The Motor Effect

Why a current in a magnetic field feels a force, how its size and direction are set, and why two parallel currents pull on each other with equal and opposite forces.

50 min · Module 6: Electromagnetism

The full teaching. Start here.

Syllabus requirementsModule 6: Electromagnetism

investigate qualitatively and quantitatively the interaction between a current-carrying conductor and a uniform magnetic field $F = l I_{⊥} B = l I B sin θ$ to establish: (ACSPH080, ACSPH081)ACSPH080 ACSPH081
conditions under which the maximum force is produced
the relationship between the directions of the force, magnetic field strength and current
conditions under which no force is produced on the conductor
conduct a quantitative investigation to demonstrate the interaction between two parallel current-carrying wires
analyse the interaction between two parallel current-carrying wires $\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$ and determine the relationship between the International System of Units (SI) definition of an ampere and Newton’s Third Law of Motion (ACSPH081, ACSPH106)ACSPH081 ACSPH106

What this actually asks for

A short topic with two formulas, and both are on the data sheet. The marks go to using them properly: the right angle, the right length, the right direction, and the right units.

The first dot point says qualitatively and quantitatively, so expect a calculation and a direction question on the same wire. The second is a practical investigation, which means you may be asked to design it. The third says analyse and names Newton's third law and the definition of the ampere, so the parallel wire formula comes with an explanation attached.

The short version

A current in a magnetic field feels a force

$F = l I_{⊥} B = l I B sin θ$

where $θ$ is the angle between the wire and the field. The force is largest at $9 0^{\circ}$ and zero when the wire runs along the field.

The force is perpendicular to both the current and the field. Use the right hand palm rule: fingers along the field, thumb along the current, and the force comes out of the palm.

Two parallel wires each sit in the other's field, so each pushes on the other with

$\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$

Currents in the same direction attract; opposite directions repel. The two forces are equal and opposite, a Newton's third law pair, whatever the currents. That equality is what let the ampere be defined by a force.

Where the force comes from

A force on moving charges, passed to the wire

AnswersA magnetic field pushes on moving charges. Why does the wire itself move?

A magnetic field exerts a force on a charge only when the charge moves across the field lines. A current is a stream of moving charges, so every electron drifting along a wire in a field is pushed sideways.

The electrons cannot leave the wire. They press against the side of the conductor, and the metal lattice holding them in transfers the push to the whole wire. So a force on countless tiny charges becomes one force on a solid object you can see move. That is the motor effect, and it is what turns the coil in every electric motor.

Derivation

The force on a current-carrying wire

Starts from

the force on each moving charge, F = qvB

Ends at

F = l I B sin θ

Holds only if

The wire is straight
The field is uniform over the length of wire inside it

The force on a current-carrying wire

3 steps

1
Start with the charge in the wire
The force on one moving charge is known from the previous topic. Here, take q to be all the charge passing along the length l.
$F = q v_{⊥} B$
2
Write the speed in terms of the wire
In a time t, the charge q travels the length l of wire in the field.
$v_{⊥} = \frac{l _{⊥}}{t} \Rightarrow F = q \frac{l _{⊥}}{t} B = \frac{q}{t} l_{⊥} B$
3
Recognise the current
Charge per unit time is the current, which is the quantity that can be measured.
$F = l I_{⊥} B = l I B sin θ$
Only the part of the current across the field, $I sin θ$ , counts. $θ$ is measured from the field.

Force on a current-carrying conductor in a magnetic fieldon the NESA formulae sheet

F = l I_{⊥} B = l I B sin θ

Symbols: $F$ size of the magnetic force on the conductorN
$l$ length of conductor inside the fieldm
$I$ current in the conductorA
$B$ magnetic field strengthT
$θ$ angle between the conductor and the field°
Valid when: The conductor is straight and the field is uniform along the length that sits inside it. The force is perpendicular to both the conductor and the field, with its direction given by the right hand palm rule.
Not valid when: The field varies along the wire, the wire is curved, or θ is measured from the normal to the field rather than from the field itself.
Rearranged: for B: $B = \frac{F}{l I s i n θ}$ for I: $I = \frac{F}{l B s i n θ}$ for \theta: $sin θ = \frac{F}{l I B}$
Where it turns up: Finding the force on one side of a motor coil
Measuring a magnetic field with a current balance
Predicting when a wire in a field feels no force at all
Where marks go missing: Using the whole length of the wire rather than the length inside the field
Using cos θ, or measuring θ from the perpendicular
Giving the force a direction along the field or along the wire

Where this comes from

The syllabus asks for the conditions for the largest force and for no force at all. Both come out of $sin θ$ . At $9 0^{\circ}$ the whole current crosses the field and the force is $l I B$ . At $0^{\circ}$ or $18 0^{\circ}$ the current runs along the field lines, crosses none of them, and there is no force, however big the current or strong the magnet.

