How to Calculate Induced EMF from a Piecewise Flux Graph

Imagine a 100-turn coil sitting in a changing magnetic field. The magnetic flux through each turn increases, stays constant for a while, and then decreases.

At which point is the induced EMF largest? You might choose the moment when the magnetic flux reaches its maximum. That sounds reasonable, but it is wrong. The induced EMF depends on how quickly the flux changes, not how large the flux is.

This is exactly what a piecewise magnetic-flux-time graph allows you to calculate.

To find the induced EMF, calculate the gradient of each straight-line section separately, multiply by the number of turns, and apply Faraday’s law:

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

Here:

• \(\mathcal{E}\) is the induced EMF, measured 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 interval, measured in seconds (s).

Let’s see how this works.

Suppose the illustrative flux graph has the following values:

The coil has 100 turns.

There are three straight-line sections:

Notice that the middle section has zero induced EMF, even though the magnetic flux is at its maximum.

The first and third sections produce equal EMF magnitudes because their gradients have equal magnitudes. Their signs are opposite because the flux changes in opposite directions.

The key idea is simple: the gradient of a magnetic-flux-time graph gives the rate of change of magnetic flux. Faraday’s law converts that gradient into an induced EMF.

However, before predicting the direction of an induced current, there is another important detail: you must define the loop and its orientation.

01Why changing magnetic flux produces an EMF

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

Faraday wanted to investigate whether magnetism could produce an electric current without a battery directly connected to the circuit.

His experiments revealed something important: a magnetic field alone was not enough. The magnetic flux linking a circuit had to change.

Faraday’s iron-ring experiment

Faraday constructed an apparatus using an iron ring with two separate insulated wire coils wound around it.

The first coil, called the primary coil, was connected to a battery and a switch. The second coil, called the secondary coil, was connected to a sensitive galvanometer, which detects electric current.

The two coils were electrically separate. There was no conducting connection between them.

What did Faraday expect to happen?

When the switch was closed, current would begin flowing through the primary coil. This would create a changing magnetic field in the iron ring, changing the magnetic flux linking the secondary coil.

If this changing flux induced an EMF, the galvanometer should briefly deflect.

That is what happened.

But the more interesting observations came next.

1. Switch closed: The galvanometer briefly deflected as the primary current increased and the magnetic flux changed.
2. Switch left closed: The galvanometer returned to zero, even though the primary coil still carried current and the iron ring remained magnetised.
3. Switch opened: The galvanometer briefly deflected in the opposite direction as the magnetic flux decreased.

The experiment showed that a steady magnetic flux does not produce a sustained induced EMF. It is the change in flux that matters.

Faraday also investigated the effect of moving a bar magnet into and out of a coil connected to a galvanometer. Moving the magnet into the coil produced a deflection, holding it stationary produced no continuing deflection, and withdrawing it produced a deflection in the opposite direction.

What does this experiment actually measure?

There is an important distinction between what Faraday’s galvanometer detected and what we calculate from a flux graph.

The galvanometer detected induced current in the closed secondary circuit. It did not directly display the induced EMF.

An induced EMF can exist even when the circuit is open. However, without a complete conducting path, there is no sustained current.

This distinction becomes important when solving HSC questions. Always calculate the induced EMF first. Only calculate the induced current if the circuit is closed and you have enough information about its resistance.

02Before calculating: define the loop and magnetic change

Faraday’s law gives you the magnitude and sign of the induced EMF, but the sign only makes sense once you have chosen a reference direction.

Consider a flat wire loop in a magnetic field.

Magnetic flux measures how much magnetic field passes through the area enclosed by the loop. For a uniform magnetic field, it is calculated using:

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

where:

• \(B\) is magnetic field strength in teslas (T).
• \(A\) is the area enclosed by the loop in square metres (m²).
• \(\theta\) is the angle between the magnetic field and the normal to the loop.

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

For a piecewise flux graph, you are already given the magnetic flux, so you do not need to calculate \(BA\cos\theta\) again.

