Magnetic Flux vs Flux Linkage: The HSC Physics Difference

Place a single wire loop and a 50-turn coil in the same uniform magnetic field. Give every turn the same area and orientation.

Before calculating anything, predict: does the 50-turn coil have 50 times the magnetic flux?

No. The magnetic flux through each individual turn is exactly the same. What is 50 times larger is the flux linkage, because it accounts for the flux through all 50 turns.

That is the essential difference:

• Magnetic flux (\(\Phi\)) measures the magnetic field passing through one turn.
• Flux linkage (\(\lambda\)) is the signed sum of the magnetic flux through every turn in a coil.

For a coil in which every turn experiences the same magnetic flux:

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

where \(N\) is the number of turns.

The distinction becomes particularly important when you work with area vectors, different coil orientations, and electromagnetic induction.

01The difference between magnetic flux and flux linkage

The two quantities are closely related, but they describe different things.

The important point is that adding turns does not increase the magnetic flux through an individual turn.

Imagine a coil with 30 identical turns. If each turn experiences a flux of \(0.002\,\text{Wb}\), then:

• Magnetic flux through each turn: \(0.002\,\text{Wb}\).
• Flux linkage of the entire coil: \(30(0.002)=0.060\,\text{Wb-turn}\).

The individual turns have not somehow collected more magnetic field. There are simply 30 separate flux contributions to add together.

A common mistake is to calculate \(BA\cos\theta\), then call the answer flux linkage without considering the number of turns.

Always establish whether the question wants the flux through one turn or the total flux linkage.

02Why the area vector matters

Before multiplying magnetic flux by the number of turns, you need to calculate the flux through one turn correctly.

For a flat loop in a uniform magnetic field:

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

Here:

• \(B\) is magnetic flux density, measured in tesla (T).
• \(A\) is the area enclosed by one turn, measured in \(\text{m}^2\).
• \(\theta\) is the angle between the magnetic field and the area vector.

The area vector is an imaginary vector perpendicular to the surface of the loop.

Its magnitude equals the area of the loop:

\[
|\vec A|=A
\]

Its direction is perpendicular to the plane of the loop.

This is important because the magnetic field does not contribute equally to the flux at every angle.

Why the angle is measured from the normal

Suppose the magnetic field points directly through the loop.

The field is parallel to the area vector, so:

\[
\theta=0^\circ
\]

Therefore:

\[
\Phi=BA\cos0^\circ=BA
\]

This gives the maximum positive flux for the chosen normal direction.

Now rotate the loop until the magnetic field runs parallel to its surface.

The field is perpendicular to the area vector:

\[
\theta=90^\circ
\]

Therefore:

\[
\Phi=BA\cos90^\circ=0
\]

The magnetic field still exists, but its component perpendicular to the loop’s surface is zero.

If a question gives the angle between the magnetic field and the plane of the coil, you must convert it before using the cosine formula.

For example, if the field makes an angle of \(30^\circ\) with the plane:

\[
\theta=90^\circ-30^\circ=60^\circ
\]

You would use \(\cos60^\circ\), not \(\cos30^\circ\).

For a more detailed explanation of this geometry, see Magnetic Flux for HSC Physics: Using Φ = BA cos θ.

The direction of the area vector also determines the sign

An area vector has a direction, not just a magnitude.

We choose a positive normal direction for a loop. For a current-carrying coil, the right-hand rule provides a consistent convention: curl your fingers in the chosen current direction, and your thumb indicates the positive area-vector direction.

A magnetic field pointing along this normal produces positive flux.

A field pointing in the opposite direction produces negative flux.

This becomes particularly important when different turns have different orientations.

For example:

\[
\Phi=BA\cos180^\circ=-BA
\]

The flux has the same magnitude as when the field points along the normal, but its sign has reversed.

When calculating flux linkage, these signs must be retained.

03Why flux linkage is the sum of the flux through every turn

Consider a conventional coil containing \(N\) tightly wound, parallel turns.

If the magnetic field is uniform and every turn has the same area and orientation, each turn experiences the same magnetic flux.

For example, a five-turn coil has:

\[
\Phi_1=\Phi_2=\Phi_3=\Phi_4=\Phi_5
\]

The total flux linkage is the sum of these contributions:

\[
\lambda=\Phi_1+\Phi_2+\Phi_3+\Phi_4+\Phi_5
\]

Since all five values are equal:

\[
\lambda=5\Phi
\]

More generally:

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

This is the standard relationship used in many HSC Physics coil problems.

However, notice the assumption: every turn must experience the same signed magnetic flux.

If different turns experience different magnetic fields, have different areas, or have different orientations, you cannot automatically multiply one turn’s flux by the total number of turns.