F = lIB sin θ

Only the part of the current across the field produces a force. At 90° all of it does; along the field, none of it does.

Current4.0 A

Length in the field0.25 m

Field strength0.30 T

Angle to the field30 °

Force at this angle: 0.150 N
Current component across the field, I sin θ: 2.00 A
Maximum force, at 90°: 0.300 N

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Worked example3 marks

One wire at three angles

A $0.25$ m length of wire carrying $4.0$ A lies in a uniform $0.30$ T field. Find the force on it when it makes an angle with the field of $9 0^{\circ}$ , then $3 0^{\circ}$ , then $0^{\circ}$ .

At right angles

This is the maximum. Doing it first gives a reference for the others.

$F = l I B sin 9 0^{\circ} = 0.25 \times 4.0 \times 0.30 \times 1 = 0.30 N$

At 30°

Only the current component across the field counts, I sin 30° = 2.0 A.

$F = 0.25 \times 4.0 \times 0.30 \times sin 3 0^{\circ} = 0.15 N$

Along the field

The zero force case needs a reason, not just a number.

$F = 0.25 \times 4.0 \times 0.30 \times sin 0^{\circ} = 0$

Answer

The force is $0.30$ N at $9 0^{\circ}$ , $0.15$ N at $3 0^{\circ}$ and zero along the field. The largest force needs the wire perpendicular to the field; no force happens when the wire is parallel to it.

Is that answer sensible?

$sin 3 0^{\circ} = 0.5$ , so the $3 0^{\circ}$ force should be exactly half the maximum, and it is. If the answers went the other way, with the largest force along the field, $cos$ was used in place of $sin$ .

CheckpointAnswer before reading on.

A straight wire carries a current of $3.0$ A. A $0.40$ m length of it lies at right angles to a uniform magnetic field of $0.25$ T.

Calculate the size of the force on the wire.

Your answerN

Give it to 2 significant figures.

Hint 1Which formula links force, length, current and field?

Hint 2At right angles, $sin θ = 1$ , so the whole current is across the field.

Hint 3 $F = l I B = 0.40 \times 3.0 \times 0.25$

CheckpointAnswer before reading on.

The same wire, $0.40$ m long in the field and carrying $3.0$ A, is turned so that it makes an angle of $4 0^{\circ}$ with the $0.25$ T field.

Calculate the new force on the wire.

Your answerN

Give it to 2 significant figures.

Hint 1Only the component of the current across the field produces a force.

Hint 2That component is $I sin θ$ , where $θ$ is measured from the field direction.

Hint 3 $F = 0.40 \times 3.0 \times 0.25 \times sin 4 0^{\circ}$

CheckpointAnswer before reading on.

A wire carrying a large current passes between the poles of a strong magnet, running straight from the north pole to the south pole.

Which statement about the force on the wire is correct?

AThere is no force, because the current is parallel to the field.
BThere is a large force, because both the current and the field are large.
CThe wire is pushed towards the south pole, along the field.
DThe wire is attracted to whichever pole is nearer.

Hint 1Which way do the field lines between the poles run?

Hint 2Compare the direction of the current with the direction of the field.

Hint 3What is $sin θ$ when the current and the field are parallel?

Exam question

Straightforward · about 3 min

2 marks

A straight conductor carrying a constant current is placed in a uniform magnetic field. State the condition for the force on it to be as large as possible, and the condition for there to be no force, referring to $F = l I B sin θ$ .

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1What is $θ$ measured between?

Hint 2Where does $sin θ$ reach its largest value?

Hint 3Where is $sin θ$ zero?

Where students lose marks on this one

Saying the force is maximum when the wire points along the field lines.
Why it happens: Picturing the wire following the field, as if the field pulled it along.
The force comes from current crossing the field. Along the field there is nothing to cross.

Written for this site.

Which way the force acts

The force is perpendicular to the current and to the field. In three dimensions that leaves one line, and the right hand palm rule picks the end:

Fingers point along the field, north to south.
Thumb points along the conventional current, positive to negative.
The palm pushes in the direction of the force.

