How to Find the North Pole of a Solenoid Using Current

Two solenoids can be connected to identical batteries and still have opposite north poles. The difference is the direction in which their wire is wound.

To find which end of a solenoid is north, use the right-hand grip rule: curl the fingers of your right hand in the direction of conventional current around the turns. Your thumb points along the magnetic field inside the solenoid and towards its north pole.

There is also a quick end-on test:

• Anticlockwise conventional current, viewed directly at an end: that end is north.
• Clockwise conventional current, viewed directly at an end: that end is south.

The important detail is your viewpoint. Let’s work through how to use this rule reliably, including what changes when you reverse the current or look at the coil from the opposite end.

01The right-hand grip rule

A solenoid is a coil of wire wound into a series of loops. When conventional current flows through it, each loop produces a magnetic field.

The fields from the individual turns combine to produce a stronger magnetic field along the solenoid’s central axis.

To find the direction of this field, use your right hand:

1. Point your fingers around the coil in the direction of conventional current.
2. Curl your fingers as though you are gripping the solenoid.
3. Your thumb points along the magnetic field inside the coil.
4. The end towards which your thumb points is the north pole.

Notice that your thumb points along the axis of the solenoid, not along the wire itself.

The quicker end-on method

You don’t always need to draw a hand. If you are looking directly at one end of a solenoid, you can determine its pole simply by checking whether the conventional current travels clockwise or anticlockwise around the visible turns.

This is exactly the same right-hand grip rule, just viewed from the end rather than from the side.

Why does this work?

Think of each turn of wire as a tiny magnetic loop. Each loop contributes a magnetic field along the solenoid’s axis.

When the turns are wound consistently, their fields reinforce one another inside the coil. The right-hand grip rule tells you which way this combined field points.

A solenoid also has the same basic field pattern as a bar magnet:

• Inside the solenoid, the magnetic field travels from south to north.
• Outside the solenoid, the magnetic field travels from north to south.

This means the north pole is the end where the internal magnetic field emerges.

For a more detailed explanation of how the individual turns combine to produce the magnetic field, see Magnetic Field Inside a Solenoid: HSC Physics Explained.

02Be careful about your viewing direction

This is one of the easiest places to lose marks.

Imagine a solenoid with its north pole on the left and its south pole on the right.

If you stand at the left end and look directly into the coil, you see anticlockwise conventional current.

Now walk around to the right end.

The current appears clockwise.

Has the current reversed? No. You are simply looking at the same coil from the opposite direction.

The physical north and south poles have not changed.

This is why you should always establish which end you are viewing before applying the clockwise-anticlockwise rule.

A useful distinction:

• Reversing the current physically swaps the north and south poles.
• Changing your viewing position reverses the apparent clockwise or anticlockwise direction, but does not change the poles.

03Always use conventional current

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

Conventional current is defined as the direction positive charge would move. In an ordinary external circuit, it travels from the positive terminal of the power supply towards the negative terminal.

Electrons in a metal conductor drift in the opposite direction.

This matters because some Physics questions draw electron-flow arrows rather than conventional-current arrows.

If a question shows electrons travelling anticlockwise when viewed from one end, conventional current travels clockwise from that same viewpoint.

You must reverse the direction before applying the right-hand grip rule.

For a more detailed explanation of this distinction, see Electron Flow vs Conventional Current in HSC Magnetic Questions.

What happens when the current reverses?

Suppose a solenoid initially has anticlockwise conventional current when viewed from its left end.

The left end is north.

Now reverse the battery connections. Conventional current travels in the opposite direction around every turn.

The current now appears clockwise from the left end, so the left end becomes south.

The two poles have swapped.

This is a physical change to the magnetic field, not simply a change in how the coil looks from a different viewpoint.

04Don’t confuse field direction rules with force rules

There are two different questions you might be asked in magnetic field problems:

1. Which direction does a current produce a magnetic field?
2. Which direction does a magnetic field exert a force on a current-carrying conductor?

These require different rules.

Fleming’s left-hand rule uses three perpendicular directions:

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

It predicts the force acting on a current-carrying conductor in an external magnetic field.

It does not directly identify the north pole of a solenoid.

A common mistake is to see a current direction, reach for Fleming’s left hand, and try to use the thumb to identify the north pole. That is solving a different problem.

For solenoid polarity, use the right-hand grip rule.

05Checking the poles with a compass

You can also identify the poles experimentally using a compass.

