Why Faster Magnet Movement Produces a Larger Induced EMF

Move the same bar magnet along the same path through the same coil. In one trial, the movement takes 0.80 s. In another, it takes just 0.20 s. Why can the second trial produce a much larger induced EMF?

The answer is that the magnetic flux through each turn of the coil changes by the same amount, but in a shorter time. Faraday’s law tells us that induced EMF depends on the rate of change of magnetic flux linkage. If the same flux change happens four times faster, the average induced EMF is four times larger.

Before going further, make one prediction. Suppose the magnet moves quickly into a coil and then stops. Does the EMF remain large because the magnet is now very close to the coil?

No. Once the magnet stops, the magnetic flux stops changing, so the induced EMF falls to zero. A large magnetic field is not enough. What matters is how quickly the magnetic flux through the coil changes.

01The experiment that revealed the relationship

The relationship between magnet movement and induced EMF comes from Michael Faraday’s electromagnetic induction experiments in 1831.

Faraday wanted to investigate whether a magnet could produce an electrical effect without a battery directly connected to the circuit.

How Faraday’s apparatus worked

Consider a bar magnet, a coil of insulated copper wire, and a sensitive galvanometer.

The apparatus is arranged as follows:

• The coil is held stationary.
• The bar magnet is positioned along the central axis of the coil.
• The two ends of the coil are connected by wires to a galvanometer.
• The magnet can be pushed into or pulled out of the coil.

The galvanometer is a sensitive instrument that detects small electric currents and indicates their direction.

What should happen when the magnet moves?

There were two possible explanations worth considering.

The first possibility was that the magnetic field itself produced an electric current. If this were true, a stationary magnet sitting inside the coil should produce a continuous current.

The second possibility was that a changing magnetic flux produced an induced EMF. This explanation predicts that an electrical effect occurs while the magnet moves, but disappears when the magnet stops.

Faraday’s observations supported the second explanation.

The important observation is that the galvanometer only deflects while the magnetic flux is changing.

The Royal Institution’s historical record of Faraday’s generator describes this magnet-through-coil arrangement.

What the original experiment could not establish

Faraday’s experiment demonstrated electromagnetic induction, but it was not a precise investigation of how magnet speed affects induced EMF.

There were two limitations.

First, moving a magnet by hand makes it difficult to control its speed accurately. One movement might take twice as long as another, even if the starting and ending positions appear identical.

Second, a galvanometer measures current, not EMF directly. Its needle movement depends on the current, circuit resistance, and the instrument’s response.

To investigate the relationship between speed and induced EMF quantitatively, we need a more controlled experiment.

02A controlled experiment: changing the magnet’s speed

Keep the coil stationary and mount the bar magnet on a straight track.

Mark a starting position and an ending position. The magnet must follow exactly the same path in every trial, with the same pole facing the coil.

Connect a high-input-resistance voltage probe across the coil. This allows us to measure the induced terminal voltage while drawing negligible current.

Now repeat the movement at different speeds.

For a fair comparison, keep the following quantities constant:

• The same bar magnet.
• The same coil and number of turns.
• The same starting and ending positions.
• The same magnet orientation.
• The same path of movement.

Only change the time taken to complete the movement.

This is important because using the same starting and ending positions means the magnetic flux change is approximately the same in each trial.

Comparing two speeds

Consider the following illustrative experimental data.

A 200-turn coil experiences an increase in magnetic flux of \(0.80\text{ mWb}\) per turn as the magnet moves between the marked positions.

Illustrative data, not measurements from a real experiment.

The fast movement produces four times the average induced EMF.

Why?

The magnetic flux change is identical. The only difference is the time taken to produce that change.

For the slow movement:

\[ |\varepsilon_{\mathrm{avg}}| =\frac{200(0.80\times10^{-3})}{0.80} \]

\[ =0.20\text{ V} \]

For the fast movement:

\[ |\varepsilon_{\mathrm{avg}}| =\frac{200(0.80\times10^{-3})}{0.20} \]

\[ =0.80\text{ V} \]

The second movement produces four times the average induced EMF.