However, you still need to understand the orientation of the loop.

Choose a positive direction

Imagine looking directly at a circular coil, with its normal pointing out of the page towards you.

Using the right-hand grip rule, anticlockwise current is defined as positive. Clockwise current is negative.

This gives you a consistent way to interpret the sign of the induced EMF.

• Positive EMF: acts in the positive, anticlockwise direction.
• Negative EMF: acts in the negative, clockwise direction.
• Zero EMF: no induced current in an ideal purely resistive closed circuit.

If you reverse your chosen normal, all the mathematical signs reverse too. The physical prediction does not change.

EMF and current are not the same thing

This is one of the easiest places to lose marks.

An induced EMF is the electrical potential difference produced by changing magnetic flux. Current is the flow of charge through a conducting path.

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

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

where:

• \(I\) is current in amperes (A).
• \(\mathcal{E}\) is induced EMF in volts (V).
• \(R\) is total circuit resistance in ohms (Ω).

If the circuit is open, the induced EMF can still exist, but the current is zero in the ideal circuit model.

For the HSC calculations below, we assume a purely resistive circuit and neglect self-inductance unless stated otherwise.

03How to calculate induced EMF from each section of a piecewise graph

Now we can build a reliable method.

Step 1: Check what the vertical axis represents

Look carefully at the graph label.

There are two possibilities.

Case 1: The graph shows magnetic flux per turn, \(\Phi\).

Use:

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

You must multiply the gradient by the number of turns.

Case 2: The graph shows magnetic flux linkage, \(N\Phi\).

Flux linkage already includes the number of turns.

Use:

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

Do not multiply by \(N\) again.

This is a common source of incorrect answers. Read the vertical axis before substituting anything.

Step 2: Identify the endpoints of each straight section

A piecewise graph contains multiple sections with different gradients.

Treat each straight section as a separate calculation.

For each section, identify:

• The initial time, \(t_1\).
• The final time, \(t_2\).
• The initial flux, \(\Phi_1\).
• The final flux, \(\Phi_2\).

Then calculate:

\[ \text{Gradient}= \frac{\Phi_2-\Phi_1}{t_2-t_1} \]

Use webers and seconds when substituting.

Remember:

\[ 1\text{ mWb}=1\times10^{-3}\text{ Wb} \]

Step 3: Apply Faraday’s law

Multiply the gradient by the number of turns and include the negative sign.

\[ \mathcal{E}=-N\times\text{Gradient} \]

Repeat this for every straight section.

For a piecewise-linear graph, the induced EMF is constant within each section because the gradient is constant.

Step 4: Interpret the sign

The negative sign in Faraday’s law comes from Lenz’s law.

Lenz’s law states that the induced current produces a magnetic field that opposes the change in magnetic flux that caused it.

For a chosen positive normal:

Here, increasing and decreasing refer to the signed numerical value of flux.

For example, flux changing from \(-0.20\text{ mWb}\) to \(+0.10\text{ mWb}\) is increasing, even though it crosses zero.

The induced current does not automatically reverse just because the flux changes direction.

Step 5: Calculate current only if required

If the question asks for induced current and provides the total circuit resistance, use:

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

Keep the sign.

A negative current means the actual current flows opposite to your chosen positive direction.

What happens at the corners of the graph?

At a corner, the gradient changes abruptly.

This means the induced EMF changes from one value to another.

For example, a graph might have:

• A gradient of \(+0.004\text{ Wb/s}\) before a corner.
• A gradient of zero after the corner.

The induced EMF changes from \(-0.004N\text{ V}\) to zero.

Strictly, the derivative at the exact corner of an ideal piecewise-linear graph is undefined. In an actual physical experiment, the transition occurs over a finite time.

For HSC calculations, use the gradient of the relevant section rather than trying to assign one value to the exact corner.

For a broader introduction to interpreting the graph itself, see How to Read a Magnetic-Flux-Time Graph to Calculate EMF.

