How to Read the Gradient of a Flux-Linkage Graph

A flux-linkage-time graph gives you the induced emf directly from its gradient. The important detail is the minus sign: induced emf equals the negative gradient of the flux-linkage graph.

If the gradient is \(+3.0\ \text{V}\), the induced emf is \(-3.0\ \text{V}\). If the gradient is \(-3.0\ \text{V}\), the induced emf is \(+3.0\ \text{V}\). The magnitude of the emf is simply the magnitude of the gradient.

\[ \boxed{\varepsilon=-\frac{\Delta\lambda}{\Delta t}} \]

Here, \(\varepsilon\) is induced emf in volts, \(\lambda\) is flux linkage in weber-turns (Wb-turn), and \(t\) is time in seconds.

That gives you the calculation. But to use it confidently in an HSC question, you also need to understand what flux linkage represents, how the coil’s orientation affects it, and when to use a tangent rather than a straight-line gradient.

01What is flux linkage?

Start with a coil sitting in a magnetic field.

The magnetic flux through one turn depends on three things:

• The magnetic field strength, \(B\), measured in tesla (T).
• The area of the coil, \(A\), measured in square metres (\(\text{m}^2\)).
• The angle between the magnetic field and the coil’s area vector.

The area vector is an imaginary vector pointing perpendicular to the plane of the coil. It is not a vector along the wire or across the surface.

This distinction matters because the angle in the magnetic flux equation is measured from the area vector, not from the plane of the coil.

The magnetic flux through one turn is:

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

Flux linkage accounts for every turn in the coil. For a coil with \(N\) turns:

\[ \boxed{\lambda=N\Phi=NBA\cos\theta} \]

Where:

• \(\lambda\) = flux linkage (Wb-turn).
• \(N\) = number of turns in the coil.
• \(B\) = magnetic field strength (T).
• \(A\) = area of one turn (\(\text{m}^2\)).
• \(\theta\) = angle between the magnetic field and the area vector.

Notice that the number of turns is already included in flux linkage.

This becomes important when reading graphs. If the vertical axis is labelled flux linkage, do not multiply your gradient by \(N\) again.

Why the angle changes the flux linkage

Consider three possible coil orientations.

The negative value at \(180^\circ\) comes from our choice of positive area-vector direction. It represents flux in the opposite direction.

There is another common complication in HSC questions: sometimes the angle is measured from the plane of the coil.

If the angle between the magnetic field and the coil’s plane is \(\alpha\), then:

\[ \theta=90^\circ-\alpha \]

For an acute angle:

\[ \Phi=BA\sin\alpha \]

For example, if the field makes an angle of \(30^\circ\) with the plane of the coil, the angle to the area vector is \(60^\circ\).

Using \(\cos30^\circ\) would give the wrong answer. You need \(\cos60^\circ\).

For a more detailed treatment of this particular trap, see How to Calculate Magnetic Flux When the Angle Is to the Coil’s Plane.

02Why does the gradient give induced emf?

Imagine rotating a coil between the poles of a magnet.

As the coil rotates, its area vector changes direction relative to the magnetic field. Consequently, the magnetic flux through each turn changes.

The total flux linkage changes too.

Faraday’s law describes the relationship between this change and the induced emf:

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

Since \(\lambda=N\Phi\), we can write:

\[ \boxed{\varepsilon=-\frac{\Delta\lambda}{\Delta t}} \]

This is the relationship you use when reading a flux-linkage-time graph.

The gradient measures how quickly flux linkage changes. Faraday’s law tells us that the induced emf has the opposite sign.

What would an experiment show?

A model school investigation can test this relationship using a rotating coil.

Picture a rectangular coil mounted on a shaft between the north and south poles of a magnet. The shaft rotates the coil in the magnetic field.

The two ends of the coil connect to slip rings attached to the shaft. Stationary brushes make electrical contact with the rings and connect the rotating coil to a voltage sensor.

An angle encoder measures the shaft’s position. With the magnetic field strength, coil area, and number of turns known, the flux linkage can be calculated at each recorded time.

The data logger can then compare the measured emf with the negative gradient of the calculated flux-linkage graph.

