How Magnetic Flux Changes When Field Strength and Area Both Change

An electromagnet is turned up while a coil is stretched to enclose a larger area. Both changes should increase magnetic flux. But do they simply add together?

No. Magnetic flux depends on the product of magnetic flux density and enclosed area, with an additional adjustment for the coil’s orientation. If the magnetic field increases by 50% and the coil’s area doubles, the flux becomes three times its original value, provided the orientation stays unchanged.

If the coil also rotates, however, the result can be quite different. The increase in field strength and area might be partly cancelled, completely cancelled, or even outweighed by the change in orientation.

The key relationship is:

\[ \boxed{\Phi_B=BA\cos\theta} \]

Understanding how these three factors interact is the difference between correctly comparing two coils and simply assuming that a stronger field or a bigger coil must produce more flux.

01What happens when magnetic field strength and coil area both change?

Start with a flat coil sitting entirely inside a uniform magnetic field.

Three quantities determine the magnetic flux through each turn:

• \(B\): magnetic flux density, measured in tesla (T).
• \(A\): the area enclosed by one turn, measured in square metres (\(\mathrm{m^2}\)).
• \(\theta\): the angle between the magnetic field and the coil’s area vector.

For a uniform magnetic field:

\[ \Phi_B=BA\cos\theta \]

The magnetic flux \(\Phi_B\) is measured in webers (Wb).

The important point is that these factors multiply. Increasing both \(B\) and \(A\) does not mean adding their percentage increases.

For example, suppose:

• Magnetic flux density increases by 50%.
• Coil area increases by 100%.
• Coil orientation remains unchanged.

The new magnetic flux is:

\[ \Phi_{\mathrm{new}} =(1.5B)(2A)\cos\theta \]

Therefore:

\[ \boxed{\Phi_{\mathrm{new}}=3\Phi_{\mathrm{old}}} \]

The flux triples, representing a 200% increase.

Why does the angle matter?

Imagine holding a rectangular sheet of cardboard in the rain.

If the sheet faces the falling rain directly, it catches a large amount of water. If you turn it sideways, much less rain passes through its surface.

Magnetic flux behaves in a similar way. A larger coil provides more area for the magnetic field to pass through, but the orientation determines how much of the field actually passes through that area.

The mathematical representation of this orientation is the area vector.

The area vector is an imaginary vector with two properties:

1. Its magnitude equals the area enclosed by the coil.
2. Its direction is perpendicular to the plane of the coil.

We can write:

\[ \vec A=A\hat n \]

Here, \(\hat n\) is a unit vector pointing perpendicular to the coil.

This means the magnetic flux equation can also be written as a dot product:

\[ \Phi_B=\vec B\cdot\vec A \]

The cosine appears because only the component of the magnetic field parallel to the area vector contributes to flux.

Three important orientations

The angle is measured from the area vector, not from the flat surface of the coil.

Notice the surprising case at \(90^\circ\).

The coil can be completely surrounded by a strong magnetic field and still experience zero magnetic flux if the field runs parallel to its plane.

At \(120^\circ\), the flux is negative. This does not mean the field has become weaker. It means the field points partly opposite to the chosen positive direction of the area vector.

For questions asking only for flux magnitude, use:

\[ |\Phi_B|=BA|\cos\theta| \]

For questions involving direction or induced emf, keep track of the sign.

What if the angle is measured from the coil’s plane?

Some questions give the angle between the magnetic field and the flat surface of the coil instead.

If that angle is \(\alpha\), the angle to the area vector is:

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

Therefore:

\[ |\Phi_B|=BA\sin\alpha \]

Don’t substitute an angle measured from the plane directly into the cosine equation.

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

02Comparing simultaneous changes in field strength and area

Now we can build a reliable method for comparing two different conditions.

Suppose a coil initially has magnetic flux:

\[ \Phi_1=B_1A_1\cos\theta_1 \]

After changing the magnetic field and coil area, its flux becomes:

\[ \Phi_2=B_2A_2\cos\theta_2 \]

Dividing the two equations gives:

\[ \boxed{ \frac{\Phi_2}{\Phi_1} = \frac{B_2}{B_1} \frac{A_2}{A_1} \frac{\cos\theta_2}{\cos\theta_1} } \]

This is the most useful comparison equation for this topic.

