Why Reversing Magnet Motion Reverses Induced EMF

Push the north pole of a bar magnet into a coil connected to a galvanometer. The needle deflects in one direction. Now pull the same magnet back out. The needle deflects in the opposite direction.

The magnet has not changed, and neither has the coil. So why does the induced EMF reverse?

The reason is that reversing the magnet’s motion reverses the direction of change in magnetic flux through the loop. With the coil stationary and the same magnetic pole facing it, this reverses the sign of the rate of change of flux, so Faraday’s law predicts an induced EMF in the opposite direction.

If the magnet moves along the same path at the same speed, the magnitude of the induced EMF stays the same. Only its direction changes.

01What Faraday’s experiment actually showed

This behaviour was demonstrated by Michael Faraday in 1831, during his investigations into electromagnetic induction.

His experiment is particularly useful because it shows that the direction of the magnet’s motion matters, not simply whether a magnet is present.

Reconstructing the experiment

Faraday’s apparatus consisted of a hollow paper cylinder wrapped with eight separate helices of copper wire. The helices were wound in the same direction and insulated from one another using twine and calico.

The ends of the wires were gathered into two bundles, one at each end of the cylinder. These bundles were connected by long copper wires to a galvanometer, an instrument that detects electrical current.

A cylindrical bar magnet was then inserted into the open end of the cylinder, moving along its central axis.

What should happen when Faraday pushes the magnet into the cylinder and then pulls it out?

If magnet movement produces an induced current, the galvanometer should respond while the magnet moves. If reversing the motion reverses the electrical effect, the needle should move in the opposite direction.

That is exactly what Faraday recorded.

On 17 October 1831, his diary describes the galvanometer needle moving when the magnet was pushed into the coil, then moving in the opposite direction when the magnet was withdrawn. The needle returned to its resting position after each movement.

You can read the original experimental entry in Faraday’s diary.

This observation established two important points:

• Moving the magnet relative to the coil produces an induced electrical effect.
• Reversing the magnet’s motion reverses the direction of that effect.

Faraday also tested a tin-foil sleeve around a paper cylinder. That arrangement produced no detectable galvanometer response. He suggested that the continuous conducting surface provided a return path that prevented the effect from registering in the same way as it did with the separate wire helices.

This was an important experimental limitation: the arrangement of the conducting material and its connections affected what the galvanometer could detect.

In modern physics, we explain the successful experiment using magnetic flux, Faraday’s law, and Lenz’s law.

02Why reversing the motion reverses the induced EMF

Imagine the coil as a hoop held in a stream. The magnetic field passing through the hoop represents the amount of magnetic flux through the loop.

Moving the magnet closer changes how much magnetic field passes through the coil. Moving it away reverses that change.

The analogy has a limit: magnetic field lines are not actual streams of material. Magnetic flux is a signed physical quantity that describes the magnetic field passing through a surface.

The important distinction is between:

• The direction of the external magnetic field.
• Whether the magnetic flux through the loop is increasing or decreasing.

These are not the same thing.

First, establish the magnetic field direction

Consider a bar magnet with its north pole facing a stationary coil.

Magnetic field lines leave the north pole and enter the south pole. Near the north pole, the external magnetic field points away from the magnet.

Therefore, if the magnet is to the left of the coil, the magnetic field through the coil points towards the right.

For this explanation, we will use the following sign convention:

• Positive magnetic flux points away from the magnet, through the coil.
• Positive current is clockwise when viewed from the magnet side.
• Negative current is anticlockwise when viewed from the magnet side.

The positive current direction follows the right-hand grip rule and corresponds to an area normal pointing away from the magnet.

Keeping this convention fixed makes it much easier to compare the two situations.

Case 1: The north pole moves towards the coil

As the north pole approaches, the magnetic field through the coil becomes stronger.

The positive magnetic flux therefore increases.

The coil responds by producing an induced magnetic field that opposes this increase. Its induced field points towards the magnet, against the increasing external flux.

