How to Read a Magnetic-Flux-Time Graph to Calculate EMF

Two identical coils each experience a change in magnetic flux per turn from \(6.0\,\text{mWb}\) to \(2.0\,\text{mWb}\). In coil A, the change takes \(0.10\,\text{s}\). In coil B, it takes \(0.40\,\text{s}\).

Before calculating, predict which coil produces the larger induced EMF. Is it four times larger, or does the fact that both coils experience the same flux change make their EMFs equal?

Coil A produces four times the EMF magnitude. Both coils experience the same change in flux, but A experiences it in one-quarter of the time. Its graph is four times as steep.

The key idea is simple: induced EMF depends on how quickly magnetic flux changes, not just on how much it changes. Once you can read the gradient of a magnetic-flux-time graph, you can calculate the induced EMF directly.

01What the graph is telling you

A magnetic-flux-time graph has two axes:

• The vertical axis shows magnetic flux, represented by \(\Phi_B\).
• The horizontal axis shows time, represented by \(t\).

Magnetic flux describes how much magnetic field passes through a loop, taking the loop’s orientation into account. Its SI unit is the weber (Wb).

For a graph-reading question, the important distinction is between the height of the graph and its gradient.

The height tells you the flux at a particular instant. The gradient tells you how quickly that flux is changing.

Imagine a graph sitting at a constant flux of \(+5.0\,\text{mWb}\). The flux is not zero, but it is not changing either. The induced EMF is therefore zero.

Now imagine the graph dropping steeply. Flux is changing rapidly, so the magnitude of the induced EMF is larger.

This gives you an immediate rule:

• A horizontal section means zero induced EMF.
• A straight, sloping section means constant induced EMF magnitude.
• A steeper section means a larger induced EMF magnitude.

The graph can also sit below the time axis. That means the flux is negative relative to the chosen reference direction. It does not automatically mean the induced EMF is negative or that the induced current must reverse.

02From gradient to induced EMF

The gradient of a graph is its vertical change divided by its horizontal change. For a magnetic-flux-time graph,

\[ \text{gradient}=\frac{\Delta\Phi_B}{\Delta t} \]

Here:

• \(\Delta\Phi_B\) is the change in magnetic flux, in webers (Wb).
• \(\Delta t\) is the elapsed time, in seconds (s).

Always calculate the change as final value minus initial value:

\[ \Delta\Phi_B=\Phi_{B,f}-\Phi_{B,i} \]

This preserves the sign of the gradient, which becomes important when determining direction.

Faraday’s law relates the rate of change of flux to the induced EMF in a coil:

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

In this equation, \(\varepsilon\) is the induced EMF in volts (V), and \(N\) is the number of turns in the coil.

The minus sign represents Lenz’s law. It tells you about the polarity of the induced EMF, not its magnitude.

If the question asks only for the magnitude, use:

\[ \boxed{|\varepsilon|=N|\text{gradient}|} \]

Because \(1\,\text{Wb}=1\,\text{V\,s}\), a gradient measured in \(\text{Wb\,s}^{-1}\) has units of volts. The factor \(N\) accounts for the contribution of all the turns in the coil.

One important detail: this formula assumes the graph gives magnetic flux through each turn. If the vertical axis already shows flux linkage, \(N\Phi_B\), its gradient already includes the number of turns. Do not multiply by \(N\) again.

03Flat, sloping, and piecewise graph sections

The easiest way to read a graph is to break it into straight sections and calculate each section separately.

1. A horizontal section

If the flux remains constant, then

\[ \Delta\Phi_B=0 \]

Therefore,

\[ |\varepsilon|=N\left|\frac{0}{\Delta t}\right|=0 \]

It makes no difference whether the constant flux is positive, negative, or large in magnitude. If it does not change, there is no induced EMF from that flux.

2. A straight, sloping section

A straight line has a constant gradient. The flux changes at a constant rate, so the induced EMF magnitude is also constant over that section.

