How to Find Induced Current Direction When a Magnet Enters a Coil

A bar magnet is pushed north pole first into a wire coil connected to a galvanometer. The needle deflects. Reverse the magnet, and the needle deflects in the opposite direction.

But which way does the current actually flow around the coil?

For a north pole entering the coil, the induced current is anticlockwise when viewed from the magnet side. For a south pole entering, it is clockwise from the magnet side.

The reliable way to work this out is to use Lenz’s law and the right-hand grip rule. The important part is getting the magnetic change and your viewing direction clear before predicting the current.

01Start by identifying the magnetic change

Imagine holding a coil in front of you, with a bar magnet moving towards it.

The magnet produces a magnetic field through the area enclosed by the coil. As the magnet approaches, the magnetic field through the coil becomes stronger, so the magnetic flux changes.

Your first job is to identify two things:

• Which magnetic pole is approaching the coil?
• Is the magnetic flux through the coil increasing or decreasing?

Don’t predict the current yet. First establish what is happening to the magnetic field.

What does the coil do when the magnet approaches?

The coil responds by producing its own magnetic field. According to Lenz’s law, this induced magnetic field opposes the change in magnetic flux that produced it.

This is the central idea:

• A north pole approaching the coil makes the near face of the coil act like a north pole.
• A south pole approaching the coil makes the near face act like a south pole.

Why? Like poles repel. The induced magnetic field creates a magnetic effect that opposes the approaching magnet.

Think of the coil as resisting the magnet’s approach, rather than simply copying the magnet’s magnetic field.

02Convert the coil’s magnetic pole into a current direction

Now use the right-hand grip rule.

Imagine gripping the coil with your right hand:

• Curl your fingers in the direction of conventional current.
• Your thumb points towards the coil’s north pole.

This gives you a direct relationship between the current and the magnetic pole of the coil.

If you are looking directly at one face of the coil:

• Anticlockwise current makes that face a north pole.
• Clockwise current makes that face a south pole.

The viewing direction is essential. Clockwise from one side is anticlockwise from the other.

The quick decision table

Assume you are looking at the coil from the side where the magnet approaches.

This table assumes a closed coil circuit and a changing magnetic flux. The current directions are conventional current, not electron flow.

The first two rows answer the title question. The other two are important because moving the magnet away reverses the direction of the induced current.

03How Faraday established electromagnetic induction

The relationship between a moving magnet and an induced current was established through Michael Faraday’s experiments in 1831.

His problem was straightforward: could a magnetic field produce an electric current without connecting the circuit to a battery?

The first apparatus: two separate coils

Faraday began with an iron ring. He wound two insulated wire coils around opposite sides of it.

• The primary coil was connected to a battery and a switch.
• The secondary coil was connected to a galvanometer.
• The two circuits were electrically separate.

When Faraday closed the switch, current began flowing through the primary coil. This produced a changing magnetic field in the iron ring, which passed through the secondary coil.

What should happen if changing magnetic flux induces current?

The galvanometer should briefly deflect.

That is what Faraday observed. More importantly, the needle returned to zero while the primary current continued flowing. When he opened the switch, the needle briefly deflected in the opposite direction.

The experiment revealed that a steady magnetic field was not enough. The magnetic flux linking the secondary coil had to change.

The revised experiment: move a permanent magnet

The iron-ring experiment established electromagnetic induction, but it did not directly answer how a moving permanent magnet would induce a current.

Faraday therefore tested a simpler arrangement.

He connected a stationary coil to a galvanometer, then moved a bar magnet along the coil’s central axis.

He observed three things:

1. Pushing the magnet into the coil produced a galvanometer deflection.
2. Holding the magnet stationary produced no continuing deflection.
3. Pulling the magnet out produced a deflection in the opposite direction.

This experiment gives us the observations needed to understand the direction of induced current.

Heinrich Lenz later formulated the direction rule in 1834, now known as Lenz’s law.

The important distinction is that Faraday demonstrated the induction effect, while Lenz’s law explains the direction of the induced current.

04Keep induced EMF and induced current separate

Before working through numerical examples, make sure you understand the difference between these two quantities.

Induced EMF is the potential difference produced by changing magnetic flux. It is measured in volts (V).