What should happen?

The model predicts that:

• When flux linkage reaches a maximum or minimum, its graph has a horizontal tangent. The induced emf should be zero.
• When flux linkage changes rapidly, the graph becomes steep. The induced emf has a large magnitude.
• When flux linkage increases, the induced emf is negative under the chosen sign convention.
• When flux linkage decreases, the induced emf is positive.

A first attempt using a hand-turned coil would have a problem: the rotational speed could vary between measurements. That makes it difficult to compare the voltage and the rate of change of flux linkage accurately.

A motor-driven coil with a synchronised angle encoder improves the investigation. It provides more reliable timing and makes it possible to calculate flux linkage at known instants.

The motor does not automatically guarantee a uniform magnetic field, though. That assumption must still be checked.

The important observation is that the voltage follows the rate of change of flux linkage, not simply the amount of flux linkage present.

This is why the gradient is the quantity you need.

03How to read the gradient correctly

There are two situations you need to recognise in an HSC question.

Situation A: A straight-line section

If the flux-linkage graph contains a straight-line section, calculate its gradient using two points on that line.

\[ \text{Gradient}= \frac{\text{Change in flux linkage}} {\text{Change in time}} \]

Then apply Faraday’s law:

\[ \varepsilon=-\text{Gradient} \]

The gradient is constant along a straight-line section, so the induced emf is constant over that interval.

Situation B: A curved section

A curved graph has a changing gradient.

If the question asks for the induced emf at one particular instant, you need the gradient of the tangent at that time.

Do not use the gradient between two distant points on the curve. That gives an average rate of change, not the instantaneous rate.

If you need to refresh how to calculate gradients from straight and curved graphs, the broader method is covered in HSC Physics Graphs: How to Use Gradient and Area.

Check the units before calculating

Flux linkage is measured in Wb-turn, and time is measured in seconds.

Therefore:

\[ \frac{\text{Wb-turn}}{\text{s}}=\text{V} \]

The units work out directly because turns are dimensionless in the SI system.

Be particularly careful when the graph uses milliseconds or milliwебers.

Useful conversions include:

\[ 1\ \text{mWb}=10^{-3}\ \text{Wb} \]

\[ 1\ \text{ms}=10^{-3}\ \text{s} \]

An interesting shortcut follows:

\[ \frac{1\ \text{mWb-turn}}{1\ \text{ms}} =1\ \text{V} \]

The prefixes cancel.

However, if the graph uses mWb-turn per second, the gradient is in millivolts, not volts.

One more distinction: flux versus flux linkage

Check the vertical-axis label before applying Faraday’s law.

If the graph shows magnetic flux per turn rather than flux linkage, you still need to account for the number of turns.

This distinction is explored further in How to Read a Magnetic-Flux-Time Graph to Calculate EMF.

04Worked examples

05Questions and solutions

The following questions build from straightforward graph reading to situations where the magnetic field strength and coil orientation change together.

Question 1

A straight section of a flux-linkage-time graph runs between the following points:

• At \(t=2.0\ \text{ms}\), flux linkage is \(-8.0\ \text{mWb-turn}\).
• At \(t=6.0\ \text{ms}\), flux linkage is \(+4.0\ \text{mWb-turn}\).

Calculate the induced emf over this interval, including its sign.

Solution 1

The induced emf is \(\boxed{-3.0\ \text{V}}\).

The flux linkage increases, so the graph has a positive gradient. Faraday’s law requires the induced emf to have the opposite sign.

Step 1: Calculate the change in flux linkage.

\[ \Delta\lambda=4.0-(-8.0)=12.0\ \text{mWb-turn} \]

Step 2: Calculate the change in time.

\[ \Delta t=6.0-2.0=4.0\ \text{ms} \]

Step 3: Calculate the gradient.

\[ \text{Gradient} =\frac{12.0}{4.0} =+3.0\ \text{mWb-turn/ms} \]

Converting the units:

\[ \text{Gradient}=+3.0\ \text{V} \]

Step 4: Apply Faraday’s law.

\[ \begin{aligned} \varepsilon&=-3.0\\ &=\boxed{-3.0\ \text{V}} \end{aligned} \]

The negative sign indicates the induced emf’s polarity relative to the chosen reference direction.