Instead, use the more general relationship:

\[
\boxed{\lambda=\sum_{k=1}^{N}\Phi_k}
\]

Each turn contributes its own signed magnetic flux.

For a standard, tightly wound coil in a uniform field, this simplifies to \(N\Phi\).

For a more complicated winding, the individual contributions must be considered separately.

What unit should you use for flux linkage?

Magnetic flux is measured in webers (Wb).

Flux linkage is commonly expressed in weber-turns (Wb-turn).

The word ‘turn’ makes it clear that the quantity accounts for the number of linked turns. Strictly speaking, turn count is dimensionless in SI, so the dimensional unit is equivalent to the weber.

For HSC calculations, writing Wb-turn is useful because it distinguishes flux linkage from the flux through one turn.

04How an experiment demonstrates the difference

A useful way to investigate flux linkage is to compare the induced emf from one turn with the induced emf from several turns.

Consider an electromagnet with two opposing pole pieces. The gap between them produces an approximately uniform magnetic field.

Place a rectangular coil inside this gap. Mount it on a rotating shaft so that its orientation can be changed.

Connect the coil terminals to a high-impedance voltage sensor, which sends its readings to an oscilloscope.

What should the experiment show?

Start with a one-turn loop.

Rotate it from an orientation where its area vector is parallel to the magnetic field to an orientation where the area vector makes an angle of \(60^\circ\) with the field.

Repeat the same movement using a ten-turn coil with identical turn areas.

Keep the following conditions constant:

• Magnetic field strength.
• Area of each turn.
• Initial and final orientations.
• Time taken to rotate the coil.
• Position within the magnetic field.

Predict what happens to the measured emf.

The ten-turn coil should produce ten times the average induced emf of the single-turn loop, provided each turn experiences the same change in magnetic flux.

Why?

The induced emf depends on the rate of change of flux linkage:

\[
\boxed{\varepsilon_{\text{avg}}=-\frac{\Delta\lambda}{\Delta t}}
\]

Here, \(\varepsilon_{\text{avg}}\) is the average induced emf in volts, \(\Delta\lambda\) is the change in flux linkage, and \(\Delta t\) is the time interval in seconds.

For identical turns:

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

The number of turns multiplies the change in flux linkage, not the flux through each individual turn.

Why the experimental design matters

Suppose you initially compare two separate coils, one with a single turn and another with ten turns.

If they have different areas or sit in different parts of the magnetic field, the comparison is not reliable. A difference in emf could result from the area or field strength rather than the number of turns.

A better design uses one ten-turn winding with an electrical tap after the first turn.

You can measure the voltage across the first turn, then across all ten turns, without changing the physical coil arrangement.

This controls the geometry much more effectively.

The predicted result is a tenfold increase in average emf. If the actual ratio differs, check whether the field is sufficiently uniform, whether the turns have comparable areas, and whether the coil’s motion was identical.

The experiment does not measure magnetic flux directly. It measures induced emf, which responds to the rate of change of flux linkage.

05Worked examples

06Questions and solutions

Question 1

A circular coil contains 25 turns. Each turn has an area of \(3.0\times10^{-3}\,\text{m}^2\). The coil is placed in a uniform magnetic field of \(0.40\,\text{T}\).

The angle between the magnetic field and the area vector is \(60^\circ\).

Calculate:

a) The magnetic flux through one turn.

b) The flux linkage of the entire coil.

Solution 1

The magnetic flux is \(6.0\times10^{-4}\,\text{Wb}\) per turn, and the total flux linkage is \(1.5\times10^{-2}\,\text{Wb-turn}\).

Part a: Magnetic flux through one turn

Use:

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

Substitute:

\[
\Phi=(0.40)(3.0\times10^{-3})\cos60^\circ
\]

\[
=(0.40)(3.0\times10^{-3})(0.5)
\]

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

This is the flux through one individual turn.

Part b: Flux linkage

Since all 25 turns have the same area and orientation:

\[
\lambda=N\Phi
\]

\[
=25(6.0\times10^{-4})
\]

\[
\boxed{\lambda=1.5\times10^{-2}\,\text{Wb-turn}}
\]

The flux linkage is 25 times the flux through one turn because all 25 turns contribute equally.

Question 2

A 20-turn rectangular coil has an area of \(1.5\times10^{-3}\,\text{m}^2\) per turn.

It is placed in a uniform magnetic field of \(0.40\,\text{T}\). Initially, the magnetic field makes an angle of \(30^\circ\) with the plane of the coil.

The coil is then flipped through \(180^\circ\) in \(0.12\,\text{s}\).

