How to Calculate Magnetic Flux Through a Coil Partly in a Field

A rectangular coil can have half its wire inside a magnetic field without having half its maximum magnetic flux. The important quantity is not how much wire is inside the field. It is how much of the area enclosed by the coil lies inside the field, and the angle at which the magnetic field passes through that area.

For a flat coil partly inside a uniform magnetic field, the calculation is:

\[ \boxed{\Phi = BA_{\text{in}}\cos\theta} \]

Here, \(A_{\text{in}}\) is the area of the coil’s enclosed surface that is actually inside the magnetic field, and \(\theta\) is the angle between the magnetic field and the coil’s area vector (the normal to its surface).

If the coil has \(N\) identical turns, its magnetic flux linkage is:

\[ \boxed{N\Phi = NBA_{\text{in}}\cos\theta} \]

This method assumes the magnetic field is uniform over the area inside the field region and negligible outside it. If the field outside is not negligible, its contribution must also be included.

01Start with the area inside the magnetic field

Imagine a rectangular coil that is 20 cm wide and 10 cm high. It sits across the boundary of a uniform magnetic field, with only 6 cm of its width inside the field.

Which area should you use?

• The entire area of the coil?
• Half the area?
• Only the section inside the magnetic field?

The third option is correct.

The coil’s full area is:

\[ A = 0.20\times0.10 =0.020\ \text{m}^2 \]

However, only 0.06 m of its width is inside the field. Therefore, the relevant area is:

\[ A_{\text{in}}=0.06\times0.10 \]

\[ \boxed{A_{\text{in}}=0.0060\ \text{m}^2} \]

Think of the coil as a window frame. The wire forms the frame, but magnetic flux is associated with the area inside it, like the glass in a window.

If the field only occupies part of that window, only the area inside the field contributes.

The analogy has one limitation: magnetic flux also depends on direction. A magnetic field running along the plane of the window contributes no flux through it, even if the entire window sits in the field.

What if the boundary cuts across the coil at an angle?

You still calculate the actual area of the enclosed surface inside the magnetic field.

For a rectangular overlap, this might involve finding the dimensions of the smaller rectangle. If the boundary creates a triangle or another shape, calculate that shape’s area instead.

Do not use the length of wire inside the magnetic field as a substitute for area.

There is another important assumption here. If the field extends outside the region shown in the question, that external field also contributes to the flux. For multiple regions with different uniform fields, calculate each signed contribution and add them.

02Use the area vector to find the correct angle

Getting the overlap area right is only half the calculation.

You also need the correct angle.

This is where a common HSC mistake appears: students sometimes use the angle between the magnetic field and the plane of the coil when the formula requires the angle between the magnetic field and the area vector.

What is an area vector?

An area vector is an imaginary arrow that:

• Is perpendicular to the surface of the coil.
• Has a magnitude equal to the area of that surface.
• Points in one of the two possible directions normal to the surface.

It is often called the surface normal.

For a flat coil, the area vector points straight out of the plane of the coil, rather than along the wire.

Magnetic flux depends on the component of the magnetic field parallel to this area vector.

The relationship is:

\[ \boxed{\Phi=BA_{\text{in}}\cos\theta} \]

where:

• \(\Phi\) = magnetic flux through one turn, measured in webers (Wb).
• \(B\) = magnetic flux density, measured in tesla (T).
• \(A_{\text{in}}\) = area inside the magnetic field, measured in square metres (\(\text{m}^2\)).
• \(\theta\) = angle between the magnetic field and the area vector.

The three important orientations

The angle determines how much of the magnetic field passes through the coil’s surface.

Notice what happens when the magnetic field is parallel to the coil’s plane.

You might predict that the flux should be large because the field is passing through the region occupied by the coil. But it actually has no component perpendicular to the surface.

Therefore:

\[ \Phi=BA_{\text{in}}\cos90^\circ=0 \]

The coil can be sitting in a strong magnetic field and still have zero magnetic flux through its surface.

What if the question gives the angle to the plane?

This is extremely common in exam questions.

If the magnetic field makes an angle \(\alpha\) with the plane of the coil, the angle to the area vector is:

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

Therefore, for the magnitude of the flux:

\[ \Phi=BA_{\text{in}}\sin\alpha \]

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

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

\[ \theta=60^\circ \]

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

A useful exam habit is to sketch the area vector before substituting any angles.

Does the direction of the area vector matter?

Yes, if the question requires signed flux or an induced emf.

You can choose either direction normal to the coil as your positive direction. Keep that choice consistent.

• If the magnetic field points along the chosen normal, flux is positive.
• If it points opposite to the chosen normal, flux is negative.
• If it is perpendicular to the normal, flux is zero.

If a current direction is provided, the right-hand grip rule can help establish the normal: curl your fingers in the direction of conventional current, and your thumb points along the chosen area vector.

A negative flux does not mean the field is weaker. It means the field points opposite to your chosen positive normal.