Consider a solenoid connected to a low-voltage DC supply through a switch. Place a small compass on the solenoid’s central axis, just outside one end.

The compass’s north-seeking end aligns with the local magnetic field.

This gives you a useful prediction:

• Near the north end, the compass’s north-seeking end points away from the solenoid.
• Near the south end, it points towards the solenoid.

What should the experiment show?

If the left end is north, magnetic field lines emerge from that end. A compass placed just outside it should therefore point away from the coil.

If the left end is south, the external magnetic field points towards the coil, so the compass’s north-seeking end should point towards it.

There is one experimental complication.

A compass also responds to Earth’s magnetic field. A single reading therefore measures the combined effect of the solenoid’s field and Earth’s field, not necessarily the solenoid’s field alone.

A better investigation keeps the compass in the same position and compares three conditions:

1. Current switched off.
2. Current switched on.
3. Current switched on with the supply polarity reversed.

Reversing the current reverses the solenoid’s magnetic field, while Earth’s magnetic field remains approximately unchanged.

If the compass deflection changes accordingly, this provides evidence that the solenoid is producing the field.

The compass is an independent check of your right-hand grip rule, but the current direction and winding geometry remain the information you need to predict the poles.

06Worked examples

07What if a solenoid has two competing windings?

The same reasoning can be extended to a more unusual arrangement.

Imagine two separate windings wrapped around the same cylindrical former. Both produce magnetic fields along the same axis, but their winding directions are opposite.

One winding produces a field towards the left, while the other produces a field towards the right.

You cannot simply choose one winding and ignore the other. The fields combine through superposition.

For a long air-core solenoid, the magnetic flux density is approximately

\[
B=\mu_0 nI
\]

where:

• \(B\) is magnetic flux density, measured in tesla (T).
• \(\mu_0\) is the permeability of free space, \(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\).
• \(n\) is the number of turns per metre, measured in \(\mathrm{m^{-1}}\).
• \(I\) is conventional current, measured in amperes (A).

Since \(n=N/L\), where \(N\) is the number of turns and \(L\) is the solenoid’s length, the field magnitude depends on the product of turns and current.

For two opposing windings of equal length, the stronger winding determines the direction of the resultant internal field.

If their magnetic fields are equal and opposite, they cancel in the ideal long-solenoid model.

This is an important extension of the basic rule: when multiple currents contribute to a magnetic field, determine the direction of each contribution before combining them.

08Questions and solutions

Question 1

A solenoid has its axis running north-south. You look directly at its southern end and observe conventional current travelling clockwise around the turns.

Identify:

a) The magnetic pole at the southern end.

b) The magnetic pole at the northern end.

c) The direction of the magnetic field inside the solenoid.

Solution 1

The southern end is south, the northern end is north, and the internal magnetic field points towards the north.

The reasoning follows directly from the right-hand grip rule.

Step 1: Apply the end-on rule.

Clockwise conventional current, viewed directly from an end, identifies that end as south.

Therefore, the southern end of the solenoid is south.

Step 2: Identify the opposite pole.

The other end must be north.

Step 3: Determine the internal field direction.

Magnetic field lines travel from south to north inside a solenoid.

Therefore, the internal magnetic field points from the southern end towards the northern end.

Since the solenoid is aligned north-south, the field points geographically north.

Question 2

A solenoid has two labelled ends, A and B.

Initially, end A is on the western side and end B is on the eastern side. An observer standing to the west looks directly at end A and sees conventional current travelling anticlockwise.

The solenoid is then rotated through \(180^\circ\) about a vertical axis, swapping the positions of A and B.

The solenoid and its electrical connections rotate together, so the direction of conventional current relative to the winding does not change.

A student standing at the same western position now looks at end B and sees clockwise current.

The student claims that the north and south poles must have swapped.

Is the student correct? Identify the north pole after rotation and explain why the observed current direction has changed.

Solution 2

The student is incorrect. End A remains north after the rotation, even though it has moved to the eastern side.

The key is to distinguish a change in physical current direction from a change in viewing direction.

Step 1: Identify the original north pole.

Initially, the observer looks from the west and sees anticlockwise conventional current at end A.

Therefore, A is north.

End B is south.

Step 2: Track the physical solenoid during rotation.

The solenoid rotates as a complete object, and its current direction relative to the winding remains unchanged.