Take the initial flux as positive.

Calculate:

a) The initial flux linkage.

b) The final flux linkage.

c) The magnitude of the average induced emf.

Explain why the change in flux linkage is not zero, even though the initial and final flux magnitudes are equal.

Solution 2

The initial flux linkage is \(+0.0060\,\text{Wb-turn}\), the final flux linkage is \(-0.0060\,\text{Wb-turn}\), and the average induced emf has a magnitude of \(0.10\,\text{V}\).

Part a: Initial flux linkage

The given angle is measured from the plane, not the normal.

Therefore, the angle between the magnetic field and the area vector is:

\[
\theta_i=90^\circ-30^\circ=60^\circ
\]

Calculate the flux through one turn:

\[
\Phi_i=BA\cos60^\circ
\]

\[
=(0.40)(1.5\times10^{-3})(0.5)
\]

\[
\Phi_i=3.0\times10^{-4}\,\text{Wb}
\]

Now multiply by the number of turns:

\[
\lambda_i=20(3.0\times10^{-4})
\]

\[
\boxed{\lambda_i=+0.0060\,\text{Wb-turn}}
\]

Part b: Final flux linkage

After a \(180^\circ\) flip, the area vector points in the opposite direction.

The final angle between the field and the area vector is:

\[
\theta_f=120^\circ
\]

Therefore:

\[
\Phi_f=BA\cos120^\circ
\]

\[
=(0.40)(1.5\times10^{-3})(-0.5)
\]

\[
\Phi_f=-3.0\times10^{-4}\,\text{Wb}
\]

The final flux linkage is:

\[
\lambda_f=20(-3.0\times10^{-4})
\]

\[
\boxed{\lambda_f=-0.0060\,\text{Wb-turn}}
\]

Part c: Average induced emf

Calculate the change in flux linkage:

\[
\Delta\lambda=\lambda_f-\lambda_i
\]

\[
=-0.0060-0.0060
\]

\[
\Delta\lambda=-0.012\,\text{Wb-turn}
\]

Using Faraday’s law:

\[
|\varepsilon_{\text{avg}}|
=\frac{|\Delta\lambda|}{\Delta t}
\]

\[
=\frac{0.012}{0.12}
\]

\[
\boxed{|\varepsilon_{\text{avg}}|=0.10\,\text{V}}
\]

The tempting mistake is to assume that equal initial and final flux magnitudes mean there is no change.

However, flux is a signed quantity. The final flux points in the opposite direction, so the change in flux linkage is twice the initial magnitude.

Question 3

A four-turn search coil is deliberately spread across a region where the magnetic field is non-uniform.

Each turn has an area of \(1.2\times10^{-3}\,\text{m}^2\). All four area vectors point in the same direction, parallel to their local magnetic fields.

The magnetic field strength at each turn is:

Calculate:

a) The total flux linkage.

b) The average magnetic flux through one turn.

c) Explain why using the magnetic field strength at turn 1 for all four turns would give an incorrect answer.

Solution 3

The total flux linkage is \(8.4\times10^{-4}\,\text{Wb-turn}\), and the average flux through one turn is \(2.1\times10^{-4}\,\text{Wb}\).

Part a: Total flux linkage

The four turns experience different magnetic field strengths.

Therefore, we calculate the flux through each turn separately.

Because every area vector is parallel to its local magnetic field:

\[
\Phi_k=B_kA
\]

For the first turn:

\[
\Phi_1=(0.10)(1.2\times10^{-3})
\]

\[
\Phi_1=1.2\times10^{-4}\,\text{Wb}
\]

For the second turn:

\[
\Phi_2=(0.15)(1.2\times10^{-3})
\]

\[
\Phi_2=1.8\times10^{-4}\,\text{Wb}
\]

For the third turn:

\[
\Phi_3=2.4\times10^{-4}\,\text{Wb}
\]

For the fourth turn:

\[
\Phi_4=3.0\times10^{-4}\,\text{Wb}
\]

Now add all four contributions:

\[
\lambda=\Phi_1+\Phi_2+\Phi_3+\Phi_4
\]

\[
=(1.2+1.8+2.4+3.0)\times10^{-4}
\]

\[
\boxed{\lambda=8.4\times10^{-4}\,\text{Wb-turn}}
\]

Part b: Average flux per turn

Divide the total flux linkage by the number of turns:

\[
\Phi_{\text{avg}}=\frac{\lambda}{N}
\]

\[
=\frac{8.4\times10^{-4}}{4}
\]

\[
\boxed{\Phi_{\text{avg}}=2.1\times10^{-4}\,\text{Wb}}
\]

This is the average signed flux contribution per turn.