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

The coil’s near face becomes a north pole, which repels the approaching north pole.

Case 2: The north pole moves away from the coil

Now reverse the magnet’s motion.

The north pole moves away, so the magnetic field through the coil becomes weaker.

The positive magnetic flux decreases.

The coil responds by producing an induced magnetic field in the same direction as the original external field. This helps maintain the magnetic flux that is being lost.

The induced field now points away from the magnet.

Using the right-hand grip rule, the conventional current must flow clockwise when viewed from the magnet side.

The coil’s near face becomes a south pole, attracting the north pole as it moves away.

Notice something important: the induced magnetic field does not always oppose the external magnetic field.

When the magnet approaches, the induced field opposes the external field. When the magnet retreats, the induced field supports the external field because it is opposing a decrease in flux.

This is the central idea behind Lenz’s law.

The mathematical reason: Faraday’s law

Faraday’s law expresses the relationship between the rate of change of magnetic flux and induced EMF:

\[
\varepsilon_{\mathrm{avg}}
=
-N\frac{\Delta\Phi}{\Delta t}
\]

where:

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

The negative sign represents Lenz’s law. It ensures that the induced EMF acts in a direction that opposes the change in magnetic flux.

Now compare the two situations.

For the north pole approaching the coil:

\[
\Delta\Phi>0
\]

Therefore:

\[
\varepsilon_{\mathrm{avg}}<0
\]

For the north pole moving away:

\[
\Delta\Phi<0
\]

Therefore:

\[
\varepsilon_{\mathrm{avg}}>0
\]

The sign of the induced EMF reverses because the sign of the change in magnetic flux reverses.

For the same path and equal speeds at corresponding positions, the two EMF magnitudes are equal.

More generally, the magnitude depends on how quickly the flux changes. Reversing the motion does not guarantee equal magnitudes if the magnet moves at different speeds.

Compare all four situations

There is another detail worth understanding before attempting HSC questions.

The direction of the magnet’s motion alone is not enough to predict the induced current. You must also identify which pole faces the coil.

Using our established sign convention gives the following results.

All current directions are viewed from the magnet side.

The table exposes a common misconception.

A student might memorise that moving a magnet towards a coil produces an anticlockwise current, while moving it away produces a clockwise current.

That rule only works when the same pole faces the coil.

If the south pole faces the coil instead, the external magnetic flux points in the opposite direction. The current predictions change accordingly.

Always establish the magnetic field direction before applying Lenz’s law.

03Induced EMF and induced current are not the same thing

There is one more distinction to get right.

An induced EMF is a voltage produced by electromagnetic induction. An induced current is the flow of electric charge through a conducting circuit.

They are related, but they are not interchangeable.

For a simple closed circuit with negligible self-inductance, Ohm’s law gives:

\[
I=\frac{\varepsilon}{R_{\mathrm{total}}}
\]

where:

• \(I\) is the induced current, measured in amperes (A).
• \(\varepsilon\) is the induced EMF, measured in volts (V).
• \(R_{\mathrm{total}}\) is the total circuit resistance, measured in ohms (\(\Omega\)).

Consider two situations.

Situation 1: The coil is connected in a complete circuit.

When the magnet moves, an induced EMF drives a current through the coil and galvanometer. The galvanometer needle deflects.

Reversing the magnet’s motion reverses the induced EMF, so the current also reverses in a simple resistive circuit.

Situation 2: The coil is an open circuit.

An induced EMF can still exist across the coil’s terminals when the magnet moves.

However, there is no complete conducting path for a sustained current. An ideal open circuit therefore has zero current, despite having an induced EMF.

A high-resistance voltmeter can detect the voltage.

This explains why an induced EMF does not automatically mean that a current must flow.

What happens when the magnet stops?

Suppose you push a north pole into a coil, hold it still, and then pull it out.

The sequence is:

The stationary magnet can still produce a substantial magnetic field through the coil. However, because that field is no longer changing, there is no induced EMF.