It allows you to calculate the change in flux without finding both flux values separately.

For magnitudes, use the absolute values of the cosine terms. For signed flux, retain their signs.

The ratio method requires a non-zero initial flux. If the initial angle is \(90^\circ\), calculate the two fluxes directly instead.

A quick comparison table

Assume the coil remains in a uniform magnetic field and the changes below are measured relative to its initial state.

The third row is particularly useful.

You might initially predict that doubling both the field and area produces four times the flux. That would be correct if the coil maintained its original orientation.

But rotating the coil to \(60^\circ\) reduces the cosine factor to 0.5.

The actual change is:

\[ 2\times2\times0.5=2 \]

The flux only doubles.

A stronger field and larger area do not guarantee a particular increase in flux. You must account for orientation as well.

03Flux linkage: what changes when there are multiple turns?

So far, we’ve considered the magnetic flux through one turn.

A coil containing multiple turns introduces another quantity: magnetic flux linkage.

Flux linkage is the magnetic flux through one turn multiplied by the number of turns.

\[ \boxed{\lambda=N\Phi_B} \]

Substituting the magnetic flux equation:

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

Where:

• \(\lambda\) = magnetic flux linkage, commonly expressed in Wb-turn.
• \(N\) = number of turns.
• \(B\) = magnetic flux density (T).
• \(A\) = area enclosed by each turn (\(\mathrm{m^2}\)).
• \(\theta\) = angle between the magnetic field and the area vector.

This gives us an important distinction:

• Magnetic flux describes the flux through one turn.
• Flux linkage accounts for the total contribution from all the turns.

If two coils have different numbers of turns, comparing their magnetic flux alone does not tell you which has greater flux linkage.

For two different coils:

\[ \boxed{ \frac{\lambda_2}{\lambda_1} = \frac{N_2}{N_1} \frac{B_2}{B_1} \frac{A_2}{A_1} \frac{\cos\theta_2}{\cos\theta_1} } \]

This is particularly useful when HSC questions compare coils with different numbers of turns, areas, and orientations.

04How an experiment tests the relationship

How could you test whether magnetic flux really increases in proportion to both magnetic field density and coil area?

The challenge is that a voltmeter does not measure magnetic flux directly. It measures potential difference, which depends on how quickly flux linkage changes.

A controlled experiment needs to separate these quantities.

The apparatus

Consider an electromagnet with two pole pieces facing each other. A DC power supply drives the electromagnet, producing a magnetic field in the gap.

A Hall probe is placed between the pole pieces to measure the magnetic flux density.

A flat, multi-turn test coil is mounted on an adjustable holder inside the gap. Its area and orientation can be changed, while its number of turns remains constant.

The coil is connected to a voltage sensor and data logger. The sensor records the induced voltage as the coil is withdrawn from the magnetic field.

What should the experiment show?

Suppose we keep the number of turns and orientation constant.

We can then change the magnetic field strength and coil area independently.

The prediction is straightforward:

• Increasing \(B\) by a factor of 1.5 should increase flux linkage by a factor of 1.5.
• Increasing \(A\) by a factor of 1.5 should also increase flux linkage by a factor of 1.5.
• Increasing both by a factor of 1.5 should increase flux linkage by a factor of 2.25.

The following table contains illustrative predicted values, not actual experimental measurements.

Assume \(N\) and \(\theta=0^\circ\) remain constant.

Trial D is the important comparison.

The combined increase is multiplicative:

\[ \frac{\lambda_D}{\lambda_A} = \frac{0.30}{0.20} \times \frac{0.0060}{0.0040} \]

\[ \boxed{\frac{\lambda_D}{\lambda_A}=2.25} \]

Why measuring peak voltage can give the wrong impression

There is a problem with a simple experimental design.

A student might compare the peak voltage produced when two coils are pulled out of the magnetic field. They might assume that the coil producing the larger voltage initially had greater magnetic flux.

That conclusion is not necessarily correct.

Faraday’s law relates induced emf to the rate of change of flux linkage:

\[ \epsilon_{\mathrm{avg}} = -\frac{\Delta\lambda}{\Delta t} \]

A coil withdrawn more slowly can produce a smaller average voltage even if its initial flux linkage is greater.

