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

Pull a bar magnet out of a coil with its north pole nearest the coil. Which way does the induced current flow?

The conventional current is clockwise when viewed from the magnet side. The coil’s nearest face becomes a south pole, attracting the departing north pole and opposing its motion.

If the south pole is nearest the coil instead, the current flows anticlockwise.

The important part is understanding why. Once you can track the change in magnetic flux, you can work out the current direction without relying on memorised rules.

01Start with the magnetic flux, not the current

The most common mistake in this question is to look at the magnet’s direction of motion and immediately guess the current direction.

Instead, start by identifying three things:

• Which magnetic pole is facing the coil?
• Which direction does the magnet’s magnetic field pass through the coil?
• Is the magnetic flux increasing or decreasing?

Only then should you determine the induced magnetic field and current.

What happens when a north pole leaves?

Imagine a bar magnet positioned to the left of a coil, with its north pole facing the coil.

The magnetic field outside a bar magnet travels from north to south. Therefore, near the coil, the magnetic field points away from the north pole.

In our arrangement, this means the magnetic field passes through the coil from left to right.

Now pull the magnet away.

As the distance between the magnet and coil increases, the magnetic field at the coil becomes weaker. Consequently, the magnetic flux through the coil decreases.

What should the coil do in response?

It produces its own magnetic field that tries to maintain the original magnetic flux.

This is an application of Lenz’s law:

The induced current flows in a direction that produces a magnetic field opposing the change in magnetic flux.

Notice the wording. The induced field opposes the change in flux, not necessarily the magnet’s direction of motion.

In this example:

1. The original magnetic field points to the right.
2. The magnet moves away, so the rightward magnetic flux decreases.
3. The coil produces a magnetic field pointing to the right.
4. The coil’s nearest face becomes a south pole.
5. The south face attracts the departing north pole.

This attraction opposes the magnet’s departure.

A useful check is to think about what would happen if the coil repelled the departing magnet. That would help the magnet leave faster, rather than oppose the change in flux.

What if the south pole leaves instead?

Now reverse the magnet so its south pole faces the coil.

The external magnetic field through the coil points towards the south pole. In our arrangement, this means the field points to the left.

As the magnet moves away, the leftward magnetic flux decreases in magnitude.

The coil therefore produces its own leftward magnetic field to oppose this decrease.

Its nearest face becomes a north pole, attracting the departing south pole.

The current direction also reverses.

02Use the right-hand grip rule to find the current

Once you know the direction of the induced magnetic field, you can determine the current using the right-hand grip rule.

Curl the fingers of your right hand in the direction of conventional current around the coil.

Your thumb points in the direction of the magnetic field inside the coil. It also points towards the coil’s north face.

When looking directly at the coil from one end:

• Anticlockwise current produces a north face towards you.
• Clockwise current produces a south face towards you.

For our standard arrangement, always imagine yourself standing on the magnet side of the coil, looking towards it.

The resulting directions are:

The table reveals a useful pattern.

For a single magnet whose facing pole dominates the magnetic flux:

• When the magnet approaches, the coil produces the same pole on its nearest face. The two poles repel.
• When the magnet leaves, the coil produces the opposite pole on its nearest face. The two poles attract.

These are consequences of Lenz’s law, not separate rules.

Be careful with your viewpoint. If you look at the coil from the opposite side, clockwise and anticlockwise will appear reversed. The physical current has not changed; only your viewing direction has changed.

03The experiment that explains the rule

The relationship between magnet movement and induced current comes from the classic magnet-and-coil electromagnetic induction experiment.

Faraday’s 1831 experiments established that changing magnetic conditions can produce an electrical effect. A simple classroom version makes the reasoning particularly clear.

Setting up the experiment

The apparatus consists of:

• A bar magnet.
• A stationary coil of insulated copper wire.
• Connecting wires joining the coil to a sensitive centre-zero galvanometer.

The galvanometer measures small currents and indicates their direction.

The magnet moves along the central axis of the coil. The coil remains stationary.

What should happen?

Consider two possible explanations.

The first is that the magnetic field itself produces a current. If that were true, a stationary magnet sitting inside the coil should produce a continuous current.

The second is that a changing magnetic flux produces an induced emf. This predicts a current only while the magnet moves and the flux changes.

The observations support the second explanation.

The exact left or right direction of galvanometer deflection depends on how its connecting wires are arranged.

