How to Find the Force Direction on a Current-Carrying Wire

A straight wire sits between the poles of a magnet. Current flows from left to right, while the magnetic field points into the page. You know the wire experiences a magnetic force, but which way will it move: up or down?

The answer is upwards. You can find this using Fleming’s left-hand rule, which connects three directions:

• First finger: magnetic field, \(B\).
• Second finger: conventional current, \(I\).
• Thumb: magnetic force, \(F\).

The key is to keep the magnetic field and the force separate. They are not the same thing. The magnetic field describes the environment around the wire; the force describes the push acting on the wire.

01Understand the three directions

Before using the hand rule, identify exactly what each quantity represents.

There are three different directions to keep track of:

1. The magnetic field direction, \(B\).
2. The conventional current direction, \(I\).
3. The resulting force direction, \(F\).

The force acts perpendicular to both the current and the magnetic field.

How to recognise the magnetic field direction

If you have a bar magnet, magnetic field lines outside the magnet travel from its north pole to its south pole.

In a diagram, the field might be represented by arrows or by symbols showing its direction relative to the page.

Two symbols are particularly important:

• \(\times\) means the magnetic field points into the page. Imagine looking at the tail of an arrow travelling away from you.
• \(\bullet\) means the magnetic field points out of the page towards you. Imagine looking directly at the arrowhead.

Don’t confuse these symbols with current direction. A cross or dot can represent either a magnetic field or a current, depending on what the diagram labels.

Fleming’s left-hand rule

This is the rule you need to determine the force direction on a current-carrying conductor in a magnetic field.

Hold your left hand with your thumb, first finger, and second finger pointing at right angles to one another.

Assign each finger as follows:

The important detail is that you must use conventional current, which flows from the positive terminal towards the negative terminal through the external circuit.

Electron flow is in the opposite direction. If a question gives electron movement rather than conventional current, reverse that direction before applying Fleming’s rule.

Also, don’t confuse Fleming’s left-hand rule with the right-hand grip rule. The right-hand grip rule determines the circular magnetic field around a current-carrying wire. Fleming’s left-hand rule determines the force on a current-carrying wire already inside a magnetic field. They answer different questions.

02How the experiment establishes the force direction

Why do we use this particular rule? It comes from the observed behaviour of a current-carrying conductor placed in a magnetic field.

Consider a simple motor-effect experiment designed to investigate the relationship between current direction, magnetic field direction, and force.

Setting up the experiment

Place the two poles of a horseshoe magnet facing each other, leaving a gap between them. The north pole is on the left, and the south pole is on the right.

Suspend a straight conducting wire vertically through the gap. Connect the wire to a low-voltage DC power supply using flexible leads. Include a switch and an ammeter in the circuit so that the current can be controlled and monitored.

The wire must be free to move slightly into or out of the page. The section between the magnet poles is the active conductor length.

The magnetic field points horizontally from left to right, from north to south.

What should happen when the current is switched on?

Suppose the current flows upwards through the wire.

The magnetic field points from left to right. Applying Fleming’s left-hand rule predicts that the force acts into the page.

The suspended conductor should therefore deflect away from the observer.

Now reverse the current while leaving the magnet in exactly the same orientation.

The force should reverse and act out of the page. The conductor should deflect towards the observer.

This is an important observation: reversing the current reverses the magnetic force, provided the magnetic field remains unchanged.

Testing whether the magnetic field also matters

There is another way to reverse the force.

Keep the current flowing upwards, but swap the magnet poles so that the magnetic field points from right to left.

The force also reverses.

Finally, reverse both the current and the magnetic field. The force returns to its original direction.

The ideal directional outcomes are summarised below.

These are the predicted directional outcomes for the stated arrangement, rather than claimed measurements from an actual experiment.

The pattern establishes three useful relationships:

• Reversing the current alone reverses the force.
• Reversing the magnetic field alone reverses the force.
• Reversing both together leaves the force direction unchanged.

Why the experimental arrangement matters

A wire resting on a bench might experience friction that prevents it from moving freely. Stiff connecting leads could also pull the wire sideways, creating an unwanted force.