At the instant the magnet reverses direction, its speed passes through zero if it turns around smoothly. For a stationary coil, the induced EMF also passes through zero before changing polarity.

The key is not whether magnetic flux exists. It is whether that flux is changing.

04Worked examples

The following examples use illustrative flux readings. They are not measurements from Faraday’s historical experiment.

05Questions and solutions

Question 1

A 40-turn coil is connected to a \(3.0\,\Omega\) resistor.

A north pole faces the stationary coil and moves towards it. The magnetic flux per turn increases from \(+0.10\,\mathrm{mWb}\) to \(+0.34\,\mathrm{mWb}\) in \(0.080\,\mathrm{s}\).

Calculate:

(a) The magnitude and sign of the average induced EMF.

(b) The magnitude and direction of the conventional induced current, viewed from the magnet side.

Assume negligible self-inductance.

Solution 1

The signed average induced EMF is \(-0.12\,\mathrm{V}\), and the conventional current is \(0.040\,\mathrm{A}\), flowing anticlockwise when viewed from the magnet side.

First, calculate the change in flux:

\[
\begin{aligned}
\Delta\Phi
&=0.34-0.10\\
&=+0.24\,\mathrm{mWb}\\
&=+2.4\times10^{-4}\,\mathrm{Wb}
\end{aligned}
\]

Apply Faraday’s law:

\[
\begin{aligned}
\varepsilon_{\mathrm{avg}}
&=-N\frac{\Delta\Phi}{\Delta t}\\
&=-40\frac{2.4\times10^{-4}}{0.080}\\
&=-0.12\,\mathrm{V}
\end{aligned}
\]

The negative sign indicates an anticlockwise induced EMF.

Now calculate the current:

\[
\begin{aligned}
I&=\frac{|\varepsilon|}{R}\\
&=\frac{0.12}{3.0}\\
&=0.040\,\mathrm{A}
\end{aligned}
\]

The current flows anticlockwise because the coil must produce a magnetic field opposing the increasing external flux.

Question 2

A 100-turn coil has a total circuit resistance of \(5.0\,\Omega\).

A bar magnet has its north pole facing the coil. The coil remains stationary.

The magnetic flux per turn is:

• \(+0.12\,\mathrm{mWb}\) when the magnet is \(10\,\mathrm{cm}\) from the coil.
• \(+0.52\,\mathrm{mWb}\) when the magnet is \(2.0\,\mathrm{cm}\) from the coil.

The magnet moves from \(10\,\mathrm{cm}\) to \(2.0\,\mathrm{cm}\) in \(0.20\,\mathrm{s}\). It then reverses direction and returns to \(10\,\mathrm{cm}\) in \(0.50\,\mathrm{s}\).

Calculate:

(a) The signed average induced EMF during the approach.

(b) The signed average induced EMF during the retreat.

(c) The magnitude and direction of the current during each interval.

Explain why the two EMF magnitudes are different even though the magnet travels the same path in opposite directions.

Solution 2

The average induced EMF is \(-0.20\,\mathrm{V}\) during the approach and \(+0.080\,\mathrm{V}\) during the retreat.

The current is \(0.040\,\mathrm{A}\) anticlockwise during the approach and \(0.016\,\mathrm{A}\) clockwise during the retreat.

(a) Approach

Calculate the change in flux:

\[
\begin{aligned}
\Delta\Phi
&=0.52-0.12\\
&=+0.40\,\mathrm{mWb}\\
&=+4.0\times10^{-4}\,\mathrm{Wb}
\end{aligned}
\]

Apply Faraday’s law:

\[
\begin{aligned}
\varepsilon_1
&=-100\frac{4.0\times10^{-4}}{0.20}\\
&=-0.20\,\mathrm{V}
\end{aligned}
\]

(b) Retreat

The magnet returns along the same path, so the flux change has the opposite sign.

\[
\begin{aligned}
\Delta\Phi
&=0.12-0.52\\
&=-0.40\,\mathrm{mWb}\\
&=-4.0\times10^{-4}\,\mathrm{Wb}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
\varepsilon_2
&=-100\frac{-4.0\times10^{-4}}{0.50}\\
&=+0.080\,\mathrm{V}
\end{aligned}
\]

(c) Current during each interval

For the approach:

\[
I_1=\frac{0.20}{5.0}
=0.040\,\mathrm{A}
\]

The negative EMF means anticlockwise current.