Diagrams show directions in and out of the page with two symbols. A cross, $\otimes$ , is the tail of an arrow going away from you, into the page. A dot, $⊙$ , is its point coming towards you.

Worked example2 marks

Finding the direction

A wire lying on a bench carries current towards the east. A magnet provides a horizontal field pointing north. Which way is the force on the wire? What changes if the current is reversed?

Set the fingers along the field

The field is the one direction the hand is fixed by, so it goes first.

Right hand flat, fingers pointing north, away from you if you face north.

Turn the hand so the thumb follows the current

Rotating about the fingers keeps them along the field while the thumb finds the current.

To point the thumb east with the fingers still north, the palm must face up.

Reverse one quantity

Reversing the current or the field, but not both, reverses the force.

With the current flowing west, the palm faces down.

Answer

The force is vertically upwards. With the current reversed it is downwards, pressing the wire into the bench.

Is that answer sensible?

The force is perpendicular to both east and north, so it must be up or down. A horizontal answer here is always wrong, whatever the hand says.

CheckpointAnswer before reading on.

A horizontal wire carries current towards the north. It lies in a magnetic field that points vertically downwards.

In which direction is the force on the wire?

AWest
BEast
CDown
DNorth

Hint 1Use the right hand palm rule: fingers along the field, thumb along the current.

Hint 2Point your fingers at the floor and your thumb north.

Hint 3The force comes out of your palm.

CheckpointAnswer before reading on.

A wire in a magnetic field experiences a force to the left. The current is reversed and the magnet is turned round so the field is reversed.

What is the force now?

AThe same size, to the left
BThe same size, to the right
CZero, because the two reversals cancel the force
DTwice as large, to the right

Hint 1Reversing one quantity reverses the force.

Hint 2What does reversing it a second time do?

Hint 3The sizes of $I$ and $B$ have not changed.

Measuring the force: a current balance

The force is often small, a fraction of a newton. The simulation shows one way to see it. A rod hangs on two light conducting threads in a vertical field. When current flows along the rod, the magnetic force is horizontal, and the rod swings aside until its weight, the tension and the magnetic force balance.

Speed

Current5.0 A

Field strength0.20 T

Rod length in the field0.10 m

Rod mass20 g

F = lIB sin θ

The same situation drawn as a graph, shown because this page has no JavaScript to animate it.

The rod is seen end on. A cross means current into the screen, a dot means current out of it. The field points up the screen.

Reverse the current and the rod swings the other way. Set the field to zero and nothing happens, whatever the current. The final angle obeys $tan φ = F / m g$ , so measuring it measures the force.

CheckpointAnswer before reading on.

A $15$ g metal rod hangs horizontally from two light conducting threads. $8.0$ cm of the rod lies in a vertical magnetic field of $0.25$ T. When a current of $4.0$ A flows through the rod, it swings sideways and comes to rest with the threads at an angle to the vertical.

Calculate this angle.

Your answer°

Give it to 3 significant figures.

Hint 1At rest, three forces act on the rod: its weight, the tension, and the magnetic force.

Hint 2The magnetic force is horizontal because the field is vertical and the rod is horizontal.

Hint 3The tension balances both, so $tan φ = \frac{F}{m g}$ .

CheckpointAnswer before reading on.

In a current balance, a $5.0$ cm length of wire carries $2.0$ A at right angles to a magnetic field. The balance shows a force of $1.2 \times 1 0^{- 2}$ N on the wire.

Calculate the magnetic field strength.

Your answerT

Give it to 2 significant figures.

Hint 1Rearrange $F = l I B sin θ$ for $B$ .

Hint 2The length must be in metres.

Hint 3 $B = \frac{1.2 \times 1 0 ^{- 2}}{0.050 \times 2.0}$

Two parallel wires

Each wire sits in the other's field

AnswersNeither wire is a magnet. Why do they push or pull on each other?

A current creates a magnetic field around it, in circles centred on the wire. Its strength a distance $r$ from a long straight wire is $B = \frac{μ _{0} I}{2 π r}$ , which you met in Module 4.

Put a second wire alongside, parallel to the first. It is now a current in a magnetic field, the first wire's field, so it feels a motor effect force. The first wire sits in the second wire's field in the same way. Each wire pushes on the other, and neither needs a magnet.