A better design suspends the conductor using flexible leads and keeps it clear of the magnet poles. The magnet remains fixed while the current or magnetic field is reversed independently.

The ammeter helps check that the current magnitude remains constant between trials.

These controls matter because you want to attribute a change in the wire’s movement to the variable you deliberately changed, rather than to friction, a different current, or a mechanical obstruction.

Estimating the size of the force

The direction experiment can also be connected to a force calculation.

For a straight conductor in a uniform magnetic field, the magnitude of the magnetic force is:

\[ F=BIL\sin\theta \]

Here, \(B\) is the magnetic flux density, \(I\) is the current, \(L\) is the active conductor length inside the field, and \(\theta\) is the angle between the current and magnetic field directions.

Consider illustrative experimental values:

• Magnetic field strength: \(0.25\text{ T}\)
• Current: \(2.0\text{ A}\)
• Active wire length: \(0.10\text{ m}\)
• Angle between current and field: \(90^\circ\)

Substituting:

\[ \begin{aligned} F &= BIL\sin\theta\\ &= (0.25)(2.0)(0.10)\sin90^\circ\\ &=0.050\text{ N} \end{aligned} \]

The magnetic force is \(0.050\text{ N}\).

To understand the scale, compare this with the weight of a small mass:

\[ m=\frac{F}{g} \]

\[ m=\frac{0.050}{9.8} =0.0051\text{ kg} \]

This is approximately \(5.1\text{ g}\).

So the magnetic force is comparable in magnitude to the weight of a mass of about \(5.1\text{ g}\). This is an illustrative scale estimate, not a measured deflection or a claim about a particular apparatus.

Notice that the equation calculates the force magnitude. Fleming’s left-hand rule determines its direction. You need both pieces of information to describe the magnetic force completely.

03A reliable method for finding force direction

In an HSC question, don’t immediately reach for your left hand. First make sure you have identified the correct conductor segment and magnetic field direction.

Use this five-step method.

Step 1: Identify the active conductor segment.

Find the part of the wire that lies inside the magnetic field. This is the section that experiences the magnetic force from that field.

If a wire is \(0.50\text{ m}\) long but only \(0.08\text{ m}\) lies inside the field, the active length is \(0.08\text{ m}\).

Step 2: Identify the magnetic field direction.

Look for the north and south poles, or the arrows and crosses representing the field.

Remember that the field points from north to south outside a magnet.

Step 3: Identify the conventional current direction.

Follow the current through the active section. If the question gives the direction of electron flow, reverse it.

Step 4: Apply Fleming’s left-hand rule.

Position your first finger along the magnetic field and your second finger along the conventional current.

Your thumb now points in the direction of the magnetic force.

Step 5: Calculate the force magnitude if required.

Use:

\[ F=BIL\sin\theta \]

The angle \(\theta\) must be measured between the current direction and the magnetic field direction.

The equation gives the magnitude of the force, not its direction.

04The common misconception: The force follows the magnetic field

A tempting mistake is to assume that a wire is pushed in the direction of the magnetic field.

For example, imagine a horizontal wire carrying current to the right. The magnetic field also points to the right.

You might predict that the wire experiences a force to the right.

However, this is incorrect.

The current and magnetic field are parallel, meaning:

\[ \theta=0^\circ \]

Substituting into the force equation:

\[ F=BIL\sin0^\circ=0 \]

There is no magnetic force on the straight conductor from that field.

The same result occurs if the current and magnetic field point in opposite directions:

\[ \theta=180^\circ \]

Since \(\sin180^\circ=0\), the magnetic force is still zero.

This gives you three useful cases to remember:

The force depends on the component of the magnetic field perpendicular to the current.

What happens when you reverse the current or field?

This is another useful way to check your answer.

Suppose a wire initially experiences a magnetic force upwards.

These relationships apply when the original force is non-zero.

A quick way to remember them is:

• One reversal changes the force direction.
• Two reversals restore the original force direction.

If your answer contradicts this pattern, check your left-hand rule again.

05Questions and solutions

The following questions progress from straightforward direction identification to problems involving experimental reasoning and multiple conductor sections.