For the retreat:

\[
I_2=\frac{0.080}{5.0}
=0.016\,\mathrm{A}
\]

The positive EMF means clockwise current.

The tempting prediction is that the EMF magnitudes should be equal because the magnet travels the same distance in both directions.

However, the travel times are different.

The magnet takes \(0.20\,\mathrm{s}\) to approach but \(0.50\,\mathrm{s}\) to retreat. The flux changes at a slower average rate during the retreat.

The direction reverses, but the magnitude does not have to remain the same.

Question 3

A coil is tested in two separate experiments.

In the first experiment, its terminals are connected to a high-resistance voltmeter. In the second, the coil is connected to a closed circuit containing a galvanometer and resistor.

In both experiments, a north pole is pushed towards the coil, held stationary for two seconds, and then pulled away.

Describe what happens to:

(a) The voltmeter reading during each stage.

(b) The galvanometer deflection during each stage.

(c) Explain why an induced EMF can exist without a sustained induced current.

Assume the voltmeter has a sufficiently high input resistance that its current is negligible.

Solution 3

The voltmeter records a voltage during magnet movement, with opposite polarities during approach and withdrawal. The galvanometer deflects in opposite directions during those two stages. Both instruments show no sustained response while the magnet is stationary.

(a) Voltmeter

When the north pole approaches the coil, the magnetic flux increases. An induced EMF develops across the coil terminals, producing a voltage reading.

When the magnet stops, the flux becomes constant. The induced EMF falls to zero, so the voltmeter reading returns to zero.

When the magnet is withdrawn, the flux decreases. The induced EMF reverses polarity, so the voltmeter reading has the opposite sign.

(b) Galvanometer

The galvanometer requires a complete conducting circuit.

During the approach, the induced EMF drives a current in one direction, deflecting the needle.

When the magnet is held stationary, the current falls to zero and the needle returns to its resting position.

During withdrawal, the induced EMF reverses. The current also reverses, causing the needle to deflect in the opposite direction.

(c) Why EMF does not require current

EMF is the induced voltage, whereas current is the movement of charge through a complete conducting path.

The open-circuit coil can develop an induced voltage even though it has no sustained current. The voltmeter detects the voltage because its input resistance is very high.

The closed circuit allows current to flow, and the galvanometer detects that current.

This is why induced EMF and induced current must be treated as separate quantities.

Question 4

Two experiments are performed using the same stationary coil and the same bar magnet.

• Experiment A: The north pole approaches the coil.
• Experiment B: The south pole moves away from the coil.

Assume the magnet has symmetrical poles, and both experiments use the same path and speed.

Using the sign convention established in this article:

(a) State whether the signed external magnetic flux increases or decreases in each experiment.

(b) Determine the sign of the induced EMF in each experiment.

(c) Determine the conventional current direction, viewed from the magnet side.

(d) Are the induced EMFs in the same direction or opposite directions? Explain why.

Solution 4

The induced EMFs have the same sign in both experiments. Both produce anticlockwise conventional current when viewed from the magnet side.

This result is less obvious than it first appears because both the magnetic pole and the direction of motion have changed.

(a) Change in external flux

In Experiment A, the north pole approaches the coil.

The external magnetic field points away from the magnet, so the positive magnetic flux increases.

Therefore:

\[
\Delta\Phi_A>0
\]

In Experiment B, the south pole moves away from the coil.

The external magnetic field points towards the magnet, so the magnetic flux is negative.

As the south pole retreats, the magnitude of this negative flux decreases. The signed flux therefore becomes less negative.