A steep line gives a larger EMF than a shallow line, provided the coils have the same number of turns.

The sign of the gradient tells you whether the signed flux is increasing or decreasing. Its absolute value tells you how steep the line is for the magnitude calculation.

3. A piecewise graph

A piecewise graph has sections with different gradients. The induced EMF can change from one section to the next.

Do not calculate one gradient using the endpoints of the entire graph if the question asks for the EMF during individual sections. That would give an average rate over the whole interval and could hide much larger EMFs during short, steep sections.

Predict this: If a sloping line crosses the horizontal axis, must the induced EMF momentarily become zero?

No. The horizontal axis marks zero flux, not zero gradient. If the line is still sloping as it crosses the axis, the induced EMF is still non-zero.

06Magnitude and direction are different jobs

Calculating the magnitude of an induced EMF is a gradient problem. Determining the direction of the induced current is a separate physics problem.

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

For example, suppose the applied magnetic flux through a loop points into the page.

• If that flux into the page is increasing, the induced magnetic field points out of the page.
• If that flux into the page is decreasing, the induced magnetic field points into the page.

Once you know the direction of the induced magnetic field, you can use the right-hand grip rule to determine the current direction, provided the coil’s orientation is given.

This is where a common misconception causes trouble. Students sometimes assume that a negative flux value means the current must flow in the opposite direction. It does not.

Imagine the flux increasing from \(-5.0\,\text{mWb}\) to \(-1.0\,\text{mWb}\). The flux is negative, but it is increasing because it is becoming less negative. Its gradient is positive.

The relevant information is how the flux changes, not whether the graph sits above or below the horizontal axis.

If the signed flux and the positive loop direction are defined, the sign of the gradient determines the induced EMF polarity relative to that convention. However, a flux-time graph alone does not tell you whether the current is clockwise or anticlockwise unless the coil’s orientation and reference direction are also specified.

There is one more distinction worth keeping clear: an induced EMF does not guarantee that a current flows. The circuit must be closed for an induced current to flow, and the current magnitude depends on the circuit’s electrical properties. For more on this, see Induced EMF vs Current: Avoid a Common HSC Physics Mistake.

07A reliable exam method

When a question gives you a magnetic-flux-time graph, use this sequence.

1. Check the vertical axis. Is it magnetic flux per turn, or flux linkage?
2. Identify the interval. If the graph is piecewise, work on one straight section at a time.
3. Read the endpoints. Calculate final flux minus initial flux, and convert mill webers to webers.
4. Calculate the gradient. Divide the flux change by the elapsed time, then multiply its magnitude by \(N\).
5. Handle direction separately. If the question asks for current direction, apply Lenz’s law and use the given orientation.

For a curved graph, the gradient between two points gives the average rate of change over that interval. The gradient of the tangent at a particular instant gives the instantaneous rate of change.

The distinction matters because the EMF at an instant depends on the local gradient, not the average gradient over a much longer interval.

08Questions and solutions

Question 1

An 80-turn coil experiences a uniform decrease in magnetic flux per turn from \(3.0\,\text{mWb}\) to \(1.0\,\text{mWb}\) over \(0.050\,\text{s}\).

Calculate the magnitude of the induced EMF.

Solution 1

The induced EMF magnitude is \(3.2\,\text{V}\).

First, calculate the change in flux:

\[ \begin{aligned} \Delta\Phi_B &=(1.0-3.0)\,\text{mWb}\\ &=-2.0\times10^{-3}\,\text{Wb} \end{aligned} \]

The gradient is

\[ \begin{aligned} \text{gradient} &=\frac{-2.0\times10^{-3}\,\text{Wb}} {0.050\,\text{s}}\\ &=-0.040\,\text{Wb\,s}^{-1} \end{aligned} \]

Now multiply the gradient magnitude by the number of turns:

\[ |\varepsilon|=80(0.040)=3.2\,\text{V} \]

The result is a constant \(3.2\,\text{V}\) magnitude because the flux changes at a constant rate. The negative gradient indicates decreasing signed flux; it is not a negative EMF magnitude.