Induced current is the flow of electric charge through a conducting circuit. It is measured in amperes (A).

Faraday’s law gives the induced EMF:

\[ \mathcal{E}=-N\frac{\Delta\Phi}{\Delta t} \]

where:

• \(\mathcal{E}\) is induced EMF in volts (V).
• \(N\) is the number of turns in the coil.
• \(\Delta\Phi\) is the change in magnetic flux through one turn, measured in webers (Wb).
• \(\Delta t\) is the time taken for that change, measured in seconds (s).

The negative sign represents Lenz’s law: the induced EMF acts in a direction that produces a magnetic field opposing the change in flux.

For magnitude calculations, we can write:

\[ |\mathcal{E}|= N\frac{|\Delta\Phi|}{\Delta t} \]

If the circuit is closed and its total resistance is known, Ohm’s law gives the induced current:

\[ I=\frac{|\mathcal{E}|}{R} \]

Here, \(I\) is current in amperes (A), and \(R\) is the total circuit resistance in ohms (\(\Omega\)).

There is an important catch.

An open coil can have an induced EMF, but it cannot sustain a current because there is no complete conducting path.

This is why you should never automatically equate induced EMF with induced current.

In the following calculations, assume the coil has a complete circuit, the stated resistance is the total circuit resistance, and self-inductance is neglected. Use conventional current throughout.

05Worked examples

06Common mistakes when finding current direction

There are four mistakes worth watching for in HSC questions.

Mistake 1: Thinking Lenz’s law always opposes the magnetic field

Lenz’s law opposes the change in magnetic flux, not necessarily the original magnetic field.

Consider a north pole moving away from a coil.

The magnetic flux through the coil decreases. The coil responds by producing a magnetic field in the same direction as the original magnetic field. This helps maintain the decreasing flux.

That is why the coil’s near face becomes south and attracts the departing north pole.

Mistake 2: Forgetting to state your viewing direction

Clockwise and anticlockwise are not absolute directions.

A current that looks clockwise from the magnet side looks anticlockwise from the opposite side.

Always write something like:

“The induced conventional current is anticlockwise when viewed from the magnet side.”

That removes ambiguity from your answer.

Mistake 3: Assuming a magnetic field always produces a current

A stationary magnet can produce a strong magnetic field through a coil without inducing a current.

If the magnet and coil remain stationary relative to each other, and the magnetic field is constant, the magnetic flux does not change.

Therefore, the induced EMF is zero.

Remember: a magnetic field is not enough. You need changing magnetic flux.

Mistake 4: Assuming the current direction depends only on magnet motion

This is particularly important when a magnet passes completely through a coil.

The magnet can continue moving in the same direction while the magnetic flux first increases and then decreases.

When that happens, the induced current reverses, even though the magnet has not reversed its motion.

The next question explores this more carefully.

07A reliable method for exam questions

When you see a magnet moving near a coil, work through these steps in order.

1. Define your viewpoint. State which side of the coil you are looking from.
2. Identify the magnetic pole. Is the north or south pole facing the coil?
3. Identify the flux change. Is the magnetic flux through the loop increasing or decreasing?
4. Apply Lenz’s law. Determine the direction of the magnetic field produced by the coil.
5. Identify the coil’s near pole. Is it north or south?
6. Apply the right-hand grip rule. Convert the coil’s magnetic polarity into a current direction.
7. Check the circuit. Is there a complete conducting path for current to flow?

If a question asks for the current magnitude as well, calculate the induced EMF using Faraday’s law, then use Ohm’s law.

For questions involving a magnetic flux-time graph, the gradient is particularly important. See How to Calculate Induced EMF from a Piecewise Flux Graph for the calculation method.

08Questions and solutions

Question 1

A bar magnet is held with its south pole facing a circular coil. The magnet is pushed towards the coil along its central axis.

Determine:

a) The magnetic pole induced on the face nearest the magnet.

b) The direction of conventional current when viewed from the magnet side.

Solution 1

The near face of the coil becomes a south pole, and the induced conventional current is clockwise when viewed from the magnet side.

The south pole of the magnet is approaching, so the magnetic flux through the coil increases in the direction of the magnet’s field.