The underlying physical explanation is covered in Why Induced EMF Depends on Changing Magnetic Flux.

04Worked examples

05Common mistakes when calculating induced EMF

Before attempting exam questions, make sure you can avoid these five errors.

06Questions and solutions

The following questions progress from a straightforward HSC-style calculation to more demanding problems involving circuit resistance, flux linkage, and the relationship between induced EMF and total charge.

All numerical data are illustrative.

Question 1

A 40-turn coil is placed in a changing magnetic field. Its positive normal points out of the page, and positive current is defined as anticlockwise.

The coil is connected to an \(8.0\,\Omega\) resistor.

The magnetic flux per turn follows this piecewise graph:

Calculate:

a) The induced EMF during each interval.

b) The induced current during each interval.

c) The direction of the induced current during the first interval.

Solution 1

The induced EMF is \(-0.12\text{ V}\) during the first interval and zero during the second interval. The corresponding currents are \(-0.015\text{ A}\) and zero.

Part a: Calculate the induced EMF.

During the first interval, the flux changes from \(+0.20\text{ mWb}\) to \(+0.50\text{ mWb}\).

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

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

The time interval is \(0.10\text{ s}\).

Applying Faraday’s law:

\[ \begin{aligned} \mathcal{E} &=-40\frac{3.0\times10^{-4}}{0.10}\\ &=-0.12\text{ V} \end{aligned} \]

During the second interval, the flux remains constant.

\[ \mathcal{E}=-40\frac{0}{0.20} \]

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

Part b: Calculate the induced current.

Using Ohm’s law:

First interval:

\[ I=\frac{-0.12}{8.0} \]

\[ I=-0.015\text{ A} \]

Second interval:

\[ I=\frac{0}{8.0}=0\text{ A} \]

Part c: Determine the current direction.

The magnetic flux increases in the positive-normal direction during the first interval.

The induced magnetic field must oppose this increase, so it points into the page.

Using the right-hand grip rule, a clockwise current produces a magnetic field into the page.

Therefore, the induced current is clockwise.

The negative sign in the calculated current confirms that it flows opposite to the chosen positive direction.

Question 2

Two separate coils, A and B, have the same area and orientation. They are placed in identical changing magnetic fields, so the magnetic flux per turn is the same for both coils.

The positive normal points out of the page.

The illustrative flux per turn decreases linearly from \(+0.40\text{ mWb}\) to \(-0.20\text{ mWb}\) over \(0.15\text{ s}\).

The coils have the following properties:

Both circuits are closed.

Calculate:

a) The induced EMF in each coil.

b) The induced current in each coil.

c) Which coil produces the greater EMF and which produces the greater current.

Explain why the coil with more turns does not necessarily produce the greater current.

Solution 2

Coil B produces twice the EMF of coil A, but coil A produces twice the current. The difference comes from their circuit resistances.

Part a: Calculate the induced EMF.

First calculate the change in flux per turn:

\[ \Delta\Phi=(-0.20-0.40)\times10^{-3} \]

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

The time interval is \(0.15\text{ s}\).

The flux gradient is:

\[ \frac{\Delta\Phi}{\Delta t} = \frac{-6.0\times10^{-4}}{0.15} \]

\[ =-4.0\times10^{-3}\text{ Wb/s} \]

For coil A:

\[ \mathcal{E}_A=-100(-4.0\times10^{-3}) \]

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

For coil B:

\[ \mathcal{E}_B=-200(-4.0\times10^{-3}) \]

\[ \boxed{\mathcal{E}_B=+0.80\text{ V}} \]

Part b: Calculate the induced current.

For coil A:

\[ I_A=\frac{0.40}{5.0} \]

\[ \boxed{I_A=0.080\text{ A}} \]

For coil B:

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

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

Part c: Compare the results.

Coil B produces the greater EMF because it has twice as many turns.

However, its resistance is four times greater.

Although the EMF doubles, the resistance increases by a factor of four. Consequently, its current is halved.

Both currents flow in the positive direction, which is anticlockwise.