Question 1

A straight horizontal conductor has a length of \(0.25\text{ m}\) inside a uniform magnetic field of \(0.40\text{ T}\).

The magnetic field points into the page, and the conductor carries a conventional current of \(3.0\text{ A}\) from left to right.

a. Determine the direction of the magnetic force.

b. Calculate the magnitude of the force.

Solution 1

The conductor experiences an upward magnetic force with a magnitude of \(0.30\text{ N}\).

Part a: Direction

The magnetic field points into the page, and the conventional current points to the right.

Using Fleming’s left-hand rule:

• First finger: into the page.
• Second finger: right.
• Thumb: upwards.

Therefore, the magnetic force acts upwards.

Part b: Magnitude

The current is perpendicular to the magnetic field, so:

\[ \theta=90^\circ \]

Using:

\[ F=BIL\sin\theta \]

Substitute the values:

\[ \begin{aligned} F&=(0.40)(3.0)(0.25)\sin90^\circ\\ &=0.30\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F=0.30\text{ N upwards}} \]

The magnetic field produces an upward force on the active \(0.25\text{ m}\) section of the conductor.

Question 2

A straight horizontal conductor carries a conventional current of \(3.0\text{ A}\) from left to right.

A uniform magnetic field of \(0.40\text{ T}\) points \(120^\circ\) anticlockwise from the direction of the current. The active conductor length is \(0.25\text{ m}\).

a. Determine the direction of the magnetic force.

b. Calculate its magnitude.

c. The current is now reversed, while the magnetic field remains unchanged. State the new force direction and magnitude.

Solution 2

The original force acts out of the page with a magnitude of approximately \(0.26\text{ N}\). Reversing the current makes the force act into the page, without changing its magnitude.

Part a: Original force direction

The current points to the right, and the magnetic field points upwards and to the left.

Applying Fleming’s left-hand rule gives a force out of the page.

Part b: Force magnitude

The angle between the current and magnetic field is \(120^\circ\).

Use:

\[ F=BIL\sin\theta \]

Substitute:

\[ \begin{aligned} F&=(0.40)(3.0)(0.25)\sin120^\circ\\ &=(0.40)(3.0)(0.25)(0.866)\\ &=0.2598\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F\approx0.26\text{ N}} \]

The force acts out of the page.

An important detail is that \(\sin120^\circ=\sin60^\circ\). Both angles give the same force magnitude, but you must retain the actual current and field directions when applying the hand rule.

Part c: Reverse the current

Reversing the current while keeping the magnetic field unchanged reverses the force direction.

The force now acts into the page.

Its magnitude remains approximately \(0.26\text{ N}\), because the angle between the new current direction and the magnetic field is \(60^\circ\), and:

\[ \sin60^\circ=\sin120^\circ \]

The force changes direction but not magnitude.

Question 3

A student suspends a straight conducting wire between the poles of a horseshoe magnet.

Initially:

• The north pole is on the left.
• The south pole is on the right.
• Conventional current flows upwards through the wire.
• The wire deflects into the page.

The student claims that reversing the magnet poles will not change the force direction because the current remains unchanged.

a. Explain why this prediction is incorrect.

b. Predict the force direction when the magnet poles are reversed but the current remains upwards.

c. Predict the force direction when both the magnet poles and the current are reversed.

d. Identify one important experimental control and explain why it is needed.

Solution 3

The student’s prediction is incorrect. Reversing the magnetic field alone reverses the force, while reversing both the field and current restores the original force direction.

Part a: Explain the error

The magnetic force depends on both the current direction and the magnetic field direction.

Changing either one reverses the force, even if the other remains unchanged.

The student has incorrectly assumed that the force depends only on the current.

Part b: Reverse the magnetic field only

Initially, the magnetic field points from left to right, and the force acts into the page.

After swapping the magnet poles, the field points from right to left.

The current still flows upwards.

Applying Fleming’s left-hand rule gives a force out of the page.

Part c: Reverse both directions

Now reverse the current so that it flows downwards, while keeping the reversed magnet orientation.