Lenz’s law requires the coil to oppose this increase. Its near face must therefore become a south pole, creating repulsion between the magnet and coil.

Using the right-hand grip rule, a clockwise current produces a south pole on the face viewed by the magnet.

This gives the required current direction.

Question 2

A 80-turn coil is connected to a resistor with a total resistance of \(3.0\,\Omega\).

A south pole facing the coil is pulled away. During the movement, the magnetic flux magnitude per turn decreases from \(0.50\,\text{mWb}\) to \(0.20\,\text{mWb}\) over \(0.15\,\text{s}\).

Determine:

a) The direction of the induced current, viewed from the magnet side.

b) The magnitude of the induced EMF.

c) The magnitude of the induced current.

Solution 2

The induced current is anticlockwise when viewed from the magnet side, with a magnitude of approximately \(0.053\,\text{A}\).

Step 1: Determine the induced magnetic field.

The south pole is moving away, so the magnetic flux magnitude decreases.

The coil tries to maintain the original magnetic flux. Its near face must become north to attract the departing south pole.

Step 2: Determine the current direction.

The right-hand grip rule tells us that anticlockwise conventional current produces a north pole on the near face.

Therefore, the current is anticlockwise.

Step 3: Calculate the change in flux.

\[ \begin{aligned} |\Delta\Phi| &=(0.50-0.20)\times10^{-3}\\ &=3.0\times10^{-4}\,\text{Wb} \end{aligned} \]

Step 4: Calculate the induced EMF.

\[ \begin{aligned} |\mathcal{E}| &=N\frac{|\Delta\Phi|}{\Delta t}\\ &=\frac{80(3.0\times10^{-4})}{0.15}\\ &=0.16\,\text{V} \end{aligned} \]

Step 5: Calculate the current.

\[ \begin{aligned} I&=\frac{|\mathcal{E}|}{R}\\ &=\frac{0.16}{3.0}\\ &=0.0533\,\text{A} \end{aligned} \]

Therefore, the induced current is approximately \(0.053\,\text{A}\), anticlockwise when viewed from the magnet side.

Notice that the south pole is moving away, not towards the coil. If you had based your answer on the pole alone, you would have predicted the wrong direction.

Question 3

A 120-turn coil has an open circuit and is connected to an ideal, high-resistance voltmeter.

A north pole approaches the coil, then stops. After the magnet has been stationary for several seconds, a switch is closed to complete the coil circuit.

Explain:

a) Whether an induced EMF exists while the magnet approaches.

b) Whether an induced current flows while the circuit is open.

c) What happens when the magnet stops.

d) Whether closing the switch after the magnet has stopped produces an induced current.

Solution 3

An induced EMF exists while the magnet approaches, but there is no significant induced current in the open circuit. Once the magnet stops, the induced EMF falls to zero, and closing the switch afterwards does not produce an induced current.

Step 1: Consider the approaching magnet.

As the north pole approaches, the magnetic flux through the coil changes.

Faraday’s law tells us that an induced EMF is produced.

The voltmeter can detect the potential difference across the open terminals.

Step 2: Consider the open circuit.

Although an induced EMF exists, the circuit is incomplete.

There is no sustained current in the ideal open-circuit model.

This is the distinction between producing an EMF and having charges flow around a complete circuit.

Step 3: Consider what happens when the magnet stops.

Once the magnet becomes stationary, the magnetic flux through the coil stops changing.

Therefore:

\[ \mathcal{E}=-N\frac{\Delta\Phi}{\Delta t}=0 \]

The voltmeter reading returns to zero.

Step 4: Close the switch.

Closing the switch completes the conducting path, but it does not restore the changing magnetic flux.

Because the magnet remains stationary, the induced EMF remains zero and no induced current flows.

If the switch had been closed while the magnet was still approaching, an anticlockwise current would have flowed when viewed from the magnet side.

The important lesson is that a complete circuit is necessary for current, but completing a circuit does not automatically create an EMF.

Question 4

A 50-turn coil has a total resistance of \(5.0\,\Omega\). It is viewed from the left.