The important observation is that reversing the magnet’s movement reverses the deflection.

Why the original demonstration has limitations

Moving the magnet by hand is useful for demonstrating induction, but it makes quantitative comparisons difficult.

For example, pulling the magnet out quickly produces a different rate of flux change from pulling it out slowly.

A more controlled investigation would:

1. Secure the coil in a fixed position.
2. Move the magnet along a guide with measured starting and ending positions.
3. Control the time taken to complete the movement.
4. Use a high-resistance voltage probe to measure the induced emf.
5. Repeat using a known circuit resistance and an ammeter to measure the resulting current.

This separates two quantities that are often confused: induced emf and induced current.

04Induced emf is not the same as induced current

An induced emf is a voltage generated around a circuit due to changing magnetic flux.

An induced current is the actual flow of electric charge through a closed circuit.

A changing magnetic flux can produce an induced emf even when the circuit is open. However, an open circuit cannot sustain a current.

For a closed circuit, the current also depends on the total circuit resistance.

Faraday’s law

The average induced emf is calculated using:

\[
\mathcal{E}_{\mathrm{avg}}
=
-N\frac{\Delta\Phi_B}{\Delta t}
\]

Where:

• \(\mathcal{E}_{\mathrm{avg}}\) = average induced emf, in volts (V).
• \(N\) = number of turns in the coil.
• \(\Delta\Phi_B\) = change in magnetic flux through one turn, in webers (Wb).
• \(\Delta t\) = time taken for the change, in seconds (s).

The negative sign represents Lenz’s law.

For a uniform magnetic field passing through a flat loop, magnetic flux is:

\[
\Phi_B=BA\cos\theta
\]

Here, \(B\) is magnetic field strength in tesla (T), \(A\) is the area of the loop in square metres (m²), and \(\theta\) is the angle between the magnetic field and the normal to the loop.

For a real bar magnet, the field is not uniform across the coil. The flux expression is therefore a simplified model, but the reasoning about the direction and change of the total flux still applies.

A consistent sign convention

For the calculations below, use this arrangement:

• Magnet on the left.
• Coil on the right.
• Positive magnetic flux points from the magnet through the coil towards the right.
• Positive conventional current is clockwise when viewed from the magnet side.

The right-hand grip rule confirms that positive current produces a rightward magnetic field.

Therefore:

Once the emf is known, calculate current using Ohm’s law:

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

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

The current direction follows the sign of the induced emf, assuming a simple closed circuit with positive resistance.

Remember: a larger induced emf does not automatically mean a larger current. Resistance matters too.

05Worked examples

06Questions and solutions

Question 1

A bar magnet’s north pole faces a coil and is pulled away from it.

An observer stands on the opposite side of the coil to the magnet and looks towards the magnet.

What direction does the observer see the conventional induced current flowing? Identify the magnetic pole of the coil’s face nearest the magnet.

Solution 1

The observer sees an anticlockwise current, and the coil’s face nearest the magnet becomes south.

From the magnet side, the current is clockwise because the north pole is moving away and the coil must produce a south face to attract it.

However, the observer is looking from the opposite side.

Reversing your viewing position makes clockwise appear anticlockwise. The physical current has not changed direction.

This is why you should always state which side of the coil you are viewing.

Question 2

A bar magnet has its south pole facing a coil. The magnet is pulled away from the coil, and the coil is connected to a galvanometer.

A student argues:

“The magnet is moving away, so the induced magnetic field must point away from the magnet.”

Explain why this prediction is incorrect.

Determine:

(a) The direction of the induced magnetic field through the coil.

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

(c) What happens to the induced emf and current if the magnet is held stationary.

Solution 2

The induced field points towards the magnet, the current is anticlockwise when viewed from the magnet side, and the induced emf and current become zero when the magnet stops.

The student’s prediction confuses opposing a change in magnetic flux with opposing the magnet’s motion.

First, consider the original magnetic field.

A south pole produces a magnetic field pointing towards itself. Therefore, the magnetic field through the coil points towards the magnet, or leftwards.

As the magnet moves away, this leftward magnetic flux decreases in magnitude.

According to Lenz’s law, the coil must produce a magnetic field pointing leftwards to oppose that decrease.

(a) Induced magnetic field

The induced field points leftwards, towards the magnet.

(b) Conventional current

A leftward field means the coil’s nearest face must become north.