Consider two illustrative trials:

The larger coil initially has twice the flux linkage, yet it produces a smaller average voltage because it takes longer to leave the field.

The first design confuses the amount of flux linkage with the rate at which it changes.

Improving the experiment

A better approach is to record the complete voltage-time signal rather than comparing only peak voltages.

The signed area under the voltage-time graph represents the change in flux linkage:

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

A calibrated fluxmeter can also measure this total change directly.

The experiment should also control several other factors:

1. Keep the number of turns constant when investigating area.
2. Measure the magnetic field using the Hall probe whenever the electromagnet current changes.
3. Check that the magnetic field is reasonably uniform across the entire coil, not just at its centre.
4. Keep the coil orientation fixed when investigating the separate effects of \(B\) and \(A\).
5. Ensure the entire coil is inside the field before withdrawing it.

The uniform-field condition matters. If the magnetic field varies significantly across the coil, multiplying one field reading by the entire area is no longer accurate.

Instead, divide the surface into smaller regions and add their contributions:

\[ \Phi_B=\sum_i B_iA_i\cos\theta_i \]

If the coil is only partly inside the magnetic field, use the area actually within the field, assuming the field outside is negligible. See How to Calculate Magnetic Flux Through a Coil Partly in a Field for that situation.

05Worked examples

06Questions and solutions

Question 1

A single-turn rectangular coil has an enclosed area of \(30\,\mathrm{cm^2}\). It is placed in a uniform magnetic field of \(0.40\,\mathrm{T}\), with its area vector at \(60^\circ\) to the field.

The coil is expanded to \(45\,\mathrm{cm^2}\), and the magnetic flux density increases to \(0.60\,\mathrm{T}\). Its orientation remains unchanged.

Calculate:

a) The initial magnetic flux.

b) The final magnetic flux.

c) The factor by which the flux changes.

Solution 1

The final magnetic flux is \(1.35\times10^{-3}\,\mathrm{Wb}\), which is 2.25 times the initial flux.

Step 1: Convert the areas into square metres.

\[ A_1=30\times10^{-4} =0.0030\,\mathrm{m^2} \]

\[ A_2=45\times10^{-4} =0.0045\,\mathrm{m^2} \]

Step 2: Calculate the initial flux.

\[ \begin{aligned} \Phi_1 &=B_1A_1\cos60^\circ\\ &=0.40\times0.0030\times0.5\\ &=\boxed{6.0\times10^{-4}\,\mathrm{Wb}} \end{aligned} \]

Step 3: Calculate the final flux.

\[ \begin{aligned} \Phi_2 &=B_2A_2\cos60^\circ\\ &=0.60\times0.0045\times0.5\\ &=\boxed{1.35\times10^{-3}\,\mathrm{Wb}} \end{aligned} \]

Step 4: Calculate the factor of increase.

\[ \frac{\Phi_2}{\Phi_1} = \frac{0.60}{0.40} \times \frac{0.0045}{0.0030} \]

\[ =1.5\times1.5 \]

\[ \boxed{\frac{\Phi_2}{\Phi_1}=2.25} \]

The flux increases by a factor of 2.25, equivalent to a 125% increase.

The angle does not affect the ratio because it stays constant throughout the change.

Question 2

Two different coils, A and B, are placed entirely inside the same uniform magnetic field of \(0.50\,\mathrm{T}\).

Their properties are:

Calculate the magnetic flux through one turn of each coil and their respective flux linkages.

Determine which coil has the greater flux linkage and calculate the factor by which it exceeds the other.

Solution 2

Coil A has the greater flux linkage, approximately 1.73 times that of coil B.