Derivation

The force between parallel wires

Starts from

the field of one wire acting on the current in the other

Ends at

\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}

Holds only if

The wires are long, straight and parallel
The separation is small compared with the length

The force between parallel wires

3 steps

1
The field of wire 1 at wire 2
Wire 2 is a distance r from wire 1.
$B_{1} = \frac{μ _{0} I _{1}}{2 π r}$
2
The motor effect force on wire 2
Wire 1's field circles around it, so at wire 2 it is perpendicular to wire 2 and sin θ = 1.
$F = l I_{2} B_{1} = l I_{2} \frac{μ _{0} I _{1}}{2 π r}$
3
Divide by the length
For long wires the force grows with the length, so the force per metre is the useful quantity.
$\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$

Force between two parallel current-carrying wireson the NESA formulae sheet

\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}

Symbols: $F$ force on a length l of either wireN
$l$ length of wire consideredm
$μ_{0}$ permeability of free space, 4π × 10⁻⁷T m A⁻¹
$I_{1}, I_{2}$ currents in the two wiresA
$r$ perpendicular distance between the wiresm
Valid when: The wires are long, straight and parallel, and their separation is small compared with their length. Currents in the same direction attract; opposite directions repel.
Not valid when: The wires are short compared with their separation, not parallel, or r is measured as anything other than the centre to centre distance between them.
Rearranged: for r: $r = \frac{μ _{0} I _{1} I _{2} l}{2 π F}$ for I_{2}: $I_{2} = \frac{2 π r F}{μ _{0} I _{1} l}$
Where it turns up: Forces between conductors in cables and busbars
Stating the historical definition of the ampere
Showing that the forces on the two wires form a Newton's third law pair
Where marks go missing: Squaring r, as if the force were inverse square
Leaving r in centimetres
Giving the wire with the larger current the larger force

Where this comes from

Two parallel wires with currents into the page

1Wire 1, current into the page
The cross means the current flows away from you, into the page. With the right hand grip rule, the thumb points into the page and the fingers curl clockwise, which is the direction of the field around this wire.
2The field of wire 1
The field lines are circles centred on the wire, weaker further out. The strength falls as $1/ r$ , which is why the force between the wires is $1/ r$ rather than inverse square.
3Wire 1 field at wire 2
At the position of wire 2, to the right of wire 1, the clockwise field points straight down the page. It is perpendicular to wire 2, so all of wire 2's current crosses it.
4The force on wire 2
Palm rule on wire 2: fingers down the page along $B_{1}$ , thumb into the page along $I_{2}$ . The palm faces left, towards wire 1. Same direction currents attract.
5The force on wire 1
Repeat the reasoning for wire 1 in wire 2's field and the force points right, towards wire 2. It has the same size, $\frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$ per metre, because the formula is symmetric in the two currents.
6Wire 2, current into the page
Reverse this current, drawing a dot instead of a cross, and the palm rule gives a force to the right. Wires with opposite currents repel.

Seen end on. The field of the left wire circles clockwise. At the right wire it points down the page, and the palm rule then gives a force towards the left wire.

F / l

The force falls as one over the distance, not one over the distance squared. A negative current in the second wire means it runs the opposite way.

Current in wire 120 A

Current in wire 220 A

Separation5 cm

Force per metre: 1.60 × 10⁻³ N m⁻¹
The wires: attract (same direction)
Force on wire 1 compared with wire 2: equal size, opposite direction
At double the separation: 8.00 × 10⁻⁴ N m⁻¹

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Worked example3 marks

Forces between power cables

Two parallel cables $0.40$ m apart carry currents of $200$ A and $150$ A in opposite directions. Find the force between them over a $50$ m span, and state its direction.

Force per metre

The formula gives force per metre. Using μ₀/2π = 2 × 10⁻⁷ saves a step and a source of error.

$\frac{F}{l} = 2 \times 1 0^{- 7} \times \frac{200 \times 150}{0.40} = 1.5 \times 1 0^{- 2} N m^{- 1}$

Over the span

The question asks for a force, not a force per metre.

$F = 1.5 \times 1 0^{- 2} \times 50 = 0.75 N$

Direction

Opposite currents repel, and the force on each cable is the same size.

Each cable is pushed away from the other with $0.75$ N.

Answer

The cables repel, each with a force of $0.75$ N over the span.

Is that answer sensible?