Using the right-hand grip rule, this requires an anticlockwise conventional current when viewed from the magnet side.

(c) Magnet held stationary

Once the magnet stops, the magnetic flux through the coil remains constant.

Therefore:

\[
\Delta\Phi_B=0
\]

From Faraday’s law:

\[
\mathcal{E}=0
\]

Consequently, the induced current also becomes zero.

Importantly, the magnet’s magnetic field has not disappeared. It is still passing through the coil.

The experiment demonstrates that a magnetic field alone does not produce an induced emf in a stationary coil. A change in magnetic flux is required.

Question 3

A 150-turn coil has a total circuit resistance of 6.0 Ω. A north pole faces the coil.

The magnet undergoes three stages:

• From 0.00 s to 0.10 s, it moves away from the coil.
• From 0.10 s to 0.25 s, it remains stationary.
• From 0.25 s to 0.35 s, it moves back towards the coil.

The following illustrative data gives the signed magnetic flux per turn.

Positive flux is defined as pointing from the magnet towards the far side of the coil.

Assume the flux changes linearly during each interval.

Determine:

(a) The average induced emf during each interval.

(b) The average induced current during each interval.

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

Solution 3

The current is 0.075 A clockwise during the first interval, zero during the stationary interval, and 0.050 A anticlockwise during the final interval.

The important point is that the current reverses when the magnet changes direction.

(a) Average induced emf

First interval: 0.00-0.10 s.

The flux decreases from +0.70 mWb to +0.40 mWb.

\[
\Delta\Phi_B=-0.30\times10^{-3}\text{ Wb}
\]

Using Faraday’s law:

\[
\mathcal{E}
=
-150\frac{-0.30\times10^{-3}}{0.10}
\]

\[
\mathcal{E}=+0.45\text{ V}
\]

Second interval: 0.10-0.25 s.

The magnet is stationary, so the flux does not change.

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

Third interval: 0.25-0.35 s.

The flux increases from +0.40 mWb to +0.60 mWb.

\[
\Delta\Phi_B=+0.20\times10^{-3}\text{ Wb}
\]

Therefore:

\[
\mathcal{E}
=
-150\frac{0.20\times10^{-3}}{0.10}
\]

\[
\mathcal{E}=-0.30\text{ V}
\]

(b) Average induced current

Apply Ohm’s law to each interval.

First interval:

\[
I=\frac{0.45}{6.0}
=0.075\text{ A}
\]

Second interval:

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

Third interval:

\[
I=\frac{-0.30}{6.0}
=-0.050\text{ A}
\]

(c) Current direction

Using our sign convention:

The current reverses because the magnetic flux changes from decreasing to increasing.

A useful exam technique is to divide a changing-flux question into separate time intervals. Never assume the induced current keeps the same direction throughout an entire experiment.

Question 4

Two coils are placed in identical magnetic arrangements. In both arrangements, a north pole faces the coil and moves away.

Both coils experience the same change in magnetic flux per turn, from \(+0.50\text{ mWb}\) to \(+0.10\text{ mWb}\), over 0.20 s.

Their properties are:

Assume both coils have the same area and orientation, and the stated flux change per turn applies equally to both.

Determine:

(a) The average induced emf in each coil.

(b) The average induced current in each coil.

(c) Which coil has the greater induced emf and which has the greater current.

Explain why the results are different.

Solution 4

Coil B produces the greater induced emf, but coil A produces the greater induced current.

This happens because induced emf depends on the number of turns, while current also depends on resistance.

(a) Average induced emf

First calculate the change in flux per turn:

\[
\Delta\Phi_B
=
(0.10-0.50)\times10^{-3}
\]

\[
\Delta\Phi_B=-4.0\times10^{-4}\text{ Wb}
\]

For coil A:

\[
\mathcal{E}_A
=
-100\frac{-4.0\times10^{-4}}{0.20}
\]

\[
\mathcal{E}_A=+0.20\text{ V}
\]

For coil B:

\[
\mathcal{E}_B
=
-200\frac{-4.0\times10^{-4}}{0.20}
\]

\[
\mathcal{E}_B=+0.40\text{ V}
\]

Coil B has twice as many turns, so it produces twice the induced emf.