Consequently:

\[
\Delta\Phi_B>0
\]

(b) Induced EMF

Faraday’s law gives:

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

Since the flux change is positive in both experiments:

\[
\varepsilon_A<0
\]

and

\[
\varepsilon_B<0
\]

The induced EMFs have the same polarity.

(c) Current direction

A negative induced EMF corresponds to anticlockwise conventional current when viewed from the magnet side.

Therefore, both experiments produce anticlockwise current.

(d) Why are the directions the same?

The tempting prediction is that the current must reverse because the magnet moves in the opposite direction in Experiment B.

However, the pole facing the coil has also changed.

Changing from north to south reverses the direction of the external magnetic field. This reverses the sign of the flux itself.

The two changes cancel when determining the sign of the flux change.

This is why you cannot predict induced current direction using magnet motion alone. You must establish both the pole orientation and the direction of the flux change.

Question 5

A 120-turn coil is connected to a \(6.0\,\Omega\) resistor and faces the north pole of a bar magnet.

The magnet and coil share a common axis. The signed magnetic flux per turn at the instant considered is \(+0.30\,\mathrm{mWb}\).

Over a small range of separation, the measured flux change is:

• An increase of \(+0.030\,\mathrm{mWb}\) whenever the gap between the magnet and coil decreases by \(2.0\,\mathrm{mm}\).
• The relationship is linear over this small range.

Compare two consecutive intervals, each lasting \(0.010\,\mathrm{s}\).

Before \(t=0\):

The coil is stationary, and the magnet approaches it at \(0.20\,\mathrm{m\,s^{-1}}\).

After \(t=0\):

The magnet reverses direction and moves away from the coil at \(0.30\,\mathrm{m\,s^{-1}}\). At the same time, the coil begins moving towards the magnet at \(0.50\,\mathrm{m\,s^{-1}}\).

Assume:

• All motion is along the common axis.
• The magnetic flux depends only on the separation between the magnet and coil.
• The coil’s orientation and area remain constant.
• The total resistance remains constant.
• Self-inductance is negligible.

Determine:

(a) The signed average induced EMF and current direction before \(t=0\).

(b) The signed average induced EMF and current direction after \(t=0\), when both objects are moving.

(c) The signed average induced EMF and current direction after \(t=0\) if the coil were instead held stationary while the magnet continued moving away at \(0.30\,\mathrm{m\,s^{-1}}\).

(d) Explain why reversing the magnet’s motion does not necessarily reverse the induced EMF when the coil is also moving.

Solution 5

The induced EMF is \(-0.36\,\mathrm{V}\) before the reversal and remains \(-0.36\,\mathrm{V}\) immediately afterwards when both objects move.

However, if the coil is held stationary after the reversal, the induced EMF becomes \(+0.54\,\mathrm{V}\).

The crucial detail is that electromagnetic induction depends on the change in magnetic flux through the loop, not the magnet’s velocity considered in isolation.

(a) Before \(t=0\): magnet approaches a stationary coil

The magnet approaches at \(0.20\,\mathrm{m\,s^{-1}}\).

Over \(0.010\,\mathrm{s}\), the magnet moves:

\[
\begin{aligned}
d&=vt\\
&=0.20(0.010)\\
&=0.0020\,\mathrm{m}
\end{aligned}
\]

Therefore, the gap decreases by \(2.0\,\mathrm{mm}\).

According to the supplied calibration, the flux per turn increases by:

\[
\Delta\Phi=+0.030\,\mathrm{mWb}
\]

Convert to webers:

\[
\Delta\Phi=+3.0\times10^{-5}\,\mathrm{Wb}
\]

Apply Faraday’s law:

\[
\begin{aligned}
\varepsilon_1
&=-120\frac{3.0\times10^{-5}}{0.010}\\
&=-0.36\,\mathrm{V}
\end{aligned}
\]

Calculate the current:

\[
I_1=\frac{0.36}{6.0}
=0.060\,\mathrm{A}
\]

The current is \(0.060\,\mathrm{A}\), anticlockwise when viewed from the magnet side.