A small force, as it should be for ordinary currents. In a short circuit the currents might reach $10000$ A each, and the force per metre goes up by the factor $\frac{1 0 ^{8}}{3 \times 1 0 ^{4}}$ , to about $50$ N m⁻¹, or $2500$ N over the span. That is why the conductors in switchboards are braced.

CheckpointAnswer before reading on.

Two long parallel wires are $3.0$ cm apart. One carries $12$ A and the other carries $8.0$ A in the same direction.

Calculate the force on a $2.0$ m length of either wire, and state whether the wires attract or repel.

Your answerN

Give it to 2 significant figures.

Hint 1The data sheet formula gives force per metre, $F / l$ .

Hint 2Convert the separation to metres, and note $μ_{0} /2 π = 2 \times 1 0^{- 7}$ .

Hint 3 $\frac{F}{l} = 2 \times 1 0^{- 7} \times \frac{12 \times 8.0}{0.030}$ , then multiply by the length.

CheckpointAnswer before reading on.

Two parallel wires exert a force $F$ on each other. Both currents are doubled and the separation between the wires is also doubled.

What is the new force?

A $2 F$
B $F$
C $4 F$
D $F /2$

Hint 1The force is proportional to $I_{1} I_{2}$ and inversely proportional to $r$ .

Hint 2Doubling both currents multiplies the product by four.

Hint 3Doubling $r$ divides by two, not four.

Newton's third law and the ampere

The force on wire 2 is caused by wire 1, and the force on wire 1 by wire 2. They are an action and reaction pair, so they are equal in size and opposite in direction. That stays true even when one current is five times the other: the bigger current makes a stronger field, but it sits in a weaker one, and the product $I_{1} I_{2}$ is the same both ways round.

CheckpointAnswer before reading on.

Wire P carries $10$ A and wire Q, parallel to it, carries $2$ A. Which statement about the magnetic forces between them is correct?

AThe forces on P and Q are equal in size and opposite in direction.
BQ feels a force five times larger than P, because P's current is five times larger.
CP feels the larger force, because more current means more of the wire responds to the field.
DOnly Q feels a force, because P's field is much stronger than Q's.

Hint 1Write out $F / l$ for the force on P and for the force on Q.

Hint 2Which quantities appear in each?

Hint 3Which law of motion describes a pair of forces between two objects?

This symmetry is why the ampere could be defined using the force. From 1948 to 2019, one ampere was the steady current which, in two long straight parallel conductors $1$ m apart in a vacuum, produces a force of exactly $2 \times 1 0^{- 7}$ N on each metre of each conductor. The word "each" relies on the third law; without it the definition would have to say which wire to measure. Substituting $I_{1} = I_{2} = 1$ A and $r = 1$ m into the formula gives that force, and that is where $μ_{0} = 4 π \times 1 0^{- 7}$ came from.

Since 2019 the ampere has been defined by fixing the electron's charge exactly. The parallel wire law is unchanged, and $μ_{0}$ is now a measured value that agrees with $4 π \times 1 0^{- 7}$ to better than one part in a billion.

Exam question

Harder · about 5 min

3 marks

The ampere was defined for many years using the force between two parallel current-carrying wires. Describe that definition, and explain how it relies on Newton's third law.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1The definition fixed a separation, a force per metre and the currents.

Hint 2Substitute $I_{1} = I_{2} = 1$ A and $r = 1$ m into the formula.

Hint 3Would the definition work if the two wires felt different forces?

Where students lose marks on this one

Describing the ampere as one coulomb per second and stopping there.
Why it happens: That is the familiar relation between current and charge.
The question asks about the force definition. Give the separation, the currents and the force per metre, then connect the equal forces to the third law.

Written for this site.

Investigating parallel wires

The second dot point asks for a quantitative investigation. The main difficulty is size: at $10$ A and $1$ cm apart the force is only about $2 \times 1 0^{- 3}$ N per metre. The usual solution is to rest one wire on a sensitive electronic balance, fix the other just above it, and connect them in series so the same current flows through both. The balance reading changes as the force pushes down or pulls up.

With the same current in both wires, the force goes as $I^{2}$ , so the straight line graph to aim for is force against $I^{2}$ . Its gradient is $μ_{0} l /2 π r$ . The main hazard is heating, since these currents are much larger than in most practicals, so readings are taken in short bursts.

CheckpointAnswer before reading on.

In an investigation, two long parallel wires carry equal currents $I$ in opposite directions. A student measures the force per metre for several currents and plots $F / l$ against $I^{2}$ . The line of best fit passes through the origin with a gradient of $1.2 \times 1 0^{- 5}$ N m⁻¹ A⁻².