Question 2

Coil A has 120 turns, and its magnetic flux per turn decreases from \(+5.0\,\text{mWb}\) to \(+1.0\,\text{mWb}\) in \(0.20\,\text{s}\).

Coil B has 200 turns and experiences the same flux change over \(0.50\,\text{s}\).

A student claims that coil B must have the greater induced EMF because it has more turns. Calculate the EMF magnitude for each coil and assess the claim.

Solution 2

Coil A produces \(2.4\,\text{V}\), while coil B produces \(1.6\,\text{V}\). Coil A therefore produces an EMF magnitude 1.5 times that of coil B, so the student’s claim is incorrect.

Both coils experience a flux change of

\[ \Delta\Phi_B=-4.0\times10^{-3}\,\text{Wb} \]

For coil A, the gradient is

\[ \text{gradient}_A =\frac{-0.0040}{0.20} =-0.020\,\text{Wb\,s}^{-1} \]

Therefore,

\[ |\varepsilon_A|=120(0.020)=2.4\,\text{V} \]

For coil B, the gradient is

\[ \text{gradient}_B =\frac{-0.0040}{0.50} =-0.0080\,\text{Wb\,s}^{-1} \]

Therefore,

\[ |\varepsilon_B|=200(0.0080)=1.6\,\text{V} \]

Coil B has more turns, but its flux changes more slowly. The extra turns are not enough to compensate for the lower rate of change.

The governing relationship is \( |\varepsilon|=N|\text{gradient}| \). You must consider both the number of turns and the gradient.

Question 3

An 180-turn coil has the following magnetic-flux-per-turn graph:

• At \(t=0\), the flux is \(+2.0\,\text{mWb}\).
• At \(t=0.10\,\text{s}\), the flux is \(+6.0\,\text{mWb}\).
• The flux remains at \(+6.0\,\text{mWb}\) until \(t=0.30\,\text{s}\).
• It then decreases linearly to \(-2.0\,\text{mWb}\) at \(t=0.35\,\text{s}\).

Calculate the EMF magnitude in each of the three intervals. Explain whether the EMF becomes zero when the flux crosses zero.

Solution 3

The EMF magnitudes are \(7.2\,\text{V}\), \(0\,\text{V}\), and \(28.8\,\text{V}\). The EMF does not become zero when the flux crosses zero because the graph is still steeply sloping.

For the first interval, the flux increases by

\[ \Delta\Phi_B=4.0\times10^{-3}\,\text{Wb} \]

over \(0.10\,\text{s}\). Therefore,

\[ |\varepsilon_1| =180\left(\frac{0.0040}{0.10}\right) =7.2\,\text{V} \]

For the second interval, the flux is constant. Its gradient is zero, so

\[ |\varepsilon_2|=0\,\text{V} \]

For the third interval, the flux decreases by \(8.0\,\text{mWb}\), or \(8.0\times10^{-3}\,\text{Wb}\), in just \(0.050\,\text{s}\).

\[ |\varepsilon_3| =180\left(\frac{0.0080}{0.050}\right) =28.8\,\text{V} \]

The final gradient is much steeper than the first, giving a much larger EMF. When the flux crosses zero during this final section, the gradient remains unchanged. The EMF magnitude therefore remains \(28.8\,\text{V}\).

Question 4

A 150-turn coil has a signed magnetic-flux-time graph that rises linearly from \(-5.0\,\text{mWb}\) at \(t=0\) to \(-1.0\,\text{mWb}\) at \(t=0.20\,\text{s}\). It then rises linearly to \(+3.0\,\text{mWb}\) at \(t=0.30\,\text{s}\).

Assume the signed flux uses the same reference direction throughout. The coil is closed, its orientation remains fixed, and the usual HSC idealisation is used so that the current direction follows the induced EMF polarity.

