9.0/4.5 \times 10^{-5} = 0.040\ \text{m}$ . Check: $kI_2/r = 2.0 \times 10^{-7} \times 3.0/0.040 = 1.5 \times 10^{-5}\ \text{T}$ , matching (b).

The right-hand grip rule

Long straight wire

Flat circular coil

Long solenoid

What makes a solenoid's field stronger

The effect of a ferrous core

Forces between current-carrying conductors

Worked examples

Practice questions

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