The magnetic field points from right to left, and the current points downwards.

Both directions have been reversed, so the force returns to its original direction.

It acts into the page.

Part d: Experimental control

The current magnitude should be kept constant between trials.

An ammeter can be used to monitor the current. This matters because changing the current magnitude would also change the force magnitude, making it harder to identify the effect of reversing the magnetic field.

The wire should also remain freely suspended so that friction or mechanical contact does not interfere with its movement.

The experiment demonstrates that both the current and magnetic field determine the force direction.

Question 4

A straight conductor has an active length of \(0.15\text{ m}\) inside a uniform magnetic field of \(0.50\text{ T}\).

The magnetic field points into the page. An experiment shows that the conductor experiences a downward magnetic force of \(0.30\text{ N}\).

Determine:

a. The direction of conventional current through the conductor.

b. The magnitude of the current.

Solution 4

The conventional current flows from right to left, with a magnitude of \(4.0\text{ A}\).

Part a: Current direction

We know:

• Magnetic field: into the page.
• Magnetic force: downwards.

Work backwards using Fleming’s left-hand rule.

The thumb must point downwards, and the first finger must point into the page.

This means the second finger points towards the left.

Therefore, conventional current flows from right to left.

Part b: Current magnitude

Start with:

\[ F=BIL\sin\theta \]

The current is perpendicular to the magnetic field, so:

\[ \theta=90^\circ \]

Rearrange the equation to make current the subject:

\[ I=\frac{F}{BL\sin\theta} \]

Substitute:

\[ \begin{aligned} I&=\frac{0.30}{(0.50)(0.15)\sin90^\circ}\\ &=\frac{0.30}{0.075}\\ &=4.0\text{ A} \end{aligned} \]

Therefore:

\[ \boxed{I=4.0\text{ A}} \]

The conductor carries a conventional current of \(4.0\text{ A}\) from right to left.

This question is worth practising because it reverses the usual process. Instead of using current and magnetic field to find force, you use the known force and magnetic field to infer the current direction.

Question 5

A continuous conducting wire is bent into a U-shaped arrangement.

It starts at point A in the bottom-left corner, travels horizontally \(0.30\text{ m}\) to the right, then vertically upwards \(0.12\text{ m}\), and finally horizontally \(0.30\text{ m}\) to the left, ending at point B.

The conventional current flows from A to B.

All three straight sections lie inside magnetic fields directed into the page. However, the field is not uniform across the entire arrangement:

• The lower horizontal section experiences a field of \(0.50\text{ T}\).
• The vertical section experiences a field of \(0.50\text{ T}\).
• The upper horizontal section experiences a field of \(0.30\text{ T}\).

The current is \(2.0\text{ A}\) throughout the wire.

Assume each section experiences a uniform field of the stated magnitude, and ignore magnetic forces at the bends and fringe-field effects.

a. Determine the direction and magnitude of the magnetic force on each of the three sections.

b. Calculate the resultant magnetic force on the entire U-shaped conductor. State its direction.

c. Now consider a separate experiment in which the field is made uniform at \(0.50\text{ T}\) across all three sections. Would straightening the wire into a direct vertical conductor between A and B change the resultant magnetic force? Explain why calculating the force using the total length of the U-shaped wire would give an incorrect result.

Solution 5

The resultant force is approximately \(0.17\text{ N}\), directed \(45^\circ\) above the horizontal towards the left. The horizontal-section forces do not cancel in the stated arrangement because the magnetic field strengths are different.

This problem is more subtle than a straight-wire question. You cannot treat a bent conductor as though its entire length carries current in one direction. Each straight section must be considered separately.

Part a: Force on each section

Start with the lower horizontal section.

The current flows to the right, and the magnetic field points into the page.

Using Fleming’s left-hand rule, the force acts upwards.

The magnitude is:

\[ \begin{aligned} F_1&=BIL\sin90^\circ\\ &=(0.50)(2.0)(0.30)\\ &=0.30\text{ N} \end{aligned} \]

Therefore, the lower section experiences a force of \(0.30\text{ N}\) upwards.

Now consider the vertical section.