03Why does faster movement produce a larger EMF?

To understand the result, we need to be clear about what is changing.

Start with the magnetic flux through the loop

A coil consists of multiple turns of conducting wire. Each turn encloses an area.

Think about one turn of the coil as a circular loop. The magnetic flux through that loop describes how much magnetic field passes through its enclosed area in the direction perpendicular to the loop.

For a uniform magnetic field, magnetic flux is calculated using:

\[ \Phi=BA\cos\theta \]

Where:

• \(\Phi\) is magnetic flux, measured in webers (Wb).
• \(B\) is magnetic field strength, measured in teslas (T).
• \(A\) is the area enclosed by the loop, measured in square metres (\(\text{m}^2\)).
• \(\theta\) is the angle between the magnetic field and the normal to the loop.

The normal is an imaginary line perpendicular to the surface of the loop.

For a real bar magnet, the magnetic field is generally non-uniform across the coil. The expression above is therefore a simplified model using an appropriate average field. In many HSC problems, the magnetic flux change is supplied directly.

As the magnet approaches the coil, the magnetic field through the loop generally becomes stronger.

This means the magnetic flux changes.

When the magnet moves away, the magnetic field through the loop generally becomes weaker, so the flux changes in the opposite direction.

The coil responds to this changing flux.

Faraday’s law connects flux change to EMF

A coil with \(N\) turns has a quantity called magnetic flux linkage:

\[ \text{Flux linkage}=N\Phi \]

Here, \(N\) is the number of turns and \(\Phi\) is the magnetic flux through one turn.

Faraday’s law states that the induced EMF is proportional to the rate of change of magnetic flux linkage.

For an average EMF over a time interval:

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

Where:

• \(\varepsilon_{\mathrm{avg}}\) is average 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, in webers (Wb).
• \(\Delta t\) is the time taken for that change, in seconds (s).

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

When comparing EMF magnitudes, we use:

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

This is the equation that explains the effect of magnet speed.

If the same magnetic flux change occurs over a shorter time, the average induced EMF increases.

For example:

• Halving the time taken doubles the average EMF.
• Reducing the time to one-third produces three times the average EMF.
• Reducing the time to one-quarter produces four times the average EMF.

The key assumption is that the same coil experiences the same total change in magnetic flux.

Why the gradient matters

Imagine plotting magnetic flux against time.

The gradient of the graph tells us how quickly magnetic flux is changing.

A shallow gradient represents a slower flux change. A steep gradient represents a faster flux change.

Faraday’s law connects this gradient directly to induced EMF.

For the instantaneous value:

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

This is the more precise form of Faraday’s law.

You don’t need to use advanced calculus to understand the HSC application. For an average EMF, you can calculate the gradient using the change in flux divided by the change in time.

The important distinction is that a large magnetic flux does not necessarily produce a large EMF. A rapidly changing magnetic flux does.

04Specify the loop before predicting current

There is another detail that can cause confusion in HSC questions.

Before predicting the direction of an induced current, you need to identify the particular loop and the direction of its normal.

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

The magnetic field outside the magnet travels away from its north pole. Therefore, the magnetic field passes through the coil from left to right.

Choose the positive normal to point from the magnet towards the coil.

Using the right-hand grip rule, define positive current as clockwise when viewed from the magnet side.

Now consider three situations.

1. The north pole approaches the coil.

The rightward magnetic flux increases.

The induced EMF acts in the negative direction relative to our chosen convention. If the circuit is closed, the induced current is anticlockwise when viewed from the magnet side.

This produces a north pole on the near face of the coil, repelling the approaching north pole.

2. The magnet stops moving.

The magnetic flux becomes constant.