03Magnetic flux versus magnetic flux linkage

There is one more distinction to get right.

Magnetic flux, \(\Phi\), describes the magnetic flux through a single turn of the coil.

Magnetic flux linkage accounts for all the turns.

If a coil has \(N\) turns, and every turn has the same area and orientation, then:

\[ \boxed{\text{Flux linkage}=N\Phi} \]

Substituting the partial-area formula:

\[ \boxed{N\Phi=NBA_{\text{in}}\cos\theta} \]

The units are:

• Magnetic flux: Wb.
• Magnetic flux linkage: Wb-turn.

The turn count is dimensionless, but writing Wb-turn makes it clear that you are calculating flux linkage rather than flux through one turn.

When should you multiply by the number of turns?

For an ordinary flat coil, all turns have approximately the same area and orientation, so multiplying by \(N\) is appropriate.

If different turns experience different fluxes, the total linkage must instead be found by adding their individual fluxes.

04An experiment that tests the partial-area model

Why should only the overlap area count?

A useful way to investigate this is to move a rectangular coil across the boundary of a magnetic-field region and observe the induced voltage.

Start with the familiar demonstration: connect a coil to a sensitive galvanometer and move a bar magnet towards and away from it.

The meter deflects while the magnet moves. When the magnet is held stationary, the deflection disappears.

This demonstrates that a change in magnetic flux can produce an induced emf.

However, there is a problem if we want to investigate partial overlap quantitatively.

A bar magnet produces a non-uniform magnetic field. The field becomes weaker with distance, and its lines spread out. There is no perfectly sharp boundary separating a strong-field region from a zero-field region.

That makes it difficult to determine whether a change in the reading comes from changing overlap area or changing magnetic field strength.

Improving the experimental design

A better investigation would use an electromagnet with broad pole pieces to create a region of approximately uniform magnetic field.

The apparatus would contain:

1. Two broad electromagnet pole pieces producing a magnetic field across an air gap.
2. A Hall probe to map the field strength in the gap.
3. A rectangular multi-turn coil mounted on a guide rail.
4. Flexible connecting leads from the coil to a galvanometer or voltage data logger.

The coil moves horizontally across the field boundary while its orientation remains fixed. Its area vector is parallel to the magnetic field.

What should the experiment show?

Let:

• \(h\) = height of the rectangular coil.
• \(x\) = width of the coil currently inside the field.
• \(B\) = uniform magnetic flux density.
• \(N\) = number of turns.

When the coil first begins entering the field, the overlap area is zero.

As it moves further inside:

\[ A_{\text{in}}=hx \]

If the magnetic field is perpendicular to the coil’s plane:

\[ \Phi=Bhx \]

The flux increases as the overlap area increases.

Once the entire coil is inside the uniform field, its overlap area becomes constant. If the coil continues moving without rotating, its magnetic flux stays constant.

The predicted galvanometer or voltage data logger response is therefore:

This is the important observation: a coil can move through a magnetic field without experiencing an induced emf if its magnetic flux linkage remains constant.

When it enters or leaves the field, the overlap area changes, so the flux linkage changes.

The revised experiment gives a much cleaner test of the partial-area model. It controls the coil’s orientation and allows the magnetic field to be mapped independently.

In a real apparatus, the field boundary still has a fringe region. The ideal sharp boundary is an approximation, and this can make the measured voltage pulse less abrupt than the simple model predicts.

A small experimental calculation

Suppose the coil has:

• 25 turns.
• Height \(h=0.060\ \text{m}\).
• Magnetic field \(B=0.30\ \text{T}\).
• Constant entry speed \(v=0.10\ \text{m s}^{-1}\).

As the coil enters, its overlap width increases at \(0.10\ \text{m s}^{-1}\).

The rate at which its overlap area changes is:

\[ \frac{\Delta A_{\text{in}}}{\Delta t}=hv \]

Using Faraday’s law, the magnitude of the induced emf is:

\[ |\varepsilon|=NBhv \]

Substituting:

\[ \begin{aligned} |\varepsilon| &=25(0.30)(0.060)(0.10)\\ &=0.045\ \text{V} \end{aligned} \]

\[ \boxed{|\varepsilon|=45\ \text{mV}} \]

This is the predicted emf while the coil enters the ideal uniform-field region at constant speed.

The calculation also shows why the height of the coil matters: moving the coil a small distance \(x\) into the field adds an area of \(hx\), not simply a length of wire.

05A reliable method for HSC questions

When you encounter a coil partly inside a magnetic field, follow this sequence.

Step 1: Identify the actual overlap area.

Work out which part of the coil’s enclosed surface is inside the field.

For a rectangular coil:

\[ A_{\text{in}}=\text{width inside}\times\text{height} \]

Convert all dimensions into metres before calculating.

Step 2: Draw the area vector.

Draw a line perpendicular to the plane of the coil.

This gives you the correct reference direction for the angle.

Step 3: Identify the angle.