A common mistake is to report \(+3.0\ \text{V}\) because the graph rises. Remember that the induced emf is the negative of the gradient, not the gradient itself.

Question 2

A coil has 150 turns and an area of \(4.0\times10^{-4}\ \text{m}^2\).

It is placed in a uniform magnetic field. Over \(0.020\ \text{s}\):

• The magnetic field strength increases from \(0.30\ \text{T}\) to \(0.50\ \text{T}\).
• The angle between the magnetic field and the area vector decreases from \(60^\circ\) to \(30^\circ\).

Calculate the average induced emf, including its sign.

Solution 2

The average induced emf is approximately \(\boxed{-0.85\ \text{V}}\).

Both changes increase the flux linkage: the magnetic field becomes stronger, and the area vector becomes more aligned with the field.

Step 1: Write the flux-linkage equation.

\[ \lambda=NBA\cos\theta \]

Step 2: Calculate the initial flux linkage.

\[ \begin{aligned} \lambda_1 &=150(0.30)(4.0\times10^{-4})\cos60^\circ\\ &=0.0090\ \text{Wb-turn} \end{aligned} \]

Step 3: Calculate the final flux linkage.

\[ \begin{aligned} \lambda_2 &=150(0.50)(4.0\times10^{-4})\cos30^\circ\\ &=0.0260\ \text{Wb-turn} \end{aligned} \]

Step 4: Calculate the average gradient.

\[ \begin{aligned} \text{Gradient} &=\frac{0.0260-0.0090}{0.020}\\ &=+0.85\ \text{V} \end{aligned} \]

Step 5: Calculate the induced emf.

\[ \begin{aligned} \varepsilon_{\text{avg}} &=-0.85\\ &=\boxed{-0.85\ \text{V}} \end{aligned} \]

The induced emf is negative because flux linkage increases throughout the interval’s net change.

The important reasoning here is that you cannot consider the changing magnetic field alone. The angle also changes, so you must calculate the total flux linkage at both endpoints.

Question 3

A curved flux-linkage-time graph represents a coil rotating in a magnetic field.

At \(t=0.30\ \text{s}\), a tangent to the curve passes through these two points:

• \((0.20\ \text{s},14\ \text{mWb-turn})\)
• \((0.40\ \text{s},6\ \text{mWb-turn})\)

Calculate the instantaneous induced emf at \(t=0.30\ \text{s}\).

Solution 3

The instantaneous induced emf is \(\boxed{+0.040\ \text{V}}\).

Because the question asks for the emf at one particular instant, we need the gradient of the tangent, not the gradient between the endpoints of the entire curved section.

Step 1: Identify the two points on the tangent.

\[ \begin{aligned} (t_1,\lambda_1)&=(0.20,14)\\ (t_2,\lambda_2)&=(0.40,6) \end{aligned} \]

Step 2: Calculate the tangent’s gradient.

\[ \begin{aligned} \text{Gradient} &=\frac{6-14}{0.40-0.20}\\ &=\frac{-8}{0.20}\\ &=-40\ \text{mWb-turn/s} \end{aligned} \]

Step 3: Convert to volts.

\[ -40\ \text{mWb-turn/s} =-0.040\ \text{V} \]

Step 4: Apply Faraday’s law.

\[ \begin{aligned} \varepsilon &=-(-0.040)\\ &=\boxed{+0.040\ \text{V}} \end{aligned} \]

The positive induced emf results from the negative tangent gradient.

The trap is using two convenient points on the original curve instead of points on the tangent. Those would give an average gradient, which is not what this question asks for.

Question 4

Two identical coils, P and Q, each have:

• 200 turns.
• An area of \(2.0\times10^{-3}\ \text{m}^2\).
• A uniform magnetic field strength of \(0.25\ \text{T}\).

Both coils rotate at the same constant angular speed, changing the angle between their area vectors and the magnetic field by \(60^\circ\) over \(0.20\ \text{s}\).

However, their starting orientations differ.