The current flows upwards, and the field still points into the page.

Applying Fleming’s left-hand rule, the force acts towards the left.

Its magnitude is:

\[ \begin{aligned} F_2&=(0.50)(2.0)(0.12)\\ &=0.12\text{ N} \end{aligned} \]

The vertical section experiences a force of \(0.12\text{ N}\) to the left.

Finally, consider the upper horizontal section.

The current flows to the left, while the magnetic field points into the page.

The force therefore acts downwards.

Its magnitude is:

\[ \begin{aligned} F_3&=(0.30)(2.0)(0.30)\\ &=0.18\text{ N} \end{aligned} \]

The upper section experiences a force of \(0.18\text{ N}\) downwards.

The three results are:

Part b: Resultant force

Now combine the forces as vectors.

First, consider the vertical forces.

The lower section pushes upwards with \(0.30\text{ N}\), while the upper section pushes downwards with \(0.18\text{ N}\).

The resultant vertical force is:

\[ \begin{aligned} F_y&=0.30-0.18\\ &=0.12\text{ N upwards} \end{aligned} \]

There is also a horizontal force of \(0.12\text{ N}\) towards the left from the vertical section.

Therefore, the resultant has two perpendicular components:

• \(0.12\text{ N}\) upwards.
• \(0.12\text{ N}\) to the left.

Use Pythagoras to find the resultant magnitude:

\[ \begin{aligned} F_{\text{net}}&=\sqrt{F_x^2+F_y^2}\\ &=\sqrt{(0.12)^2+(0.12)^2}\\ &=0.1697\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F_{\text{net}}\approx0.17\text{ N}} \]

To find its direction:

\[ \tan\alpha=\frac{0.12}{0.12} \]

\[ \alpha=45^\circ \]

The resultant force acts \(45^\circ\) above the horizontal towards the left.

Part c: What changes in a uniform magnetic field?

If the entire U-shaped conductor experiences a uniform \(0.50\text{ T}\) field, the upper horizontal section now experiences a force of:

\[ \begin{aligned} F_3&=(0.50)(2.0)(0.30)\\ &=0.30\text{ N downwards} \end{aligned} \]

This exactly balances the \(0.30\text{ N}\) upward force on the lower horizontal section.

The two horizontal forces cancel.

Only the \(0.12\text{ N}\) leftward force on the vertical section remains.

Therefore:

\[ \boxed{F_{\text{net}}=0.12\text{ N leftwards}} \]

Now imagine straightening the wire into a direct vertical conductor connecting A and B.

The straight conductor has an active length of \(0.12\text{ m}\), with current flowing upwards in the same uniform magnetic field.

Its force is:

\[ \begin{aligned} F&=BIL\\ &=(0.50)(2.0)(0.12)\\ &=0.12\text{ N leftwards} \end{aligned} \]

The resultant force is unchanged.

This is the important conceptual result: in a uniform magnetic field, the forces on the additional horizontal sections cancel because they have equal lengths, equal currents, and equal magnetic field strengths, but opposite current directions.

The tempting mistake is to add all three wire lengths and calculate:

\[ F=B I(0.30+0.12+0.30) \]

That would incorrectly treat the entire conductor as though every section experienced a force in the same direction.

They don’t.

A bent conductor can experience several magnetic forces pointing in different directions. You must calculate each section’s force and combine them as vectors.

The result also shows why the uniform-field condition matters. When the field strengths differ, equal lengths carrying opposite currents do not necessarily experience equal forces.

06Where to take this next

You now have a method for finding the force direction on a straight current-carrying conductor:

1. Identify the active wire segment.
2. Establish the magnetic field direction.
3. Establish the conventional current direction.
4. Apply Fleming’s left-hand rule.
5. Calculate the magnitude separately using \(F=BIL\sin\theta\), if required.

The next useful application is understanding what happens when one current-carrying wire creates the magnetic field that acts on another.

That is the basis of Force Between Parallel Current-Carrying Wires: HSC Physics.

For more practice with the magnitude equation, active conductor length, and angled conductors, see Force on a Current-Carrying Conductor: HSC Physics Guide.