Step 1: Convert both areas into square metres.

\[ A_A=6.0\times10^{-4}\,\mathrm{m^2} \]

\[ A_B=4.0\times10^{-4}\,\mathrm{m^2} \]

Step 2: Calculate the flux through one turn of coil A.

\[ \begin{aligned} \Phi_A &=BA_A\cos30^\circ\\ &=0.50\times6.0\times10^{-4}\times0.866\\ &=\boxed{2.60\times10^{-4}\,\mathrm{Wb}} \end{aligned} \]

Step 3: Calculate the flux through one turn of coil B.

\[ \begin{aligned} \Phi_B &=BA_B\cos60^\circ\\ &=0.50\times4.0\times10^{-4}\times0.5\\ &=\boxed{1.00\times10^{-4}\,\mathrm{Wb}} \end{aligned} \]

Coil A experiences greater flux per turn because it has a larger area and a more favourable orientation.

Step 4: Calculate the flux linkages.

For coil A:

\[ \begin{aligned} \lambda_A &=N_A\Phi_A\\ &=40\times2.60\times10^{-4}\\ &=\boxed{0.0104\,\mathrm{Wb\,turn}} \end{aligned} \]

For coil B:

\[ \begin{aligned} \lambda_B &=N_B\Phi_B\\ &=60\times1.00\times10^{-4}\\ &=\boxed{0.0060\,\mathrm{Wb\,turn}} \end{aligned} \]

Step 5: Compare the flux linkages.

\[ \frac{\lambda_A}{\lambda_B} = \frac{0.0104}{0.0060} \]

\[ \boxed{\frac{\lambda_A}{\lambda_B}\approx1.73} \]

Coil B has more turns, but this does not guarantee greater flux linkage.

Its smaller area and less favourable orientation more than offset its additional turns.

This is why you should compare the complete expression \(NBA\cos\theta\), rather than judging coils by one property.

Question 3

A 30-turn rectangular coil has a total enclosed area of \(0.0060\,\mathrm{m^2}\).

It is placed in a non-uniform magnetic field. The magnetic field is perpendicular to the plane of the coil everywhere, but its strength differs across the surface.

The left half of the coil experiences a uniform field of \(0.50\,\mathrm{T}\), while the right half experiences a uniform field of \(0.30\,\mathrm{T}\).

Calculate:

a) The magnetic flux through one turn.

b) The total flux linkage.

c) The percentage error in flux linkage if a student uses the \(0.50\,\mathrm{T}\) field reading across the entire coil.

Solution 3

The correct flux linkage is \(0.072\,\mathrm{Wb\,turn}\). Using the stronger field across the whole coil overestimates the result by 25%.

Step 1: Calculate the area of each half.

The total area is:

\[ A=0.0060\,\mathrm{m^2} \]

Each half therefore has an area of:

\[ A_1=A_2=\frac{0.0060}{2} \]

\[ A_1=A_2=0.0030\,\mathrm{m^2} \]

Step 2: Calculate the flux through the left half.

The magnetic field is perpendicular to the coil, so:

\[ \theta=0^\circ \]

\[ \begin{aligned} \Phi_1 &=B_1A_1\cos0^\circ\\ &=0.50\times0.0030\times1\\ &=0.0015\,\mathrm{Wb} \end{aligned} \]

Step 3: Calculate the flux through the right half.

\[ \begin{aligned} \Phi_2 &=B_2A_2\cos0^\circ\\ &=0.30\times0.0030\times1\\ &=0.0009\,\mathrm{Wb} \end{aligned} \]

Step 4: Add the two contributions.

Both magnetic field regions point in the same direction, so their fluxes are positive and can be added.

\[ \begin{aligned} \Phi_{\mathrm{total}} &=\Phi_1+\Phi_2\\ &=0.0015+0.0009\\ &=\boxed{0.0024\,\mathrm{Wb}} \end{aligned} \]

Step 5: Calculate the flux linkage.

\[ \begin{aligned} \lambda &=N\Phi_{\mathrm{total}}\\ &=30\times0.0024\\ &=\boxed{0.072\,\mathrm{Wb\,turn}} \end{aligned} \]

Step 6: Calculate the error from using the stronger field across the entire area.

The student’s incorrect calculation would be:

\[ \begin{aligned} \lambda_{\mathrm{wrong}} &=30\times0.50\times0.0060\\ &=0.090\,\mathrm{Wb\,turn} \end{aligned} \]

The absolute error is:

\[ 0.090-0.072=0.018\,\mathrm{Wb\,turn} \]

Percentage error:

\[ \begin{aligned} \%\text{ error} &=\frac{0.018}{0.072}\times100\\ &=\boxed{25\%} \end{aligned} \]

The mistake is assuming that a single magnetic field strength represents the entire coil.