Therefore, the magnetic poles stay attached to their original physical ends.

End A remains north, and end B remains south.

Step 3: Identify the new positions.

After rotation:

• A is now on the eastern side.
• B is now on the western side.

Therefore, A is still north, but it is now located in the east.

Step 4: Explain the clockwise observation.

The observer remains in the west but is now looking at the opposite physical end.

The same current therefore appears clockwise rather than anticlockwise.

This is a change in viewpoint, not a reversal of current.

The tempting mistake is to apply the end-on rule without tracking which physical end is being viewed. Always identify the end first, then interpret the current direction from that viewpoint.

Question 3

A solenoid has its axis running east-west.

An end-on view from the eastern end shows electrons drifting anticlockwise around the turns.

A compass is positioned just outside the eastern end, on the solenoid’s central axis. When the current is switched on, the north-seeking end of the compass points west, towards the solenoid.

Assume the solenoid’s magnetic field dominates Earth’s field at the compass.

Determine:

a) The direction of conventional current when viewed from the east.

b) Which end of the solenoid is north.

c) Whether the compass observation supports your answer.

Explain why a student who uses the electron-flow arrows directly would reach the wrong conclusion.

Solution 3

The eastern end is south, the western end is north, and the compass observation is consistent with this result.

The crucial step is reversing the electron-flow direction before applying the right-hand grip rule.

Step 1: Convert electron flow into conventional current.

Electrons drift in the opposite direction to conventional current.

The electrons travel anticlockwise, so conventional current travels clockwise when viewed from the east.

Step 2: Identify the eastern pole.

Clockwise conventional current means the viewed end is south.

Therefore, the eastern end is south.

Step 3: Identify the western pole.

The western end must be north.

Step 4: Check the compass observation.

The external magnetic field enters the south pole.

Since the compass is just outside the eastern end, the magnetic field points towards the solenoid, towards the west.

The north-seeking end of the compass therefore points west.

This agrees with the given observation.

Why the incorrect approach fails:

If you treat electron flow as conventional current, you would incorrectly apply the anticlockwise rule and label the eastern end north.

That prediction would also contradict the compass observation.

The current convention is not a minor detail. Reversing it reverses the predicted magnetic poles.

Question 4

A student places a straight test wire inside a solenoid, parallel to the solenoid’s central axis.

The wire has a length of \(0.10\ \mathrm{m}\) and carries a conventional current of \(1.5\ \mathrm{A}\).

The magnetic flux density inside the solenoid is \(0.020\ \mathrm{T}\), directed along the solenoid’s axis.

The student measures no magnetic force on the test wire and concludes that the solenoid must have no magnetic field.

a) Calculate the magnetic force on the test wire.

b) Explain why the student’s conclusion is incorrect.

c) If conventional current viewed directly from end A is anticlockwise, identify the pole at end A.

Solution 4

The force on the test wire is zero, but the solenoid still has a magnetic field. End A is north.

This question tests the difference between magnetic field direction and magnetic force.

Step 1: Write the magnetic force equation.

For a straight current-carrying conductor in a magnetic field,

\[
F=BIL\sin\theta
\]

where:

• \(F\) is magnetic force in newtons (N).
• \(B\) is magnetic flux density in tesla (T).
• \(I\) is conventional current in amperes (A).
• \(L\) is conductor length in metres (m).
• \(\theta\) is the angle between the current direction and magnetic field.

Step 2: Identify the angle.

The test wire is parallel to the solenoid’s axis.

The internal magnetic field is also parallel to this axis.

Therefore, the angle between the wire’s current and the magnetic field is either \(0^\circ\) or \(180^\circ\).

In both cases, the sine of the angle is zero.

Step 3: Substitute the values.

\[
\begin{aligned}
F&=BIL\sin\theta\\
&=(0.020)(1.5)(0.10)\sin 0^\circ\\
&=0\ \mathrm{N}
\end{aligned}
\]

The magnetic force is zero.

Step 4: Explain the misconception.

A zero magnetic force does not mean a zero magnetic field.

The force depends on the angle between the current and the magnetic field. When the wire is parallel or antiparallel to the field, there is no sideways magnetic force.

The student has confused the effect of a magnetic field on a conductor with the existence of the magnetic field itself.

Step 5: Identify the north pole.

The conventional current viewed from end A is anticlockwise.

Therefore, end A is north.

This final step uses the right-hand grip rule, not Fleming’s left-hand rule.