There is no continuing change in flux, so the induced EMF becomes zero. In an ideal resistive circuit, the induced current also becomes zero.

3. The north pole moves away from the coil.

The rightward magnetic flux decreases.

The induced EMF now acts in the positive direction. The current is clockwise when viewed from the magnet side.

The near face of the coil becomes a south pole, attracting the departing north pole.

Notice that the direction of the induced current depends on the direction of the flux change, not simply on whether the magnet is moving quickly or slowly.

05Induced EMF is not the same as induced current

This distinction is essential when answering questions about faster magnet movement.

An induced EMF is the electrical energy transferred per unit charge by the induced source. It is measured in volts.

An induced current is the actual flow of electric charge through a conducting path. It is measured in amperes.

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

For a simple closed circuit with total resistance \(R\), Ohm’s law gives:

\[ I=\frac{\varepsilon}{R} \]

Where:

• \(I\) is induced current in amperes (A).
• \(\varepsilon\) is induced EMF in volts (V).
• \(R\) is the total circuit resistance in ohms (\(\Omega\)).

This gives us an important qualification.

If the total circuit resistance remains constant, a larger induced EMF produces a larger induced current.

However, if the resistance changes, the current does not necessarily increase.

A faster magnet can produce a larger EMF without producing a larger current.

06When faster magnet movement does not produce a larger EMF

So far, we have assumed that the magnet moves along the same path relative to a stationary coil.

That condition matters.

A magnet can move very quickly without producing an induced EMF if its movement does not change the magnetic flux through the loop.

Consider a magnet and coil secured together on a trolley.

The entire arrangement moves across a room at \(4.0\text{ m s}^{-1}\). The magnet and coil maintain exactly the same separation and orientation.

Does the coil experience an induced EMF?

No. The magnetic field distribution relative to the coil remains unchanged. The magnetic flux through the coil is constant.

The magnet is moving relative to the room, but it is not moving relative to the coil.

Compare this with a coil moving towards a stationary magnet.

As the coil approaches, it enters a region of stronger magnetic field. The magnetic flux through the coil changes, producing an induced EMF.

The difference is not simply how quickly something moves. It is whether the magnetic flux through the particular loop changes.

A useful comparison

There is another limitation worth remembering.

Even if two magnets move at the same speed, they might produce different EMFs if their magnetic field strengths differ.

Similarly, the same magnet moving at the same speed can produce different EMFs along different paths because the magnetic field does not change equally at every position.

The speed of the magnet is therefore not enough information by itself.

You need to know how quickly the magnetic flux through the specified loop changes.

For a broader explanation of these different arrangements, read Induction from Moving Magnets and Conductors: HSC Physics.

If you are given a magnetic flux versus time graph in an exam, the gradient is the quantity you need. See How to Calculate Induced EMF from a Piecewise Flux Graph.

07Questions and solutions

Question 1

A 120-turn coil is held stationary in front of a bar magnet.

The magnet’s north pole moves towards the coil. During the movement, the magnetic flux through each turn increases from \(0.20\text{ mWb}\) to \(0.50\text{ mWb}\) over \(0.15\text{ s}\).

Calculate the magnitude of the average induced EMF.

Solution 1

The average induced EMF is \(0.24\text{ V}\).

The magnetic flux increases by \(0.30\text{ mWb}\) per turn. We use Faraday’s law to calculate the average EMF.

Step 1: Identify the known quantities.

\[ N=120 \]

\[ \Delta\Phi=(0.50-0.20)\times10^{-3} \]

\[ \Delta\Phi=3.0\times10^{-4}\text{ Wb} \]

\[ \Delta t=0.15\text{ s} \]

Step 2: Apply Faraday’s law.

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

\[ \begin{aligned} |\varepsilon_{\mathrm{avg}}| &=\frac{120(3.0\times10^{-4})}{0.15}\\ &=0.24\text{ V} \end{aligned} \]

The average induced EMF is \(0.24\text{ V}\).