When the field is non-uniform, calculate the contribution from each region separately before adding them.

Question 4

A 75-turn coil has an enclosed area of \(40\,\mathrm{cm^2}\) and sits in a uniform magnetic field of \(0.50\,\mathrm{T}\).

Its measured positive flux linkage is \(0.090\,\mathrm{Wb\,turn}\).

Assume the angle between the magnetic field and the chosen positive area vector is initially between \(0^\circ\) and \(90^\circ\).

The coil is then modified as follows:

• The magnetic flux density increases to \(0.80\,\mathrm{T}\).
• The enclosed area decreases to \(25\,\mathrm{cm^2}\).
• The coil rotates until the angle between the magnetic field and its area vector is \(120^\circ\).

Calculate:

a) The initial angle between the magnetic field and the area vector.

b) The final signed magnetic flux linkage.

c) Whether the direction of the flux has reversed.

Solution 4

The initial angle is approximately \(53.1^\circ\), and the final signed flux linkage is \(-0.075\,\mathrm{Wb\,turn}\). The flux has reversed direction.

Step 1: Convert the initial area into square metres.

\[ A_1=40\times10^{-4} =0.0040\,\mathrm{m^2} \]

Step 2: Rearrange the flux linkage equation to find the angle.

We know:

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

Rearranging:

\[ \cos\theta=\frac{\lambda}{NBA} \]

Substitute the initial values:

\[ \cos\theta = \frac{0.090}{75\times0.50\times0.0040} \]

\[ \cos\theta=\frac{0.090}{0.150} \]

\[ \cos\theta=0.60 \]

Therefore:

\[ \boxed{\theta=\cos^{-1}(0.60)=53.1^\circ} \]

The initial angle is acute, consistent with the positive measured flux linkage.

Step 3: Convert the final area into square metres.

\[ A_2=25\times10^{-4} =0.0025\,\mathrm{m^2} \]

Step 4: Calculate the final signed flux linkage.

The new angle is \(120^\circ\).

\[ \begin{aligned} \lambda_2 &=N B_2 A_2\cos120^\circ\\ &=75\times0.80\times0.0025\times(-0.5)\\ &=75\times(-0.0010)\\ &=\boxed{-0.075\,\mathrm{Wb\,turn}} \end{aligned} \]

Step 5: Interpret the negative sign.

Initially, the magnetic field had a positive component along the chosen area vector.

After rotation to \(120^\circ\), the field has a component in the opposite direction.

The final flux linkage is therefore negative.

Its magnitude is \(0.075\,\mathrm{Wb\,turn}\), but its direction has reversed.

Notice that the field became stronger while the area became smaller. Those two changes alone would not have reversed the sign. The reversal comes from the coil’s rotation through an angle greater than \(90^\circ\).

Question 5

A flexible 50-turn rectangular coil is placed entirely inside a uniform magnetic field.

Its orientation remains fixed throughout an experiment. The angle between its area vector and the magnetic field is \(60^\circ\).

The coil’s enclosed area changes while the electromagnet is adjusted. The following values are recorded:

Assume the field and area change smoothly between these recorded states, the number of turns remains constant, and every turn experiences the same magnetic flux.

Answer the following:

a) Calculate the flux linkage at each recorded time.

b) Calculate the average induced emf from 0.00 s to 0.10 s.

c) Calculate the average induced emf from 0.10 s to 0.30 s.

d) Calculate the average induced emf over the entire 0.30 s experiment.

e) A student argues that because the initial and final flux linkages are identical, no emf was induced during the experiment. Explain precisely why this conclusion is incorrect.

Solution 5

The flux linkage increases from \(0.060\) to \(0.080\,\mathrm{Wb\,turn}\), then decreases to \(0.060\,\mathrm{Wb\,turn}\). The average induced emf is therefore non-zero during each individual interval, even though it is zero over the entire experiment.

This is an important distinction between the total change in flux linkage and the changes occurring during the process.

Step 1: Calculate the flux linkage at each recorded time.