Part c: Why using one field strength is incorrect

Using the field strength at turn 1 for all four turns assumes that the magnetic field is uniform across the entire winding.

That assumption is false.

The fourth turn experiences a magnetic field 2.5 times stronger than the first turn, so its flux contribution is also 2.5 times larger.

You can use \(N\Phi\) if \(\Phi\) represents the correctly calculated average flux per turn. You cannot simply select one turn’s flux and assume it represents every other turn.

Question 4

Two coils are placed in the same uniform magnetic field of \(0.30\,\text{T}\).

Coil A has one turn with an area of \(0.010\,\text{m}^2\).

Coil B has ten turns, each with an area of \(0.0080\,\text{m}^2\).

Both coils rotate through the same angular displacement. Their area vectors move from an angle of \(0^\circ\) to \(60^\circ\) relative to the magnetic field in \(0.10\,\text{s}\).

Calculate:

a) The magnitude of the average induced emf in coil A.

b) The magnitude of the average induced emf in coil B.

c) The ratio of the two induced emfs.

d) Explain why coil B does not produce ten times the emf of coil A.

Solution 4

Coil A produces an average emf of \(0.015\,\text{V}\), while coil B produces \(0.12\,\text{V}\). Coil B therefore produces eight times the emf of coil A, not ten times.

Part a: Induced emf in coil A

Use Faraday’s law:

\[
|\varepsilon_A|
=\frac{|\Delta\Phi_A|}{\Delta t}
\]

The flux change is:

\[
\Delta\Phi_A=BA(\cos60^\circ-\cos0^\circ)
\]

Substitute:

\[
=(0.30)(0.010)(0.5-1)
\]

\[
=-1.5\times10^{-3}\,\text{Wb}
\]

Therefore:

\[
|\varepsilon_A|
=\frac{1.5\times10^{-3}}{0.10}
\]

\[
\boxed{|\varepsilon_A|=0.015\,\text{V}}
\]

Part b: Induced emf in coil B

Each turn has an area of \(0.0080\,\text{m}^2\).

Calculate the flux change per turn:

\[
\Delta\Phi_B
=(0.30)(0.0080)(0.5-1)
\]

\[
\Delta\Phi_B=-1.2\times10^{-3}\,\text{Wb}
\]

Now calculate the change in flux linkage:

\[
\Delta\lambda_B=N\Delta\Phi_B
\]

\[
=10(-1.2\times10^{-3})
\]

\[
\Delta\lambda_B=-0.012\,\text{Wb-turn}
\]

The induced emf is:

\[
|\varepsilon_B|
=\frac{0.012}{0.10}
\]

\[
\boxed{|\varepsilon_B|=0.12\,\text{V}}
\]

Part c: Compare the induced emfs

\[
\frac{\varepsilon_B}{\varepsilon_A}
=\frac{0.12}{0.015}
\]

\[
\boxed{\frac{\varepsilon_B}{\varepsilon_A}=8}
\]

Coil B produces eight times the induced emf.

Part d: Explain the difference

The tempting prediction is that ten turns must produce ten times the emf.

That would be correct if both coils had identical turn areas and experienced the same change in magnetic flux per turn.

However, coil B has a smaller area.

Its area is:

\[
\frac{0.0080}{0.010}=0.80
\]

So each turn experiences only 80% of the flux change experienced by coil A.

The combined effect is:

\[
10\times0.80=8
\]

The number of turns matters, but it is not the only factor controlling flux linkage.

Question 5

A specially constructed figure-eight winding contains two sections connected in series.

Each section has six turns, and every turn has an area of \(1.5\times10^{-2}\,\text{m}^2\).

The two sections have opposite winding directions. Therefore, their area vectors point in opposite directions.

The left section is defined as having positive flux linkage, while the right section has negative flux linkage.

The two sections occupy separate magnetic field regions. Each field is uniform within its respective region and perpendicular to the loop surfaces. The fields point in the same physical direction, but their strengths can differ.

Initially:

\[
B_L=0.20\,\text{T}
\]

\[
B_R=0.12\,\text{T}
\]

Assume that the number of turns and their areas remain constant.

Part a

Calculate the initial net flux linkage of the complete winding.

Part b

In the first experiment, both magnetic fields increase at a constant rate of \(0.50\,\text{T s}^{-1}\) over \(0.10\,\text{s}\).

Calculate the average induced emf.

Explain why every turn experiencing a changing magnetic flux does not necessarily produce a non-zero net emf across the complete winding.

Part c

The experiment is repeated. This time, the left magnetic field increases at \(0.50\,\text{T s}^{-1}\), while the right magnetic field increases at \(0.30\,\text{T s}^{-1}\).