Calculate the EMF magnitude in each interval. Does the induced current reverse direction when the flux crosses zero? Explain.

Solution 4

The EMF magnitudes are \(3.0\,\text{V}\) and \(6.0\,\text{V}\). The induced current does not reverse direction when the flux crosses zero because the gradient remains positive throughout both intervals.

During the first interval, the flux changes by

\[ \Delta\Phi_B=+4.0\times10^{-3}\,\text{Wb} \]

over \(0.20\,\text{s}\). Thus,

\[ \begin{aligned} \text{gradient}_1 &=\frac{0.0040}{0.20}\\ &=+0.020\,\text{Wb\,s}^{-1} \end{aligned} \]

The EMF magnitude is

\[ |\varepsilon_1|=150(0.020)=3.0\,\text{V} \]

During the second interval, the flux increases by the same amount in half the time:

\[ \begin{aligned} \text{gradient}_2 &=\frac{0.0040}{0.10}\\ &=+0.040\,\text{Wb\,s}^{-1} \end{aligned} \]

Therefore,

\[ |\varepsilon_2|=150(0.040)=6.0\,\text{V} \]

The second gradient is twice as large, so the EMF magnitude doubles.

For direction, Faraday’s law gives \(\varepsilon=-N(\Delta\Phi_B/\Delta t)\). Both gradients are positive, so the signed EMF has the same polarity in both intervals. The induced current therefore keeps the same direction under the stated assumptions.

The tempting mistake is to assume that crossing from negative flux to positive flux reverses the current. It does not. The current direction is linked to the direction of flux change, not the sign of the flux itself.

Question 5

Two 100-turn coils, A and B, experience different magnetic-flux histories. Both start with a flux per turn of \(+4.0\,\text{mWb}\), finish at \(-4.0\,\text{mWb}\), and take \(0.40\,\text{s}\) to do so.

Assume each graph consists of straight sections, the signed flux uses the same reference direction, and both coils are closed circuits with identical resistance and negligible self-inductance.

• Coil A: The flux decreases uniformly from \(+4.0\,\text{mWb}\) at \(t=0\) to \(-4.0\,\text{mWb}\) at \(t=0.40\,\text{s}\).
• Coil B: The flux decreases from \(+4.0\,\text{mWb}\) to \(-2.0\,\text{mWb}\) in the first \(0.10\,\text{s}\), increases from \(-2.0\,\text{mWb}\) to \(+2.0\,\text{mWb}\) over the next \(0.20\,\text{s}\), and then decreases to \(-4.0\,\text{mWb}\) over the final \(0.10\,\text{s}\).

A student claims that because both coils have the same initial flux, final flux, number of turns, and total time, their induced EMFs must be identical.

Calculate the signed EMF in each section of coil B and the EMF in coil A. Compare their time-averaged signed EMFs and their average EMF magnitudes. Finally, determine when coil B’s current reverses direction, and whether those reversals occur when its flux crosses zero.

Solution 5

The student’s claim is incorrect. The coils have the same time-averaged signed EMF, but their instantaneous EMFs, average EMF magnitudes, and current-direction histories are different.

The crucial distinction is between the net change in flux over the full interval and the rate of change at each instant.

Step 1: Calculate the EMF in coil A.

The total flux change is

\[ \Delta\Phi_B=-8.0\times10^{-3}\,\text{Wb} \]

Since the change occurs uniformly over \(0.40\,\text{s}\),

\[ \begin{aligned} \text{gradient}_A &=\frac{-0.0080}{0.40}\\ &=-0.020\,\text{Wb\,s}^{-1} \end{aligned} \]

Therefore, the signed EMF is

\[ \varepsilon_A=-100(-0.020)=+2.0\,\text{V} \]

Coil A has a constant signed EMF of \(+2.0\,\text{V}\) throughout the interval. Its time-averaged signed EMF and average magnitude are both \(2.0\,\text{V}\).

