Why Reversing Current Reverses a Solenoid’s Poles

Swap the two leads of a DC power supply connected to a solenoid. The coil stays in exactly the same position, and none of its turns change shape, yet its north and south poles exchange places.

Why?

Before reading on, predict what happens to the magnetic field if the current reverses but its magnitude stays the same. Does the field disappear, become weaker, or point in the opposite direction?

It points in the opposite direction, but its magnitude stays the same in the ideal air-core model. Reversing the current reverses the magnetic field produced by every turn of wire. Since the fields from all the turns combine, the magnetic field inside the entire solenoid reverses. Its north and south poles therefore swap.

The key is understanding why the magnetic field from an individual loop depends on the direction of its current.

01Start with one loop of wire

Imagine a single circular loop of wire carrying conventional current. Look directly at one face of the loop.

If the current travels anticlockwise around the loop, the magnetic field through its centre points towards you. That face behaves as a north pole.

If the current travels clockwise, the magnetic field points away from you. That face behaves as a south pole.

This follows from the right-hand grip rule:

1. Curl the fingers of your right hand in the direction of conventional current.
2. Your thumb points through the centre of the loop.
3. Your thumb indicates the magnetic field direction and points towards the loop’s north face.

Notice something important: the current travels around the wire, but the magnetic field through the loop points along an entirely different direction.

The right-hand grip rule connects these two directions.

What happens when the current reverses?

Suppose the loop initially carries anticlockwise conventional current.

Its magnetic field points towards you, making the visible face north.

Now reverse the current. It travels clockwise instead.

Your right-hand thumb now points away from you. The magnetic field has reversed, and the face that was north is now south.

Nothing about the loop’s physical arrangement changed. Only the direction of current changed.

02Why this reverses a solenoid’s poles

A solenoid is a coil containing many closely spaced loops of wire.

Each turn produces a magnetic field. Inside a long solenoid, the fields from neighbouring turns reinforce one another, producing a strong magnetic field along the solenoid’s central axis.

You can think of each turn as a tiny magnetic dipole. Its north and south faces are determined by the direction of current around that turn.

This analogy is useful because it helps you picture how the turns cooperate. However, these are not permanent magnets: for an ideal air-core solenoid, the magnetic field is produced by the electric current.

Now consider what happens when you reverse the current through the entire solenoid.

• Every turn carries current in the opposite direction.
• The magnetic field produced by every turn reverses.
• The fields still reinforce one another inside the solenoid, but now point in the opposite direction.
• The combined magnetic field along the solenoid’s axis reverses.
• The north and south poles exchange positions.

This is the physical reason reversing current reverses a solenoid’s poles.

It is not that the battery somehow assigns a north pole to one end. The winding arrangement and the direction of conventional current together determine the magnetic field.

For a fuller explanation of how the individual turns combine to produce the field inside a solenoid, see Magnetic Field Inside a Solenoid: HSC Physics Explained.

03Does reversing the current change the field strength?

Not necessarily.

This is an important distinction: magnetic field direction and magnetic field magnitude are different properties.

For a long, ideal air-core solenoid, the magnetic field strength near its centre is:

\[ B=\frac{\mu_0 NI}{L} \]

where:

• \(B\) is magnetic flux density, measured in tesla (T).
• \(\mu_0\) is the permeability of free space, approximately \(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\).
• \(N\) is the total number of turns.
• \(I\) is the magnitude of conventional current, measured in amperes (A).
• \(L\) is the length of the solenoid, measured in metres (m).

For a fixed solenoid:

\[ B\propto I \]

This means that the magnetic field magnitude depends on the current magnitude.

If you reverse the current without changing its magnitude, the field strength remains unchanged, but its direction reverses.

For example:

The field has the same magnitude in both cases. Only its direction changes.

If the current magnitude also changes, the field strength changes accordingly.

04Be careful: conventional current is not electron flow

There is another detail that can easily cause mistakes in HSC Physics questions.

The right-hand grip rule uses conventional current, not electron flow.

Conventional current is defined as the direction in which positive charge would move. In an ordinary metal wire, electrons actually drift in the opposite direction.

For example, suppose a diagram shows electrons travelling clockwise when you look directly at one end of a solenoid.

Conventional current travels anticlockwise from that same viewpoint.

You must reverse the electron-flow arrow before applying the right-hand grip rule.

Do not apply the right-hand grip rule directly to an electron-flow arrow.

05Changing your viewpoint is not the same as reversing current

Here is another situation that can make a correct answer look wrong.

Imagine a solenoid with a north pole at its western end and a south pole at its eastern end.