Question 5

A long cylindrical former contains two separate, insulated solenoid windings, X and Y.

Both windings:

• Extend over the same length of \(0.40\ \mathrm{m}\).
• Share the same air core.
• Carry the same conventional current of \(0.60\ \mathrm{A}\) in series.

Winding X has 140 turns, while winding Y has 100 turns.

When viewed directly from the left end:

• Conventional current in winding X travels anticlockwise.
• Conventional current in winding Y travels clockwise.

Assume the long-solenoid approximation applies, and neglect end effects.

a) Determine the north and south poles of the combined arrangement.

b) Calculate the magnitude of the resultant magnetic flux density inside the solenoid.

c) The electrical connections to winding Y are then reversed. The current-controlled supply maintains the current at \(0.60\ \mathrm{A}\). Conventional current in winding Y now also travels anticlockwise when viewed from the left.

Determine the new north pole and calculate the new magnetic flux density.

d) Explain why reversing the connections to winding Y does not reverse the north pole in this situation.

Solution 5

The left end is north initially, and it remains north after winding Y is reconnected. The resultant field becomes six times stronger after reconnection.

The important idea is that the two windings produce magnetic fields in opposite directions initially. The resultant field depends on the difference between their contributions, not simply on the direction of either winding individually.

Step 1: Determine the field direction produced by each winding.

Winding X carries anticlockwise conventional current when viewed from the left.

Therefore, its left end is north, and its internal field points towards the left.

Winding Y carries clockwise conventional current when viewed from the left.

Therefore, its left end is south, and its internal field points towards the right.

The two magnetic fields oppose each other.

Step 2: Determine the direction of the resultant field.

Both windings have the same current and length, so their field magnitudes are proportional to their numbers of turns.

Winding X has 140 turns, while winding Y has 100 turns.

Winding X produces the stronger field.

Therefore, the resultant field points towards the left.

The left end is north, and the right end is south.

Step 3: Calculate the initial magnetic flux density.

For two opposing long-solenoid windings carrying the same current,

\[
B_{\text{net}}=
\frac{\mu_0 (N_X-N_Y)I}{L}
\]

Substitute the values:

\[
\begin{aligned}
B_{\text{net}}
&=\frac{(4\pi\times10^{-7})(140-100)(0.60)}
{0.40}\\
&=7.5\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

The initial magnetic flux density is approximately

\[
\boxed{7.5\times10^{-5}\ \mathrm{T}}
\]

Step 4: Determine what happens when winding Y is reconnected.

Reversing the connections to winding Y reverses its conventional current direction.

Its current now appears anticlockwise when viewed from the left.

Therefore, both windings now produce magnetic fields towards the left.

Their fields reinforce one another instead of opposing each other.

The left end remains north.

Step 5: Calculate the new magnetic flux density.

Since the fields now reinforce one another, add the number of turns:

\[
N_{\text{total}}=140+100=240
\]

The new field is

\[
\begin{aligned}
B_{\text{new}}
&=\frac{\mu_0(N_X+N_Y)I}{L}\\
&=\frac{(4\pi\times10^{-7})(240)(0.60)}
{0.40}\\
&=4.5\times10^{-4}\ \mathrm{T}
\end{aligned}
\]

Therefore,

\[
\boxed{B_{\text{new}}=4.5\times10^{-4}\ \mathrm{T}}
\]

Step 6: Explain why the north pole has not changed.

A tempting prediction is that reversing the current in winding Y must reverse the north pole of the entire solenoid.

That would be true if Y were the only winding.

However, winding X has not changed its current direction. It still produces a magnetic field towards the left.

Initially, X’s field was stronger than Y’s, so the resultant field pointed left.

After reconnecting Y, both fields point left, making the resultant field stronger but leaving its direction unchanged.

The ratio of the new field to the original field is

\[
\frac{B_{\text{new}}}{B_{\text{net}}}
=\frac{240}{40}=6
\]

The magnetic field is now six times stronger, but the magnetic poles remain in the same positions.

This is the key distinction: reversing one contribution to a magnetic field does not necessarily reverse the direction of the total field.

09Where to go next

Once you can identify the north and south poles of a solenoid, you can predict the magnetic field direction inside and outside it.

That direction becomes especially important when the current changes. A changing magnetic field can produce an induced voltage in another nearby coil, which is the starting point for understanding mutual induction between solenoids.