Step 2: Calculate the signed EMF in each section of coil B.

In the first \(0.10\,\text{s}\), the flux changes from \(+4.0\) to \(-2.0\,\text{mWb}\).

\[ \begin{aligned} \text{gradient}_1 &=\frac{-0.0060}{0.10}\\ &=-0.060\,\text{Wb\,s}^{-1} \end{aligned} \]

Therefore,

\[ \varepsilon_1=-100(-0.060)=+6.0\,\text{V} \]

In the middle \(0.20\,\text{s}\), the flux rises from \(-2.0\) to \(+2.0\,\text{mWb}\).

\[ \begin{aligned} \text{gradient}_2 &=\frac{+0.0040}{0.20}\\ &=+0.020\,\text{Wb\,s}^{-1} \end{aligned} \]

Therefore,

\[ \varepsilon_2=-100(+0.020)=-2.0\,\text{V} \]

In the final \(0.10\,\text{s}\), the flux falls from \(+2.0\) to \(-4.0\,\text{mWb}\).

\[ \begin{aligned} \text{gradient}_3 &=\frac{-0.0060}{0.10}\\ &=-0.060\,\text{Wb\,s}^{-1} \end{aligned} \]

Therefore,

\[ \varepsilon_3=-100(-0.060)=+6.0\,\text{V} \]

Coil B’s signed EMF is \(+6.0\,\text{V}\), then \(-2.0\,\text{V}\), then \(+6.0\,\text{V}\). Unlike coil A, its EMF changes both magnitude and polarity.

Step 3: Compare the time-averaged signed EMFs.

For coil B, calculate a time-weighted average. The negative middle section subtracts from the positive contributions.

\[ \begin{aligned} \bar{\varepsilon}_B &=\frac{(6.0)(0.10)+(-2.0)(0.20)}{0.40}\\ &\quad+\frac{(6.0)(0.10)}{0.40}\\ &=+2.0\,\text{V} \end{aligned} \]

The time-averaged signed EMF is \(+2.0\,\text{V}\) for both coils.

This equality is not a coincidence. The average signed EMF over a fixed interval is determined by the net flux change:

\[ \bar{\varepsilon} =-N\frac{\Phi_{B,f}-\Phi_{B,i}}{\Delta t} \]

Both coils have the same number of turns, the same net flux change, and the same total time. Therefore, they must have the same time-averaged signed EMF.

Step 4: Compare their average EMF magnitudes.

For coil A, the magnitude is constantly \(2.0\,\text{V}\), so its average magnitude is \(2.0\,\text{V}\).

For coil B, use the magnitude of each section’s EMF before averaging:

\[ \begin{aligned} \overline{|\varepsilon|}_B &=\frac{(6.0)(0.10)+(2.0)(0.20)}{0.40}\\ &\quad+\frac{(6.0)(0.10)}{0.40}\\ &=4.0\,\text{V} \end{aligned} \]

Coil B’s average EMF magnitude is \(4.0\,\text{V}\), twice coil A’s.

Why? When calculating a signed average, the negative middle EMF subtracts from the positive sections. When calculating the average magnitude, all three sections contribute positively.

The same net change does not mean the same EMF at every instant, or even the same average magnitude.

Step 5: Determine when coil B’s current reverses.

The current reverses at \(t=0.10\,\text{s}\) and \(t=0.30\,\text{s}\), where the gradient changes sign and the induced EMF polarity reverses.

These reversals do not occur when the flux crosses zero. The flux crosses zero within the sloping sections, where the gradient keeps its sign. At those crossings, the current direction remains unchanged.

This is the conceptual hinge: flux tells you the state of the magnetic system; the gradient tells you how that state is changing. The induced EMF responds to the change, not simply to whether the flux is positive or negative.

Once you can read the gradient and determine the EMF polarity, the next step is to apply Lenz’s law to establish the induced-current direction, then use the circuit’s resistance to calculate current magnitude when required.