An observer looks directly at the western end and sees anticlockwise conventional current.

Now that observer walks around to the eastern end.

The same current now appears clockwise.

Has the current reversed?

No. The observer has simply changed viewing direction.

This gives us an important distinction:

• Reversing the supply connections physically reverses conventional current and swaps the poles.
• Looking at the same solenoid from the opposite end reverses the apparent clockwise or anticlockwise direction, but does not change the poles.

A useful habit is to write down which end you are looking at before applying the clockwise/anticlockwise shortcut.

06Don’t confuse magnetic field rules with magnetic force rules

There are two different questions in magnetism that require different hand rules.

Question A: Which direction does a current produce a magnetic field?

Use the right-hand grip rule.

For a solenoid, curl your fingers in the direction of conventional current. Your thumb points along the internal magnetic field and towards the north pole.

Question B: Which direction does a magnetic field exert a force on a current-carrying conductor?

Use Fleming’s left-hand rule.

• First finger: magnetic field.
• Second finger: conventional current.
• Thumb: force.

These rules answer different physical questions.

Fleming’s left-hand rule does not identify the north pole of a solenoid. It predicts the force on a current-carrying conductor in an external magnetic field.

For a solenoid’s poles, use the right-hand grip rule.

07Testing the reversal with a compass

How could you check experimentally that reversing the current really reverses the solenoid’s magnetic field?

A compass provides a straightforward test.

Setting up the experiment

Use:

• A long air-core solenoid.
• A low-voltage DC power supply.
• A switch.
• An ammeter connected in series.
• A small compass.
• An insulating or non-magnetic support.

Place the solenoid horizontally, with its axis running east-west.

Position the compass just outside the western end, directly along the solenoid’s central axis. Keep the compass fixed throughout the experiment.

Ideally, arrange the solenoid’s axis perpendicular to the compass’s direction when the current is switched off. This makes the solenoid’s magnetic field easier to distinguish from Earth’s horizontal magnetic field.

What should the experiment show?

First, leave the switch open.

The compass aligns with Earth’s local magnetic field. Record this as your baseline direction.

Next, close the switch.

The current produces a magnetic field around and inside the solenoid. The compass now responds to the combined effect of Earth’s field and the solenoid’s field.

Finally, reverse the supply connections while keeping the current magnitude approximately the same.

The solenoid’s magnetic field reverses. The compass should now deflect to the opposite side of its original baseline.

Why isn’t one compass reading enough?

A compass does not respond exclusively to the solenoid.

It responds to the resultant magnetic field at its location. This includes Earth’s magnetic field and any other nearby magnetic fields.

A single reading with the current switched on therefore cannot isolate the solenoid’s contribution.

A better investigation compares three conditions:

1. Current switched off.
2. Current switched on in the original direction.
3. Current switched on in the reversed direction.

Keep the compass in exactly the same position and use the ammeter to check that the two current magnitudes are equal.

Illustrative experimental data

The following table contains hypothetical readings to demonstrate the expected pattern. These are illustrative data, not measurements from an actual experiment.

The important observation is that reversing the current changes the direction of the compass deflection.

The angles are close, but not identical. Small differences could result from compass friction, reading uncertainty, or slight misalignment.

This supports the prediction that the solenoid’s magnetic field reverses when its current reverses.

One subtlety: the compass does not necessarily turn through 180°. Earth’s magnetic field remains present, so the compass follows the resultant field rather than the solenoid’s field alone. We will explore this distinction in Question 5.

08Worked examples using experimental observations

09Important limitations of the ideal model

The explanation so far assumes an air-core solenoid.

For a real solenoid, two limitations are worth remembering.

The magnetic field is not perfectly uniform everywhere.

The equation \(B=\mu_0 NI/L\) describes the field near the centre of a sufficiently long solenoid. Near the ends, the field spreads out and becomes less uniform.

However, the right-hand grip rule still determines the overall field direction.

A ferromagnetic core can retain magnetisation.

If a solenoid contains an iron core, the core may retain some magnetisation after the current is switched off. Its magnetic response can also depend on its previous magnetic history.

Therefore, the net external magnetic field may not reverse as simply as the ideal air-core model predicts. A sufficiently strong reversed current can reverse the core’s magnetisation, but the response is not necessarily instantaneous.

For standard ideal-solenoid questions, focus on the direction of conventional current and the right-hand grip rule unless the question specifies additional magnetic effects.

10Questions and solutions

Question 1

An observer looks directly at the western end of an air-core solenoid. Conventional current travels anticlockwise around the visible turns.

Identify:

(a) The magnetic pole at the western end.