The important point is that the EMF depends on the rate of change of flux, not simply the final flux value of \(0.50\text{ mWb}\).

Question 2

A 200-turn coil is used to investigate the effect of magnet speed.

The same bar magnet follows the same path towards the coil in both trials. The magnetic flux through each turn changes by \(0.40\text{ mWb}\).

In Trial A, the movement takes \(0.40\text{ s}\), and the total circuit resistance is \(5.0\Omega\).

In Trial B, the movement takes \(0.10\text{ s}\), and an additional external resistor makes the total circuit resistance \(20.0\Omega\).

Both circuits are closed.

Calculate:

a) The average induced EMF in each trial.

b) The induced current in each trial.

c) Explain why the faster movement does not produce a larger current.

Solution 2

Both trials produce the same induced current of \(0.040\text{ A}\), even though Trial B produces four times the induced EMF.

This happens because Trial B has four times the total resistance.

Step 1: Convert the flux change into webers.

\[ \Delta\Phi=0.40\times10^{-3}\text{ Wb} \]

Step 2: Calculate the EMF in Trial A.

\[ |\varepsilon_A| =\frac{200(0.40\times10^{-3})}{0.40} \]

\[ |\varepsilon_A|=0.20\text{ V} \]

Step 3: Calculate the EMF in Trial B.

\[ |\varepsilon_B| =\frac{200(0.40\times10^{-3})}{0.10} \]

\[ |\varepsilon_B|=0.80\text{ V} \]

Trial B produces four times the average EMF.

Step 4: Calculate the current in Trial A.

\[ I_A=\frac{\varepsilon_A}{R_A} \]

\[ I_A=\frac{0.20}{5.0} \]

\[ I_A=0.040\text{ A} \]

Step 5: Calculate the current in Trial B.

\[ I_B=\frac{\varepsilon_B}{R_B} \]

\[ I_B=\frac{0.80}{20.0} \]

\[ I_B=0.040\text{ A} \]

Therefore:

\[ I_A=I_B=0.040\text{ A} \]

The faster magnet produces a larger EMF, but the increased resistance prevents the current from increasing.

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

Question 3

A bar magnet has its north pole facing a stationary circular coil.

The coil is connected to a centre-zero galvanometer. Choose the positive magnetic flux direction to point from the magnet towards the coil, and define positive current as clockwise when viewed from the magnet side.

Describe what happens during each stage:

a) The north pole moves towards the coil.

b) The magnet stops while inside the coil.

c) The north pole is withdrawn along the same path.

For each stage, explain the change in magnetic flux, the induced EMF, and the direction of induced current. Also describe the galvanometer’s response.

Solution 3

The induced current reverses direction when the magnet changes from approaching the coil to moving away from it.

This follows from Faraday’s law and Lenz’s law.

Stage A: The magnet approaches.

The magnetic field through the coil points from the magnet towards the coil.

As the north pole approaches, the positive magnetic flux increases.

Therefore:

\[ \frac{\Delta\Phi}{\Delta t}>0 \]

From Faraday’s law:

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

The induced EMF is negative relative to our chosen positive direction.

The induced current is therefore anticlockwise when viewed from the magnet side.

This produces a north pole on the near face of the coil, opposing the approaching magnet.

The galvanometer briefly deflects in one direction.

Stage B: The magnet stops.

The magnet is stationary, so the magnetic flux through the coil is constant.

Therefore:

\[ \Delta\Phi=0 \]

The induced EMF becomes zero.

For the ideal resistive circuit, the induced current also becomes zero.

The galvanometer returns to zero.

Stage C: The magnet moves away.

The positive magnetic flux decreases.

Therefore:

\[ \frac{\Delta\Phi}{\Delta t}<0 \]

The induced EMF is now positive relative to our chosen direction.

The current becomes clockwise when viewed from the magnet side.