• Coil P rotates from \(0^\circ\) to \(60^\circ\).
• Coil Q rotates from \(60^\circ\) to \(120^\circ\).

All angles are measured from the area vector, and both angles increase in the same rotational sense.

Calculate the average induced emf for each coil. Explain why the two values differ despite the identical angular speed.

Solution 4

Coil P has an average induced emf of \(\boxed{+0.25\ \text{V}}\), while coil Q has an average induced emf of \(\boxed{+0.50\ \text{V}}\).

Coil Q produces twice the average emf magnitude, even though both coils rotate through the same angle in the same time.

The reason is that magnetic flux linkage does not change linearly with the angle. It depends on the cosine of the angle.

Step 1: Calculate the maximum flux linkage.

Both coils have the same values of \(N\), \(B\), and \(A\).

\[ \begin{aligned} NBA &=200(0.25)(2.0\times10^{-3})\\ &=0.10\ \text{Wb-turn} \end{aligned} \]

Step 2: Calculate the average emf for coil P.

Its angle changes from \(0^\circ\) to \(60^\circ\).

Initial flux linkage:

\[ \lambda_1=0.10\cos0^\circ=0.10 \]

Final flux linkage:

\[ \lambda_2=0.10\cos60^\circ=0.050 \]

Therefore:

\[ \begin{aligned} \varepsilon_P &=-\frac{0.050-0.10}{0.20}\\ &=\boxed{+0.25\ \text{V}} \end{aligned} \]

Step 3: Calculate the average emf for coil Q.

Its angle changes from \(60^\circ\) to \(120^\circ\).

Initial flux linkage:

\[ \lambda_1=0.10\cos60^\circ=0.050 \]

Final flux linkage:

\[ \lambda_2=0.10\cos120^\circ=-0.050 \]

Therefore:

\[ \begin{aligned} \varepsilon_Q &=-\frac{-0.050-0.050}{0.20}\\ &=\boxed{+0.50\ \text{V}} \end{aligned} \]

Step 4: Compare the results.

\[ \frac{\varepsilon_Q}{\varepsilon_P} =\frac{0.50}{0.25}=2 \]

Coil Q has twice the average emf magnitude.

The tempting prediction is that identical angular speeds should produce identical induced emfs. But angular speed alone does not determine the rate of change of flux linkage.

The cosine relationship means that equal angular changes can produce different changes in flux linkage depending on the starting orientation.

This is exactly why reading the gradient of the flux-linkage graph is more reliable than trying to judge the emf from angular speed alone.

Question 5

Two identical coils, A and B, each have 100 turns and an area of \(2.0\times10^{-3}\ \text{m}^2\).

Their flux linkages are investigated over a \(0.20\ \text{s}\) interval.

For coil A, the magnetic field remains at \(0.20\ \text{T}\), and the angle between the field and area vector remains at \(0^\circ\).

For coil B, both the magnetic field strength and the coil’s orientation change. Its measured conditions are:

Assume a uniform magnetic field and that the measured flux-linkage graph for coil B consists of two straight-line sections joining these three readings.

A student argues:

“Both coils have the same initial and final magnetic field strength and orientation. Therefore, no emf is induced in either coil at any time.”

Evaluate this claim.

In your answer:

• Calculate the flux linkage of coil B at all three times.
• Determine the average induced emf during each half of the interval.
• Determine the average induced emf over the entire \(0.20\ \text{s}\).
• Explain why the student’s conclusion is incorrect.

Solution 5

The student’s claim is incorrect.

Coil B experiences a non-zero induced emf during both halves of the interval, even though its initial and final flux linkages are identical. Its average emf over the entire interval is zero because the two changes in flux linkage cancel.

This question tests an important distinction: zero average emf over an interval does not mean zero emf at every instant.

Step 1: Calculate the maximum flux linkage.

For coil B:

\[ \begin{aligned} NBA &=100(0.20)(2.0\times10^{-3})\\ &=0.040\ \text{Wb-turn} \end{aligned} \]

Step 2: Calculate the flux linkage at \(t=0\).