(b) The direction of the magnetic field inside the solenoid.

Solution 1

The western end is north, and the internal magnetic field points west, towards the observer.

Using the right-hand grip rule, curl your fingers anticlockwise in the direction of conventional current.

Your thumb points towards you, indicating the magnetic field direction through the centre of the solenoid.

The end towards which the internal field points is the north pole.

Therefore:

• Western end: north.
• Eastern end: south.
• Internal magnetic field: east to west.

Question 2

An ideal air-core solenoid has 900 turns and a length of 0.45 m.

Initially, it carries a conventional current of 0.25 A, producing an internal magnetic field towards the east.

The power supply connections are then reversed, and the current magnitude increases to 0.40 A.

Calculate:

(a) The initial magnetic field strength.

(b) The final magnetic field strength.

(c) The new positions of the north and south poles.

Assume the long-solenoid approximation applies at the centre.

Solution 2

The initial field is 0.628 mT east, and the final field is approximately 1.01 mT west. The poles therefore exchange positions.

(a) Initial magnetic field strength

Use:

\[ B=\frac{\mu_0 NI}{L} \]

Substitute:

\[ B=\frac{(4\pi\times10^{-7})(900)(0.25)} {0.45} \]

\[ B=6.28\times10^{-4}\ \mathrm{T} \]

Therefore:

\[ \boxed{B=0.628\ \mathrm{mT}} \]

(b) Final magnetic field strength

The current has reversed and its magnitude is now 0.40 A.

\[ B=\frac{(4\pi\times10^{-7})(900)(0.40)} {0.45} \]

\[ B=1.005\times10^{-3}\ \mathrm{T} \]

Therefore:

\[ \boxed{B=1.01\ \mathrm{mT}} \]

The final field points west because the conventional current has reversed.

Notice that the final field is stronger. This is because the current magnitude increased from 0.25 A to 0.40 A.

(c) New pole positions

Initially, the internal field points east, so:

• Eastern end: north.
• Western end: south.

After reversing the current, the internal field points west.

Therefore:

• Western end: north.
• Eastern end: south.

The poles swap because the direction of the internal magnetic field reverses.

Question 3

An air-core solenoid carries a steady conventional current.

An observer looking directly at its western end sees anticlockwise current. A second observer looks directly at its eastern end and describes the current as clockwise.

The power supply connections are then reversed without moving the solenoid.

Determine:

(a) The north and south poles before reversing the supply.

(b) The apparent direction of conventional current from each end after reversing the supply.

(c) The new north and south poles.

Explain why the two original observations do not indicate that the current has reversed.

Solution 3

Initially, the western end is north and the eastern end is south. After reversing the supply, the western end becomes south and the eastern end becomes north.

(a) Initial poles

The western observer sees anticlockwise conventional current.

Therefore, the western end is north.

The eastern end must be south because the solenoid has opposite poles at its two ends.

(b) Current direction after reversing the supply

Before reversing the supply:

• Western observer: anticlockwise.
• Eastern observer: clockwise.

After reversing the supply, conventional current travels in the opposite direction around every turn.

Therefore:

• Western observer: clockwise.
• Eastern observer: anticlockwise.

(c) New poles

From the western end, clockwise conventional current indicates a south pole.

From the eastern end, anticlockwise conventional current indicates a north pole.

Therefore:

• Western end: south.
• Eastern end: north.

The important distinction is between changing the actual current and changing the viewing direction.

The original observers were looking at the same current from opposite ends. Their clockwise and anticlockwise descriptions were both correct.

Moving around the solenoid does not reverse its current or change its poles. Reversing the power supply does.

Question 4

An HSC Physics diagram shows electrons drifting clockwise when viewed directly from the western end of a solenoid.

A student applies the right-hand grip rule directly to the electron-flow arrow and concludes that the western end is south.

(a) Determine the correct pole at the western end and the direction of the internal magnetic field.

In a separate experiment, a straight conductor carries conventional current towards the east (to the right on the page). It is placed in a uniform external magnetic field directed into the page.

(b) Determine the direction of the force on the straight conductor.

(c) Explain why the two parts require different hand rules.

Take the top of the page as north.

Solution 4

The western end is north, and the internal magnetic field points west. The separate straight conductor experiences a force towards the north.

(a) Determine the solenoid’s pole

The diagram shows electron flow, not conventional current.

In a metal conductor, conventional current travels opposite to electron drift.

Therefore:

• Electron flow: clockwise.
• Conventional current: anticlockwise.

Now apply the right-hand grip rule using anticlockwise conventional current.