The tempting assumption is that doubling the number of turns must double the current. That would only follow if the resistance remained unchanged.

The correct relationship is:

\[ I=\frac{N}{R}\left|\frac{\Delta\Phi}{\Delta t}\right| \]

for the magnitude of the current in this simple resistive circuit.

The number of turns and the total circuit resistance both matter.

Question 3

An 80-turn coil is placed in a magnetic field. Its positive normal points out of the page.

The coil is initially connected to an open switch. The switch can connect the coil to a \(10\,\Omega\) resistor.

The magnetic flux per turn changes as follows:

The switch remains open from \(0.00\text{ s}\) until \(0.20\text{ s}\). It is then closed just after \(0.20\text{ s}\), when the flux is constant.

Assume that closing the switch does not itself change the magnetic flux.

a) Calculate the induced EMF during the first interval.

b) Calculate the current during the first interval.

c) Calculate the induced EMF and current after the switch closes.

d) Explain why closing the switch does not produce a current in this situation.

Solution 3

The coil has an induced EMF of \(-0.20\text{ V}\) during the first interval, but its current is zero because the switch is open. After the switch closes, both the induced EMF and current are zero.

Part a: Calculate the induced EMF during the first interval.

The change in flux per turn is:

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

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

Using Faraday’s law:

\[ \begin{aligned} \mathcal{E} &=-80\frac{0.50\times10^{-3}}{0.20}\\ &=-0.20\text{ V} \end{aligned} \]

The induced EMF is \(-0.20\text{ V}\).

Part b: Calculate the current during the first interval.

The switch is open.

This means there is no complete conducting path through the resistor.

Therefore:

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

The induced EMF still exists across the coil terminals. An ideal high-impedance voltmeter connected across the terminals would detect the corresponding potential difference.

Part c: Calculate the EMF and current after the switch closes.

After \(0.20\text{ s}\), the magnetic flux remains constant.

Therefore:

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

Using Faraday’s law:

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

Since the switch is now closed:

\[ I=\frac{0}{10} \]

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

Part d: Explain why closing the switch does not produce current.

The tempting prediction is that closing the circuit must allow the previously induced EMF to drive a current.

However, the induced EMF was produced by the changing magnetic flux during the first interval. By the time the switch closes, that change has stopped.

The coil still has magnetic flux passing through it, but the flux is constant.

Closing the switch completes the conducting path, but it does not create a changing magnetic flux. Therefore, there is no induced EMF to drive a current.

This demonstrates why you must consider both the magnetic change and whether the circuit is closed.

Question 4

A 150-turn coil is connected to a \(4.0\,\Omega\) resistor.

A piecewise graph records the magnetic flux linkage, \(\lambda=N\Phi\), rather than the magnetic flux per turn.

The illustrative data are:

The positive normal points out of the page, and the circuit is closed.

a) Calculate the induced EMF during each interval.

b) Calculate the current during each interval.

c) Calculate the magnetic flux per turn at \(t=0.00\text{ s}\) and \(t=0.35\text{ s}\).

d) A student multiplies the gradient of the graph by 150 to calculate the EMF. Explain the error.

Solution 4

The induced EMFs are \(+0.36\text{ V}\), \(0\text{ V}\), and \(+0.60\text{ V}\). The corresponding currents are \(+0.090\text{ A}\), zero, and \(+0.150\text{ A}\).

Part a: Calculate the induced EMF.

The graph shows flux linkage:

\[ \lambda=N\Phi \]

Therefore, we use:

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

We do not multiply by the number of turns again.

For the first interval:

\[ \begin{aligned} \mathcal{E} &=-\frac{0.024-0.060}{0.10}\\ &=+0.36\text{ V} \end{aligned} \]

For the second interval, flux linkage is constant:

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

For the third interval:

\[ \begin{aligned} \mathcal{E} &=-\frac{-0.036-0.024}{0.10}\\ &=+0.60\text{ V} \end{aligned} \]

Part b: Calculate the induced current.