Use the gradient to calculate the separation of the wires.

Your answerm

Give it to 2 significant figures.

Hint 1With $I_{1} = I_{2} = I$ , the formula becomes $F / l = \frac{μ _{0}}{2 π r} I^{2}$ .

Hint 2So the gradient of $F / l$ against $I^{2}$ is $μ_{0} /2 π r$ .

Hint 3 $r = \frac{2 \times 1 0 ^{- 7}}{1.2 \times 1 0 ^{- 5}}$

Exam question

Harder · about 9 min

5 marks

Design a quantitative investigation to show how the force between two parallel current-carrying wires depends on the current in them. Include the equipment, the method, how you would control variables and ensure safety, and how you would analyse the results.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1The forces are tiny, so you need a sensitive way to measure them, such as one wire resting on an electronic balance.

Hint 2Keep the separation and the length fixed. Change the current, which can run through both wires in series.

Hint 3With the same current in both wires, what should you plot to get a straight line?

Where students lose marks on this one

Plotting force against current and expecting a straight line.
Why it happens: Forgetting that with one current in both wires, the force depends on its square.
Write the formula for the setup you have designed before choosing the axes. Linearise it so the prediction is a straight line.
Measuring the force with a spring balance or newton meter.
Why it happens: They are the usual laboratory force measurers.
Estimate the force first. At $10$ A and $1$ cm, it is about $2 \times 1 0^{- 3}$ N per metre, far below what a spring balance can read.

Written for this site.

Common mistake

Measuring θ from the perpendicular to the field, then using sin θ.

Why it happens: In other topics angles are often measured from a normal, as in refraction and flux, so the habit carries over.

In $F = l I B sin θ$ , the angle is between the wire and the field. Check with the extremes: a wire along the field feels no force, and $sin 0^{\circ} = 0$ . If the question gives the angle to the perpendicular, subtract it from $9 0^{\circ}$ first, or equivalently use cosine of the given angle.

Explaintypically 3 to 5 marks

Demands: Give the cause, or say why the relationship holds. The word "because" should appear in substance.
Shape: A chain: this happens, which causes that, which is why the observed result follows.
Loses marks: Describing the result in more detail and never naming a mechanism.

Through a marker’s eyes

3 marks

Two parallel wires carry currents of 10 A and 2 A in the same direction. Explain why the forces on the two wires are equal in size. (3 marks)

The attempt

1 out of 3

The wires attract because the currents are in the same direction. The forces are equal because of Newton's third law.

What the marker sees

Naming the third law earns a mark, but the question says explain, so it wants the mechanism. The response does not say what exerts each force or why a larger current does not produce a larger force on the other wire. Stating that the wires attract answers a question that was not asked.

The same answer, fixed

3 out of 3

Each wire is a current in the magnetic field of the other, so each feels a motor effect force. The $10$ A wire produces a field five times stronger, but it acts on a current five times smaller; the $2$ A wire's weaker field acts on the larger current.
Both forces are therefore $\frac{F}{l} = \frac{μ _{0}}{2 π} \frac{I _{1} I _{2}}{r}$ per metre, the same product of currents either way round.
The force of wire 1 on wire 2 and of wire 2 on wire 1 are an action and reaction pair, so by Newton's third law they are equal in size and opposite in direction.

The force as a cross product

The rule in the palm is a picture of a vector cross product:

$F = I l \times B$

The size of a cross product is $l B sin θ$ , which is where the sine comes from, and its direction is perpendicular to both vectors, given by the right hand. The HSC does not require the notation, but it explains why no force ever acts along the field or along the wire: a cross product has no component along either of its factors.

A magnetic force does no work on the charges

The force on each charge is perpendicular to its velocity through the field, so it does no work on it. Yet a motor clearly does work. The resolution is that the magnetic force pushes the charges sideways, the lattice pushes the wire, and the energy comes from the power supply, which has to push harder once the wire moves. That extra push is back emf, which the next topics develop.

Where this reappears

The next topic, applications of the motor effect, applies $F = l I B$ to each side of a coil to give the torque $τ = n I A B sin θ$ that drives a DC motor. The loudspeaker and the galvanometer are the same force in different shapes. And the pinch effect, where a very large current squeezes a conductor inwards because each part of it attracts every other part, is the parallel wire law applied inside a single thick conductor.