Your thumb points towards the observer.

Therefore, the western end is north, and the internal magnetic field points west.

The student’s mistake was applying the right-hand grip rule directly to electron flow.

(b) Determine the force on the straight conductor

Use Fleming’s left-hand rule.

The directions are:

• Magnetic field: into the page.
• Conventional current: east, or right.
• Force: north, or upwards on the page.

Therefore:

\[ \boxed{\text{Force is towards the north.}} \]

(c) Explain the difference between the rules

The right-hand grip rule determines the magnetic field produced by a current.

Fleming’s left-hand rule determines the force experienced by a current-carrying conductor in an existing magnetic field.

The first problem asks what magnetic field the solenoid creates. The second asks how an external magnetic field acts on a conductor.

They are different physical situations, so they require different rules.

Question 5

An air-core solenoid lies horizontally along an east-west axis.

A compass is fixed just outside the western end. The compass’s north-seeking end initially points north when the solenoid is switched off.

At the compass position:

• Earth’s horizontal magnetic field has a magnitude of 20 μT and points north.
• With the original current, the solenoid produces a magnetic field of 20 μT towards the east.
• After reversing the supply, the solenoid’s field becomes 20 μT towards the west.

Assume the two magnetic fields are perpendicular, the compass responds to their horizontal resultant, and there are no other significant magnetic fields.

Answer the following:

(a) What direction does the compass’s north-seeking end point with the original current?

(b) What direction does it point after reversing the current?

(c) Determine the magnetic pole at the western end in both situations.

(d) A student argues that because the solenoid’s poles swap, the compass must rotate through exactly 180°. Explain why this conclusion is incorrect.

Solution 5

The compass points 45° east of north before reversal and 45° west of north afterwards. The western pole changes from south to north. The compass rotates through 90°, not 180°.

This question tests an important distinction: the compass responds to the resultant magnetic field, not just the field produced by the solenoid.

(a) Original compass direction

There are two perpendicular magnetic fields:

• Earth’s field: 20 μT north.
• Solenoid’s field: 20 μT east.

The resultant points diagonally towards the north-east.

Calculate the angle using trigonometry:

\[ \tan\theta=\frac{B_{\mathrm{east}}}{B_{\mathrm{north}}} \]

\[ \tan\theta=\frac{20}{20}=1 \]

Therefore:

\[ \theta=45^\circ \]

The compass’s north-seeking end points 45° east of north.

The resultant field magnitude is:

\[ B_{\mathrm{resultant}} =\sqrt{20^2+20^2} \]

\[ B_{\mathrm{resultant}}=28.3\ \mu\mathrm{T} \]

(b) Compass direction after reversing the current

The solenoid’s magnetic field now points west instead of east.

Earth’s magnetic field remains unchanged.

The two fields are still perpendicular, so the resultant points towards the north-west.

Again:

\[ \tan\theta=\frac{20}{20}=1 \]

\[ \theta=45^\circ \]

The compass now points 45° west of north.

Its resultant magnetic field magnitude is still:

\[ B_{\mathrm{resultant}}=28.3\ \mu\mathrm{T} \]

The compass has therefore rotated through:

\[ 45^\circ+45^\circ=90^\circ \]

(c) Determine the western pole

Initially, the solenoid’s field at the western end points east, towards the coil.

Outside a solenoid, magnetic field lines travel from north to south.

Therefore, the western end is south.

After reversing the current, the solenoid’s field at the compass points west, away from the coil.

The western end is now north.

Therefore:

The solenoid’s poles have genuinely swapped.

(d) Explain why the compass does not rotate through 180°

The student’s reasoning overlooks Earth’s magnetic field.

Reversing the solenoid’s current reverses the solenoid’s magnetic field, but it does not reverse Earth’s magnetic field.

The compass responds to the combination of both fields.

Initially, the resultant points north-east. After reversing the current, the resultant points north-west.

These directions are 90° apart, even though the solenoid’s own magnetic field has reversed through 180°.

If the solenoid’s field were much stronger than Earth’s field, the compass would turn much closer to 180°. But with equal perpendicular fields, the change is exactly 90°.

This is why a compass experiment must account for Earth’s magnetic field rather than treating the compass as if it responds only to the solenoid.

The central idea is simple: reversing conventional current reverses the magnetic field produced by every turn of a solenoid. The combined internal field reverses, so the north and south poles exchange positions.

Once you can predict that change confidently, the next useful step is understanding what happens when a solenoid’s magnetic field changes near a second coil. That leads directly to electromagnetic induction and Mutual Induction Between Solenoids for HSC Physics.