The equation is:

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

The number of turns is 50, and the angle remains \(60^\circ\).

Therefore:

\[ \lambda=50BA\cos60^\circ \]

Since \(\cos60^\circ=0.5\):

\[ \lambda=25BA \]

At \(t=0.00\,\mathrm{s}\):

\[ \begin{aligned} \lambda_0 &=25\times0.20\times0.012\\ &=\boxed{0.060\,\mathrm{Wb\,turn}} \end{aligned} \]

At \(t=0.10\,\mathrm{s}\):

\[ \begin{aligned} \lambda_1 &=25\times0.40\times0.0080\\ &=\boxed{0.080\,\mathrm{Wb\,turn}} \end{aligned} \]

At \(t=0.30\,\mathrm{s}\):

\[ \begin{aligned} \lambda_2 &=25\times0.80\times0.0030\\ &=\boxed{0.060\,\mathrm{Wb\,turn}} \end{aligned} \]

The flux linkage first increases and then decreases.

Step 2: Calculate the average induced emf during the first interval.

Faraday’s law gives:

\[ \epsilon_{\mathrm{avg}} = -\frac{\Delta\lambda}{\Delta t} \]

Between 0.00 s and 0.10 s:

\[ \Delta\lambda=0.080-0.060 =0.020\,\mathrm{Wb\,turn} \]

The time interval is:

\[ \Delta t=0.10\,\mathrm{s} \]

Therefore:

\[ \begin{aligned} \epsilon_{\mathrm{avg}} &=-\frac{0.020}{0.10}\\ &=\boxed{-0.20\,\mathrm{V}} \end{aligned} \]

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

Step 3: Calculate the average induced emf during the second interval.

Between 0.10 s and 0.30 s:

\[ \Delta\lambda=0.060-0.080 =-0.020\,\mathrm{Wb\,turn} \]

The time interval is:

\[ \Delta t=0.30-0.10=0.20\,\mathrm{s} \]

Therefore:

\[ \begin{aligned} \epsilon_{\mathrm{avg}} &=-\frac{-0.020}{0.20}\\ &=\boxed{+0.10\,\mathrm{V}} \end{aligned} \]

The induced emf has reversed polarity because the flux linkage is now decreasing rather than increasing.

Step 4: Calculate the average induced emf over the entire experiment.

The initial and final flux linkages are both:

\[ \lambda_0=\lambda_2=0.060\,\mathrm{Wb\,turn} \]

Therefore:

\[ \Delta\lambda=0 \]

The total time is:

\[ \Delta t=0.30\,\mathrm{s} \]

So:

\[ \begin{aligned} \epsilon_{\mathrm{avg,total}} &=-\frac{0}{0.30}\\ &=\boxed{0\,\mathrm{V}} \end{aligned} \]

Step 5: Explain why the student’s conclusion is incorrect.

The student is confusing zero net change with no change at all.

The flux linkage initially increases by \(0.020\,\mathrm{Wb\,turn}\), then decreases by exactly the same amount.

Consequently, the total change between the starting and ending states is zero.

However, the flux linkage was not constant throughout the experiment. It changed during both intervals, producing induced emfs of opposite polarity.

You can also see this by considering the signed areas under the voltage-time graph:

First interval:

\[ (-0.20)\times0.10=-0.020\,\mathrm{Wb\,turn} \]

Second interval:

\[ (+0.10)\times0.20=+0.020\,\mathrm{Wb\,turn} \]

The two contributions cancel.

Therefore, the total signed area is zero, even though the induced emf was non-zero during both intervals on average.

This is the crucial distinction: knowing that the initial and final flux linkages are equal does not tell you that the flux linkage remained constant between them.

It only tells you that the net change over the entire interval was zero.

07Where this leads next

When both magnetic field strength and coil area change, always start by comparing the complete expression:

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

Work out the changes in field strength, enclosed area, orientation, and number of turns before deciding whether flux linkage increases or decreases.

Once you can track flux linkage between different states, the next step is using its rate of change to calculate induced emf.

That’s where Faraday’s law becomes much more interesting: a coil can have a large flux linkage but produce no induced emf if that linkage stays constant, or a relatively small flux linkage can produce a substantial emf if it changes rapidly.