The magnetic field is \(0.20\ \text{T}\), and the angle is \(0^\circ\).

\[ \begin{aligned} \lambda_0 &=100(0.20)(2.0\times10^{-3})\cos0^\circ\\ &=\boxed{0.040\ \text{Wb-turn}} \end{aligned} \]

Step 3: Calculate the flux linkage at \(t=0.10\ \text{s}\).

The field has doubled to \(0.40\ \text{T}\), but the angle has also changed to \(30^\circ\).

\[ \begin{aligned} \lambda_{0.10} &=100(0.40)(2.0\times10^{-3})\cos30^\circ\\ &=0.080(0.866)\\ &=\boxed{0.0693\ \text{Wb-turn}} \end{aligned} \]

Notice that doubling the magnetic field does not simply double the flux linkage. The angle change also affects the result.

Step 4: Calculate the flux linkage at \(t=0.20\ \text{s}\).

The magnetic field and angle have returned to their initial values.

\[ \begin{aligned} \lambda_{0.20} &=100(0.20)(2.0\times10^{-3})\cos0^\circ\\ &=\boxed{0.040\ \text{Wb-turn}} \end{aligned} \]

Step 5: Calculate the average emf during the first half.

Flux linkage increases from \(0.040\) to \(0.0693\ \text{Wb-turn}\).

\[ \begin{aligned} \varepsilon_1 &=-\frac{0.0693-0.040}{0.10}\\ &=\boxed{-0.293\ \text{V}} \end{aligned} \]

The negative sign indicates that flux linkage is increasing.

Step 6: Calculate the average emf during the second half.

Flux linkage decreases from \(0.0693\) to \(0.040\ \text{Wb-turn}\).

\[ \begin{aligned} \varepsilon_2 &=-\frac{0.040-0.0693}{0.10}\\ &=\boxed{+0.293\ \text{V}} \end{aligned} \]

The positive sign indicates that flux linkage is decreasing.

Step 7: Calculate the average emf over the entire interval.

Use the initial and final flux linkage values:

\[ \begin{aligned} \varepsilon_{\text{whole}} &=-\frac{0.040-0.040}{0.20}\\ &=\boxed{0\ \text{V}} \end{aligned} \]

Step 8: Explain why the student’s reasoning fails.

The student has correctly noticed that the initial and final flux linkages are equal.

However, this only establishes that the average emf over the entire interval is zero.

The graph is not horizontal throughout. It rises during the first half and falls during the second half. Therefore, the induced emf is non-zero during both intervals, with opposite polarities.

The two effects cancel when calculating the average over the full interval.

There is an important limitation here: if you were given only the initial and final readings, you could calculate the average emf, but you could not establish whether the instantaneous emf was zero throughout. You need information about what happens between those endpoints.

This is a useful distinction in any question involving changing magnetic field strength and changing coil orientation.

06The final checklist

When you see a flux-linkage-time graph in an HSC Physics question, work through these decisions.

1. Check the vertical axis. Is it flux linkage, \(N\Phi\), or magnetic flux, \(\Phi\)? If it is already flux linkage, do not multiply by the number of turns again.

2. Check the units. Convert milliseconds to seconds and milliwебers to webers where necessary.

3. Choose the correct gradient. Use a straight-line gradient for a linear section, a tangent for instantaneous emf, or a secant for average emf over an interval.

4. Apply the negative sign. Remember that induced emf is the negative gradient, not the gradient itself.

5. Interpret the result. A positive gradient means negative induced emf. A negative gradient means positive induced emf. A horizontal tangent means zero instantaneous emf.

6. Check the angle if you calculate flux linkage yourself. The standard cosine equation uses the angle between the magnetic field and the area vector, not the plane of the coil.

One final connection is worth remembering. If you move from a flux-linkage-time graph to an induced-emf-time graph, the relationship changes from a gradient to an area:

\[ \boxed{\Delta\lambda=-\text{signed area under the emf-time graph}} \]

So the two graphs answer different questions. The gradient of a flux-linkage graph gives induced emf, while the signed area under an emf graph gives the change in flux linkage.

Knowing which graph you have is the first step towards choosing the correct method.