Two magnetic fields pass through the coil in the same direction, from left to right:

• A north pole approaches from the left, increasing its contribution to the magnetic flux by \(0.30\,\text{mWb}\) per turn.
• A nearby electromagnet produces an additional flux of \(0.50\,\text{mWb}\) per turn. Its current is reduced, causing this contribution to decrease to \(0.10\,\text{mWb}\) per turn.

Both changes occur over \(0.10\,\text{s}\).

Determine:

a) The net change in magnetic flux per turn.

b) The direction of the induced current, viewed from the left.

c) The magnitude of the induced current.

Assume the two flux contributions add, the stated resistance is the total circuit resistance, and self-inductance is negligible.

Solution 4

The induced current is clockwise when viewed from the left, with a magnitude of \(0.010\,\text{A}\).

The important point is that Lenz’s law applies to the change in the total magnetic flux, not just the contribution from the approaching magnet.

Step 1: Calculate the change caused by the permanent magnet.

The north pole approaches, so its contribution to the rightward flux increases:

\[ \Delta\Phi_1=+0.30\,\text{mWb} \]

Step 2: Calculate the change caused by the electromagnet.

Its contribution decreases from \(0.50\,\text{mWb}\) to \(0.10\,\text{mWb}\).

Therefore:

\[ \Delta\Phi_2=0.10-0.50 =-0.40\,\text{mWb} \]

Step 3: Calculate the net change.

The two contributions must be added because they pass through the coil in the same direction.

\[ \begin{aligned} \Delta\Phi_{\text{net}} &=\Delta\Phi_1+\Delta\Phi_2\\ &=0.30-0.40\\ &=-0.10\,\text{mWb} \end{aligned} \]

The negative sign means the total rightward magnetic flux decreases.

Step 4: Determine the induced current direction.

The coil must oppose the decrease in rightward flux.

It therefore produces its own magnetic field towards the right.

Viewed from the left, this field points away from the observer. The near face of the coil must become south.

Using the right-hand grip rule, the conventional current must flow clockwise.

Step 5: Calculate the induced EMF.

Convert the net flux change into webers:

\[ |\Delta\Phi|=0.10\times10^{-3}\,\text{Wb} \]

Then:

\[ \begin{aligned} |\mathcal{E}| &=N\frac{|\Delta\Phi|}{\Delta t}\\ &=\frac{50(0.10\times10^{-3})}{0.10}\\ &=0.050\,\text{V} \end{aligned} \]

Step 6: Calculate the induced current.

\[ \begin{aligned} I&=\frac{|\mathcal{E}|}{R}\\ &=\frac{0.050}{5.0}\\ &=0.010\,\text{A} \end{aligned} \]

The final answer is a clockwise current of \(0.010\,\text{A}\), viewed from the left.

This is a useful example of why you must consider every magnetic field passing through a loop. Looking only at the approaching north pole would give the wrong current direction.

Question 5

Extension challenge

A 100-turn circular coil is connected to a \(4.0\,\Omega\) resistor. A bar magnet travels completely through the coil from left to right at constant speed.

The north pole is at the leading end of the magnet, and the south pole is at the trailing end.

The coil is viewed from the left. Define positive magnetic flux as flux directed from left to right, away from the observer.

The following illustrative measurements show the signed magnetic flux per turn:

Assume these are rounded readings from a smooth flux-time curve, with a maximum at \(t=0.20\,\text{s}\).

Answer the following:

a) Determine the direction of conventional current during the first \(0.20\,\text{s}\), viewed from the left.

b) Determine the direction of conventional current during the next \(0.20\,\text{s}\).

c) Calculate the average induced EMF and average current magnitude over each interval.

d) Determine the instantaneous induced EMF and current at \(t=0.20\,\text{s}\).

e) Explain why the current reverses even though the magnet never reverses its direction of motion.

Solution 5

The induced current is anticlockwise during the first half of the magnet’s journey and clockwise during the second half. Its average magnitude is \(0.075\,\text{A}\) over each interval, and it momentarily falls to zero when the magnetic flux reaches its maximum.

This question tests a particularly important distinction: the direction of the induced current depends on the rate of change of magnetic flux, not simply on the direction in which the magnet moves.

Step 1: Analyse the first half of the journey.