The near face of the coil becomes a south pole, attracting the departing north pole.

The galvanometer deflects in the opposite direction to Stage A.

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

The physical result is that reversing the magnet’s movement reverses the induced current.

Question 4

A 100-turn coil has a total circuit resistance of \(10\Omega\).

Two experiments are performed.

Experiment A: A bar magnet and the coil are secured together on a trolley. The entire arrangement moves across the room at \(4.0\text{ m s}^{-1}\). Their separation and orientation remain constant. Assume external magnetic fields are negligible.

Experiment B: The magnet remains stationary while the coil moves towards it. During the movement, the magnetic flux through each turn increases by \(0.30\text{ mWb}\) over \(0.050\text{ s}\).

Calculate the induced EMF and current in each experiment.

Explain why Experiment A produces a different result, even though the magnet and coil are moving at a considerable speed.

Solution 4

Experiment A produces zero induced EMF, while Experiment B produces an average induced EMF of \(0.60\text{ V}\) and a current of \(0.060\text{ A}\).

The difference is whether the magnetic flux through the coil changes.

Experiment A

The magnet and coil move together.

Their relative positions and orientation remain unchanged, so the magnetic field through the coil stays constant.

Therefore:

\[ \Delta\Phi=0 \]

Applying Faraday’s law:

\[ \varepsilon_A=-N\frac{0}{\Delta t} \]

\[ \varepsilon_A=0\text{ V} \]

Using Ohm’s law:

\[ I_A=\frac{0}{10}=0\text{ A} \]

The movement across the room does not produce an induced EMF because the magnetic flux through the coil does not change.

Experiment B

The coil moves towards the stationary magnet, causing the magnetic flux through each turn to increase.

Step 1: Convert the flux change into webers.

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

Step 2: Calculate the induced EMF.

\[ |\varepsilon_B| =\frac{100(0.30\times10^{-3})}{0.050} \]

\[ |\varepsilon_B|=0.60\text{ V} \]

Step 3: Calculate the induced current.

\[ I_B=\frac{\varepsilon_B}{R} \]

\[ I_B=\frac{0.60}{10} \]

\[ I_B=0.060\text{ A} \]

The current is \(0.060\text{ A}\), or \(60\text{ mA}\).

This experiment demonstrates why you must consider the magnetic flux through the loop rather than simply looking at an object’s speed relative to the room.

Question 5

Two coils, A and B, are tightly layered on the same former. They experience identical magnetic flux changes per turn.

The coils have the following properties:

• Coil A has 160 turns.
• Coil B has 120 turns.
• The coils are wound in opposite directions.
• They are connected in series opposition, so their induced EMFs act against each other.
• The total circuit resistance is \(8.0\Omega\).

A bar magnet approaches the coils along their common axis.

During one movement, the magnetic flux through each turn increases by \(0.50\text{ mWb}\) over \(0.25\text{ s}\).

a) Calculate the magnitude of the induced EMF in each coil.

b) Determine the net EMF across the series combination and calculate the resulting current.

c) The magnet is then moved along exactly the same path, but five times faster. Calculate the new net EMF and current.

d) A student argues that if both coils had 160 turns, the faster movement would still produce a substantial current because each coil would have an induced EMF.

Explain why this prediction is incorrect.

Assume ideal coupling, identical flux changes per turn, and a purely resistive circuit. Neglect self-inductance.

Solution 5

The net induced EMF is \(0.080\text{ V}\) in the first trial and \(0.40\text{ V}\) in the faster trial. The corresponding currents are \(0.010\text{ A}\) and \(0.050\text{ A}\). If the coils have equal numbers of turns, their induced EMFs cancel completely and the ideal circuit current is zero.

The important detail is that the coils are connected in series opposition.

Their individual induced EMFs cannot simply be added as positive magnitudes.

Step 1: Calculate the induced EMF in Coil A.