Using Ohm’s law:

First interval:

\[ I=\frac{0.36}{4.0} \]

\[ I=+0.090\text{ A} \]

Second interval:

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

Third interval:

\[ I=\frac{0.60}{4.0} \]

\[ I=+0.150\text{ A} \]

The positive signs indicate that the current flows anticlockwise.

Part c: Calculate the magnetic flux per turn.

We know:

\[ \lambda=N\Phi \]

Rearranging:

\[ \Phi=\frac{\lambda}{N} \]

At \(t=0.00\text{ s}\):

\[ \Phi=\frac{0.060}{150} \]

\[ \boxed{\Phi=+4.0\times10^{-4}\text{ Wb}} \]

At \(t=0.35\text{ s}\):

\[ \Phi=\frac{-0.036}{150} \]

\[ \boxed{\Phi=-2.4\times10^{-4}\text{ Wb}} \]

Part d: Explain the student’s error.

The student has multiplied by the number of turns twice.

The graph already represents flux linkage, which includes the number of turns:

\[ \lambda=N\Phi \]

Faraday’s law can therefore be written directly as:

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

Multiplying by 150 again would produce an incorrect EMF.

Always check whether the graph shows magnetic flux per turn or total flux linkage before applying Faraday’s law.

Question 5

A 100-turn coil is connected to a \(5.0\,\Omega\) resistor. The circuit remains closed throughout the experiment.

The positive normal points out of the page, and positive current is defined as anticlockwise.

The magnetic flux per turn follows this illustrative piecewise-linear graph:

Assume the total resistance remains constant and self-inductance is negligible, so that \(I=\mathcal{E}/R\) throughout each interval.

A student looks at the initial and final flux values and argues:

“The magnetic flux ends at exactly the value it started at. Therefore, there must have been no net induced current, and no charge was transferred through the resistor.”

Determine whether this statement is correct.

Your response must:

a) Calculate the induced EMF and current during all three intervals.

b) Calculate the net signed charge transferred through the resistor over the entire experiment.

c) Calculate the total amount of charge that passes through the resistor, regardless of direction.

d) Suppose the middle section of the flux graph is stretched to twice its original duration, while its initial and final flux values remain unchanged. The other sections are unchanged. Determine what happens to the middle-section EMF, current, and charge transferred during that section.

Solution 5

The student’s statement is incorrect. The net signed charge transferred is zero, but a total of \(0.032\text{ C}\) passes through the resistor. The current flows in both directions during the experiment.

This question requires us to distinguish between the net charge transferred and the total amount of charge that moves.

Part a: Calculate the EMF and current during all three intervals.

First interval: 0.00-0.10 s.

The flux changes from \(+0.20\text{ mWb}\) to \(+0.60\text{ mWb}\).

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

Using Faraday’s law:

\[ \begin{aligned} \mathcal{E}_1 &=-100\frac{0.40\times10^{-3}}{0.10}\\ &=-0.40\text{ V} \end{aligned} \]

The current is:

\[ I_1=\frac{-0.40}{5.0} \]

\[ I_1=-0.080\text{ A} \]

The current flows clockwise.

Second interval: 0.10-0.30 s.

The flux changes from \(+0.60\text{ mWb}\) to \(-0.20\text{ mWb}\).

\[ \Delta\Phi=-0.80\times10^{-3}\text{ Wb} \]

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

\[ \begin{aligned} \mathcal{E}_2 &=-100\frac{-0.80\times10^{-3}}{0.20}\\ &=+0.40\text{ V} \end{aligned} \]

The current is:

\[ I_2=\frac{0.40}{5.0} \]

\[ I_2=+0.080\text{ A} \]

The current flows anticlockwise.

Third interval: 0.30-0.50 s.

The flux changes from \(-0.20\text{ mWb}\) to \(+0.20\text{ mWb}\).