Both changes occur over \(0.10\,\text{s}\).

Calculate the magnitude of the average induced emf and explain why the result differs from part b.

Solution 5

The initial net flux linkage is \(0.0072\,\text{Wb-turn}\).

In the first experiment, the average induced emf is zero because the changes in flux linkage from the two sections cancel.

In the second experiment, the average induced emf has a magnitude of \(0.018\,\text{V}\).

Part a: Initial flux linkage

The two sections have opposite area-vector directions.

Therefore, their contributions must be subtracted.

The net flux linkage is:

\[
\lambda=NAB_L-NAB_R
\]

Factor out the common terms:

\[
\lambda=NA(B_L-B_R)
\]

Substitute:

\[
\lambda=6(1.5\times10^{-2})(0.20-0.12)
\]

\[
=0.09(0.08)
\]

\[
\boxed{\lambda=0.0072\,\text{Wb-turn}}
\]

The initial flux linkage is positive because the left section has the greater flux contribution.

Part b: Both magnetic fields increase at the same rate

Each magnetic field increases by:

\[
\Delta B=(0.50)(0.10)
\]

\[
\Delta B=0.050\,\text{T}
\]

The change in flux linkage is:

\[
\Delta\lambda
=NA\Delta B_L-NA\Delta B_R
\]

Factor out the common terms:

\[
\Delta\lambda=NA(\Delta B_L-\Delta B_R)
\]

Since both fields increase by the same amount:

\[
\Delta\lambda
=6(1.5\times10^{-2})(0.050-0.050)
\]

\[
\boxed{\Delta\lambda=0}
\]

Using Faraday’s law:

\[
|\varepsilon_{\text{avg}}|
=\frac{|\Delta\lambda|}{\Delta t}
\]

\[
=\frac{0}{0.10}
\]

\[
\boxed{|\varepsilon_{\text{avg}}|=0\,\text{V}}
\]

This result is quite different from what you might initially expect.

Every individual turn experiences a changing magnetic flux. However, the changes in the two sections produce opposing induced emfs.

The left section produces a change in flux linkage of:

\[
\Delta\lambda_L
=6(1.5\times10^{-2})(0.050)
\]

\[
=0.0045\,\text{Wb-turn}
\]

The right section produces an equal change in magnitude but with the opposite sign:

\[
\Delta\lambda_R=-0.0045\,\text{Wb-turn}
\]

They cancel.

There is another important detail here: the initial flux linkage was not zero. It was \(0.0072\,\text{Wb-turn}\).

However, Faraday’s law depends on the change in flux linkage, not its initial value.

The flux linkage remains constant even though both magnetic fields are increasing.

Part c: The magnetic fields increase at different rates

Now the left field increases by:

\[
\Delta B_L=(0.50)(0.10)
\]

\[
\Delta B_L=0.050\,\text{T}
\]

The right field increases by:

\[
\Delta B_R=(0.30)(0.10)
\]

\[
\Delta B_R=0.030\,\text{T}
\]

The change in net flux linkage is:

\[
\Delta\lambda
=NA(\Delta B_L-\Delta B_R)
\]

Substitute:

\[
=6(1.5\times10^{-2})(0.050-0.030)
\]

\[
=0.09(0.020)
\]

\[
\boxed{\Delta\lambda=0.0018\,\text{Wb-turn}}
\]

Now calculate the average induced emf:

\[
|\varepsilon_{\text{avg}}|
=\frac{0.0018}{0.10}
\]

\[
\boxed{|\varepsilon_{\text{avg}}|=0.018\,\text{V}}
\]

The left section now produces a greater change in flux linkage than the right section.

Their induced emfs no longer cancel completely, leaving a net emf of \(0.018\,\text{V}\).

The key mistake to avoid is treating all 12 turns as though they experience the same signed flux.

The number of turns alone does not determine the flux linkage. Their individual flux contributions, including their directions, must be considered.

07What to remember when solving HSC Physics questions

When you encounter a coil problem, work through these decisions in order.

Three common mistakes are worth checking before you finish a calculation:

1. Using the angle between the magnetic field and the coil’s plane instead of the area vector.
2. Multiplying magnetic flux by the number of turns when the turns do not all experience the same signed flux.
3. Ignoring negative flux when calculating the change in flux linkage.

Once you can distinguish flux from flux linkage, the next useful step is understanding how effectively two different coils share the same changing magnetic flux.

This becomes particularly important in transformers, where not all the magnetic flux produced by one winding necessarily links the other.

See Incomplete Flux Linkage and Leakage Flux in Transformers for how this distinction affects transformer behaviour.