(b) After \(t=0\): magnet and coil both move

This is the important part.

The magnet now moves away at \(0.30\,\mathrm{m\,s^{-1}}\).

On its own, this would increase the gap by:

\[
0.30(0.010)=0.0030\,\mathrm{m}
\]

The coil, however, moves towards the magnet at \(0.50\,\mathrm{m\,s^{-1}}\).

It reduces the gap by:

\[
0.50(0.010)=0.0050\,\mathrm{m}
\]

The net change in separation is therefore:

\[
\begin{aligned}
\Delta x
&=+0.0030-0.0050\\
&=-0.0020\,\mathrm{m}
\end{aligned}
\]

The negative sign means the gap still decreases by \(2.0\,\mathrm{mm}\).

Despite the magnet moving away, the two objects are getting closer overall.

Therefore, the flux per turn still increases by:

\[
\Delta\Phi=+3.0\times10^{-5}\,\mathrm{Wb}
\]

The induced EMF is:

\[
\begin{aligned}
\varepsilon_2
&=-120\frac{3.0\times10^{-5}}{0.010}\\
&=-0.36\,\mathrm{V}
\end{aligned}
\]

The current is:

\[
I_2=\frac{0.36}{6.0}
=0.060\,\mathrm{A}
\]

The current remains \(0.060\,\mathrm{A}\), anticlockwise when viewed from the magnet side.

Reversing the magnet’s motion has not reversed the induced EMF because the relative separation between the magnet and coil has continued to decrease at the same rate.

(c) After \(t=0\): the coil is stationary

Now remove the coil’s movement.

The magnet moves away at \(0.30\,\mathrm{m\,s^{-1}}\) for \(0.010\,\mathrm{s}\).

The gap increases by:

\[
\begin{aligned}
\Delta x
&=0.30(0.010)\\
&=0.0030\,\mathrm{m}\\
&=3.0\,\mathrm{mm}
\end{aligned}
\]

The supplied calibration tells us that a \(2.0\,\mathrm{mm}\) increase in gap would decrease flux by \(0.030\,\mathrm{mWb}\).

Therefore, a \(3.0\,\mathrm{mm}\) increase produces:

\[
\begin{aligned}
\Delta\Phi
&=-\frac{3.0}{2.0}(0.030)\\
&=-0.045\,\mathrm{mWb}\\
&=-4.5\times10^{-5}\,\mathrm{Wb}
\end{aligned}
\]

Apply Faraday’s law:

\[
\begin{aligned}
\varepsilon_3
&=-120\frac{-4.5\times10^{-5}}{0.010}\\
&=+0.54\,\mathrm{V}
\end{aligned}
\]

The current magnitude is:

\[
I_3=\frac{0.54}{6.0}
=0.090\,\mathrm{A}
\]

The current is \(0.090\,\mathrm{A}\), clockwise when viewed from the magnet side.

(d) The deeper principle

The tempting prediction is that reversing the magnet’s motion must reverse the induced EMF.

That prediction assumes the coil is stationary.

When both objects move, the change in separation depends on their relative motion. The magnet can move away while the coil moves towards it quickly enough that the gap continues to decrease.

The actual quantity controlling the induced EMF is the rate of change of magnetic flux through the loop.

The initial flux of \(+0.30\,\mathrm{mWb}\) is not what determines the EMF. What matters is how that flux changes over the specified time interval.

This is the more general way to approach electromagnetic induction: identify the loop, establish the magnetic field through it, and determine how the flux changes before predicting the induced EMF or current.

Once you can predict the direction correctly, the next step is calculating the magnitude of the induced EMF using the number of turns and the rate of change of flux.

For more practice, see Faraday’s Law for HSC Physics: Calculating Induced EMF and How to Find Induced Current Direction When a Magnet Enters a Coil.