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

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

\[ \begin{aligned} \mathcal{E}_3 &=-100\frac{0.40\times10^{-3}}{0.20}\\ &=-0.20\text{ V} \end{aligned} \]

The current is:

\[ I_3=\frac{-0.20}{5.0} \]

\[ I_3=-0.040\text{ A} \]

The current flows clockwise.

We can summarise the results:

The current clearly flows in both directions.

Part b: Calculate the net signed charge transferred.

The charge transferred during a constant-current interval is:

\[ Q=I\Delta t \]

Because current changes direction, we must retain its sign.

For the first interval:

\[ Q_1=(-0.080)(0.10) \]

\[ Q_1=-0.008\text{ C} \]

For the second interval:

\[ Q_2=(+0.080)(0.20) \]

\[ Q_2=+0.016\text{ C} \]

For the third interval:

\[ Q_3=(-0.040)(0.20) \]

\[ Q_3=-0.008\text{ C} \]

The net signed charge is:

\[ \begin{aligned} Q_{\text{net}} &=Q_1+Q_2+Q_3\\ &=-0.008+0.016-0.008 \end{aligned} \]

\[ \boxed{Q_{\text{net}}=0\text{ C}} \]

The student is correct that the net signed charge is zero.

However, that does not mean no charge moved.

Part c: Calculate the total amount of charge transferred.

To calculate the total amount of charge that passed through the resistor, regardless of direction, add the magnitudes of the charges from each interval.

\[ Q_{\text{total}}= |Q_1|+|Q_2|+|Q_3| \]

\[ Q_{\text{total}}=0.008+0.016+0.008 \]

\[ \boxed{Q_{\text{total}}=0.032\text{ C}} \]

This is the important distinction.

The positive and negative charge transfers cancel when calculating the net signed charge, but they do not disappear from the total amount of charge that moved.

There is another useful way to understand this result.

Since:

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

and:

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

we can write:

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

Rearranging for the charge transferred:

\[ Q=I\Delta t \]

\[ \boxed{Q=-\frac{N}{R}\Delta\Phi} \]

This means the signed charge transferred during an interval depends on the change in magnetic flux, provided the resistance is constant and self-inductance is negligible.

For the entire experiment, the initial and final flux are equal. Therefore, the net signed charge must be zero.

But the flux changes direction along the way, so the total amount of charge transferred is not zero.

Part d: What happens if the middle section takes twice as long?

The original middle interval lasts \(0.20\text{ s}\).

Its flux change remains:

\[ \Delta\Phi=-0.80\times10^{-3}\text{ Wb} \]

If the duration doubles:

\[ \Delta t=0.40\text{ s} \]

The new induced EMF is:

\[ \begin{aligned} \mathcal{E}_2' &=-100\frac{-0.80\times10^{-3}}{0.40}\\ &=+0.20\text{ V} \end{aligned} \]

The new current is:

\[ I_2'=\frac{0.20}{5.0} \]

\[ I_2'=+0.040\text{ A} \]

The EMF and current have both halved.

However, the charge transferred during this section is:

\[ Q_2'=I_2'\Delta t \]

\[ Q_2'=(0.040)(0.40) \]

\[ \boxed{Q_2'=+0.016\text{ C}} \]

This is exactly the same charge as before.

The current is half as large, but it flows for twice as long.

The total signed charge and total amount of charge transferred across the complete experiment therefore remain unchanged.

This is a useful result:

• Induced EMF depends on how quickly magnetic flux changes.
• Current depends on the induced EMF and circuit resistance.
• The charge transferred over a section depends on the net flux change when resistance is constant and self-inductance is negligible.

A student who only looks at the steepness of the graph might correctly predict the new EMF but incorrectly predict the total charge transferred.

That is why it is important to distinguish between a rate of change and the total change itself.

Once you can calculate induced EMF from a supplied piecewise graph, the next step is handling questions where the graph is not provided. Instead, you are given a magnetic field, a coil’s area, and its orientation, and must first calculate the magnetic flux using \(\Phi=BA\cos\theta\). You can then construct the flux-time relationship and apply exactly the same gradient method.