(b) Average induced current

For coil A:

\[
I_A=\frac{0.20}{2.0}
\]

\[
I_A=0.10\text{ A}
\]

For coil B:

\[
I_B=\frac{0.40}{8.0}
\]

\[
I_B=0.050\text{ A}
\]

(c) Compare the results

Both coils have clockwise current because the north pole is moving away in both arrangements.

However, coil B has four times the resistance and only twice the induced emf.

Consequently, its current is half that of coil A.

This is an important distinction: Faraday’s law determines the induced emf, while Ohm’s law determines the resulting current.

Question 5

A coil is placed between two bar magnets. Both magnets are aligned along the same horizontal axis.

The arrangement is:

• The left magnet has its north pole facing the coil and moves away from it.
• The right magnet has its south pole facing the coil and moves towards it.

Both magnets produce magnetic fields pointing to the right through the coil.

The coil has 200 turns and a total circuit resistance of 4.0 Ω.

The following illustrative data gives the individual signed flux contributions per turn.

Assume that both flux contributions change linearly within each 0.10 s interval.

Determine:

(a) The average induced emf during each 0.10 s interval.

(b) The average induced current during each interval, including its direction when viewed from the left side.

(c) The average induced emf over the complete 0.20 s interval.

A student argues:

“The north pole on the left is moving away for the entire experiment, so the induced current must remain clockwise.”

Explain precisely why this argument is incorrect.

Solution 5

The induced current is clockwise during the first interval and anticlockwise during the second interval. Its magnitude is 0.050 A in both intervals.

The average induced emf over the entire 0.20 s is zero.

The student’s argument is incorrect because Lenz’s law responds to the change in the total magnetic flux through the coil, not the movement of just one magnet.

(a) Average induced emf during each interval

First interval: 0.00-0.10 s.

The total flux decreases from +0.80 mWb to +0.70 mWb.

\[
\Delta\Phi_B
=
(0.70-0.80)\times10^{-3}
\]

\[
\Delta\Phi_B=-1.0\times10^{-4}\text{ Wb}
\]

Using Faraday’s law:

\[
\mathcal{E}_1
=
-200\frac{-1.0\times10^{-4}}{0.10}
\]

\[
\mathcal{E}_1=+0.20\text{ V}
\]

Second interval: 0.10-0.20 s.

The total flux increases from +0.70 mWb to +0.80 mWb.

\[
\Delta\Phi_B
=
(0.80-0.70)\times10^{-3}
\]

\[
\Delta\Phi_B=+1.0\times10^{-4}\text{ Wb}
\]

Therefore:

\[
\mathcal{E}_2
=
-200\frac{1.0\times10^{-4}}{0.10}
\]

\[
\mathcal{E}_2=-0.20\text{ V}
\]

(b) Average induced current and direction

First interval:

\[
I_1=\frac{0.20}{4.0}
\]

\[
I_1=+0.050\text{ A}
\]

The positive current is clockwise when viewed from the left.

Second interval:

\[
I_2=\frac{-0.20}{4.0}
\]

\[
I_2=-0.050\text{ A}
\]

The negative current is anticlockwise when viewed from the left.

(c) Average emf over the complete interval

The total flux starts at +0.80 mWb and finishes at +0.80 mWb.

Therefore:

\[
\Delta\Phi_B=0
\]

Applying Faraday’s law over the complete 0.20 s:

\[
\mathcal{E}_{\mathrm{avg}}
=
-200\frac{0}{0.20}
\]

\[
\boxed{\mathcal{E}_{\mathrm{avg}}=0\text{ V}}
\]

But this does not mean there was no induced current.

During the first half, the left magnet’s decreasing flux dominates, so the total flux decreases.

During the second half, the right magnet’s increasing flux dominates, so the total flux increases.

The induced current therefore reverses direction.

The crucial distinction is that the left magnet continues moving away throughout the experiment, but its contribution is not the only contribution to the total flux.

This also reveals a subtle point about Faraday’s law: identical initial and final flux values establish zero average emf over the full interval. They do not establish zero instantaneous emf throughout that interval.

A student who looks only at the left magnet’s movement misses the changing contribution from the right magnet.

07Where to go next

Once you can determine current direction from a changing magnetic flux, the next step is to calculate the magnitude of induced emf from a flux-time graph.

Read How to Calculate Induced EMF from a Piecewise Flux Graph to practise finding the rate of change of flux over different time intervals.

The same Faraday’s law also explains how a rotating coil produces an alternating emf. This is the foundation of DC generators: how rotating coils produce direct current.