From \(0.00\,\text{s}\) to \(0.20\,\text{s}\), the magnetic flux increases from \(+0.20\,\text{mWb}\) to \(+0.80\,\text{mWb}\).

The positive flux is directed away from the observer.

Because this flux is increasing, Lenz’s law requires the coil to produce a magnetic field towards the observer.

The near face of the coil therefore becomes north.

Using the right-hand grip rule, the induced conventional current is anticlockwise when viewed from the left.

Step 2: Calculate the average induced EMF during the first half.

The change in flux per turn is:

\[ \begin{aligned} \Delta\Phi &=(0.80-0.20)\times10^{-3}\\ &=6.0\times10^{-4}\,\text{Wb} \end{aligned} \]

The time interval is \(0.20\,\text{s}\).

Using Faraday’s law:

\[ \begin{aligned} |\mathcal{E}_{\text{avg}}| &=N\frac{|\Delta\Phi|}{\Delta t}\\ &=\frac{100(6.0\times10^{-4})}{0.20}\\ &=0.30\,\text{V} \end{aligned} \]

The average current magnitude is:

\[ \begin{aligned} I_{\text{avg}} &=\frac{0.30}{4.0}\\ &=0.075\,\text{A} \end{aligned} \]

Step 3: Analyse the second half of the journey.

From \(0.20\,\text{s}\) to \(0.40\,\text{s}\), the magnetic flux decreases from \(+0.80\,\text{mWb}\) to \(+0.20\,\text{mWb}\).

The important detail is that the flux is still positive. Its direction has not reversed. Its magnitude is decreasing.

The coil now tries to maintain the original rightward magnetic flux.

Its induced magnetic field therefore points away from the observer, towards the right.

The near face becomes south, so the conventional current is clockwise when viewed from the left.

Step 4: Calculate the average induced EMF and current during the second half.

The magnitude of the flux change is again:

\[ |\Delta\Phi|=6.0\times10^{-4}\,\text{Wb} \]

Therefore:

\[ \begin{aligned} |\mathcal{E}_{\text{avg}}| &=\frac{100(6.0\times10^{-4})}{0.20}\\ &=0.30\,\text{V} \end{aligned} \]

The average current magnitude is:

\[ \begin{aligned} I_{\text{avg}} &=\frac{0.30}{4.0}\\ &=0.075\,\text{A} \end{aligned} \]

The current direction has reversed, even though the average current magnitude is unchanged.

Step 5: Consider the instant of maximum magnetic flux.

At \(t=0.20\,\text{s}\), the magnetic flux reaches its maximum value.

Because the flux-time curve is smooth, its gradient is zero at this instant.

Faraday’s law gives:

\[ \mathcal{E}=-N\frac{d\Phi}{dt} \]

At the maximum:

\[ \frac{d\Phi}{dt}=0 \]

Therefore:

\[ \mathcal{E}=0\,\text{V} \]

For the assumed purely resistive circuit:

\[ I=0\,\text{A} \]

The current momentarily reaches zero before reversing direction.

Step 6: Explain why the current reverses.

This is the most important part of the question.

The magnet continues travelling from left to right throughout the experiment. However, the magnetic flux through the coil does not continuously increase.

Initially, the magnet approaches and the flux increases. After the flux reaches its maximum, it begins decreasing as the magnet continues through and beyond the coil.

Lenz’s law responds to this reversal in the change of flux.

The first half requires an induced field opposing an increase in rightward flux. The second half requires an induced field maintaining a decreasing rightward flux.

The two induced magnetic fields point in opposite directions, so the induced current reverses.

A student who predicts current direction using only the magnet’s direction of motion will miss this reversal.

The deeper lesson is that the magnet’s motion is not itself the deciding quantity. The change in magnetic flux through the chosen loop is.

09Where to go next

You can now determine the direction of induced current from the physical arrangement of a magnet and coil, rather than relying on a memorised current-direction diagram.

The next useful step is understanding what happens when the coil itself rotates.

In a DC generator, the magnetic flux through the rotating coil changes continuously. This causes the induced EMF to reverse during each half-turn. A split-ring commutator then changes the external connections to produce a unidirectional output.

The same ideas you have used here – magnetic flux, Faraday’s law, Lenz’s law, and current direction – provide the foundation for understanding that process.