The flux change is:

\[ \Delta\Phi=0.50\times10^{-3}\text{ Wb} \]

Using Faraday’s law:

\[ |\varepsilon_A| =\frac{160(0.50\times10^{-3})}{0.25} \]

\[ |\varepsilon_A|=0.32\text{ V} \]

Step 2: Calculate the induced EMF in Coil B.

\[ |\varepsilon_B| =\frac{120(0.50\times10^{-3})}{0.25} \]

\[ |\varepsilon_B|=0.24\text{ V} \]

The individual EMFs are therefore \(0.32\text{ V}\) and \(0.24\text{ V}\).

However, the coils are connected in series opposition.

Their EMFs act in opposite directions, so the net EMF is the difference:

\[ |\varepsilon_{\mathrm{net}}| =|\varepsilon_A-\varepsilon_B| \]

\[ =0.32-0.24 \]

\[ |\varepsilon_{\mathrm{net}}|=0.080\text{ V} \]

Step 3: Calculate the circuit current.

\[ I=\frac{\varepsilon_{\mathrm{net}}}{R} \]

\[ I=\frac{0.080}{8.0} \]

\[ I=0.010\text{ A} \]

The current is \(10\text{ mA}\).

Step 4: Calculate the EMFs when the magnet moves five times faster.

The same magnetic flux change occurs in one-fifth of the original time.

\[ \Delta t_{\mathrm{fast}}=\frac{0.25}{5} \]

\[ \Delta t_{\mathrm{fast}}=0.050\text{ s} \]

Calculate the new EMF in Coil A:

\[ |\varepsilon_{A,\mathrm{fast}}| =\frac{160(0.50\times10^{-3})}{0.050} \]

\[ |\varepsilon_{A,\mathrm{fast}}|=1.60\text{ V} \]

Calculate the new EMF in Coil B:

\[ |\varepsilon_{B,\mathrm{fast}}| =\frac{120(0.50\times10^{-3})}{0.050} \]

\[ |\varepsilon_{B,\mathrm{fast}}|=1.20\text{ V} \]

Because the coils are still connected in series opposition:

\[ |\varepsilon_{\mathrm{net,fast}}| =1.60-1.20 \]

\[ |\varepsilon_{\mathrm{net,fast}}|=0.40\text{ V} \]

Step 5: Calculate the new current.

\[ I_{\mathrm{fast}}=\frac{0.40}{8.0} \]

\[ I_{\mathrm{fast}}=0.050\text{ A} \]

The faster movement produces a current of \(50\text{ mA}\).

Step 6: Consider what happens if both coils have 160 turns.

If both coils have identical numbers of turns and experience the same flux change per turn, their individual induced EMFs have equal magnitudes.

However, they still act in opposite directions.

Therefore:

\[ \varepsilon_{\mathrm{net}} =\varepsilon_A-\varepsilon_B \]

\[ \varepsilon_{\mathrm{net}}=0 \]

The net circuit EMF is zero, so the ideal circuit has no induced current.

This remains true even if the magnet moves five times faster.

Each coil individually experiences a larger induced EMF, but the two EMFs cancel when connected in series opposition.

The tempting mistake is to assume that two induced EMFs must add together because both coils experience changing magnetic flux.

They do not necessarily add. Their relative winding directions and electrical connections determine the net EMF.

This is why identifying the actual loop and circuit is essential before predicting an induced current.

08Where this leads next

The central relationship is:

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

Faster magnet movement produces a larger average induced EMF when the same magnetic flux change occurs through the same loop in less time.

But three conditions must remain clear:

1. The magnetic flux through the specified loop must change.
2. The rate of that change determines the magnitude of the induced EMF.
3. The induced current also depends on the circuit resistance and the way the EMFs combine.

The next useful step is learning to interpret magnetic flux versus time graphs. Once you can recognise that the graph’s gradient represents the rate of change of magnetic flux, you can calculate induced EMF even when the magnet changes speed during its movement.