Check whether the question gives the angle between the magnetic field and:

• The area vector, in which case use cosine.
• The plane of the coil, in which case use sine for the flux magnitude.

Step 4: Calculate magnetic flux per turn.

\[ \Phi=BA_{\text{in}}\cos\theta \]

Include the correct sign if the question requires it.

Step 5: Calculate flux linkage if required.

\[ N\Phi=NBA_{\text{in}}\cos\theta \]

Do not multiply by the number of turns unless the question requires flux linkage or an induced emf calculation.

If the magnetic field is produced by a long solenoid rather than given directly, determine \(B\) first. The method is covered in Magnetic Field Inside a Solenoid: HSC Physics Explained.

06Worked examples

07The contrasting case: A coil inside a field with zero flux

Consider a rectangular coil that is partly inside a strong uniform magnetic field.

This time, the magnetic field runs parallel to the plane of the coil.

A student might reason that the field is passing through the region occupied by the coil, so there must be substantial flux.

But the field is parallel to the surface, not perpendicular to it.

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

\[ \theta=90^\circ \]

Therefore:

\[ \Phi=BA_{\text{in}}\cos90^\circ \]

\[ \boxed{\Phi=0\ \text{Wb}} \]

Compare this with a magnetic field perpendicular to the coil’s plane.

This is why you should never assume that a coil experiences maximum flux simply because it is inside a magnetic field.

The overlap area and the field orientation must both be considered.

08Questions and solutions

Question 1

A 25-turn rectangular coil is 0.18 m wide and 0.12 m high.

It is partly inside a uniform magnetic field of strength \(0.40\ \text{T}\). Only 0.075 m of its width is inside the field.

The magnetic field is perpendicular to the plane of the coil and negligible outside the field region.

Calculate:

(a) The magnetic flux through one turn.

(b) The magnetic flux linkage of the coil.

Solution 1

The magnetic flux through one turn is \(3.60\times10^{-3}\ \text{Wb}\), and the flux linkage is \(9.00\times10^{-2}\ \text{Wb-turn}\).

Step 1: Calculate the overlap area.

\[ \begin{aligned} A_{\text{in}}&=0.075\times0.12\\ &=0.0090\ \text{m}^2 \end{aligned} \]

Step 2: Identify the angle.

The field is perpendicular to the coil’s plane, so:

\[ \theta=0^\circ \]

Step 3: Calculate flux per turn.

\[ \begin{aligned} \Phi&=BA_{\text{in}}\cos\theta\\ &=0.40(0.0090)\cos0^\circ\\ &=0.00360\ \text{Wb} \end{aligned} \]

Therefore:

\[ \boxed{\Phi=3.60\times10^{-3}\ \text{Wb}} \]

Step 4: Calculate flux linkage.

\[ \begin{aligned} N\Phi&=25(0.00360)\\ &=0.0900\ \text{Wb-turn} \end{aligned} \]

\[ \boxed{N\Phi=9.00\times10^{-2}\ \text{Wb-turn}} \]

The key was using the 0.075 m overlap width rather than the coil’s full width of 0.18 m.

Question 2

A 50-turn rectangular coil is 0.16 m wide and 0.10 m high.

Only 0.10 m of its width is inside a uniform magnetic field of strength \(0.25\ \text{T}\).

The magnetic field makes an angle of \(40^\circ\) with the plane of the coil.

Calculate:

(a) The magnitude of the magnetic flux through one turn.

(b) The magnetic flux linkage.

A student uses \(\cos40^\circ\) in their calculation. Explain why this is incorrect.

Solution 2

The flux per turn is approximately \(1.61\times10^{-3}\ \text{Wb}\), and the flux linkage is approximately \(8.03\times10^{-2}\ \text{Wb-turn}\).

The student’s use of \(\cos40^\circ\) is incorrect because the given angle is measured from the plane of the coil, not its normal.

Step 1: Calculate the overlap area.

\[ \begin{aligned} A_{\text{in}}&=0.10\times0.10\\ &=0.010\ \text{m}^2 \end{aligned} \]

Step 2: Convert the angle.

\[ \theta=90^\circ-40^\circ \]

\[ \theta=50^\circ \]

Step 3: Calculate the flux per turn.

\[ \begin{aligned} \Phi&=BA_{\text{in}}\cos\theta\\ &=0.25(0.010)\cos50^\circ\\ &=0.001607\ \text{Wb} \end{aligned} \]

Therefore:

\[ \boxed{\Phi=1.61\times10^{-3}\ \text{Wb}} \]

Step 4: Calculate the flux linkage.

\[ \begin{aligned} N\Phi&=50(0.001607)\\ &=0.08035\ \text{Wb-turn} \end{aligned} \]

\[ \boxed{N\Phi\approx8.03\times10^{-2}\ \text{Wb-turn}} \]

The tempting mistake is to substitute the given \(40^\circ\) directly into the cosine formula. The formula requires the angle to the normal, which is \(50^\circ\).

Question 3

Two 20-turn rectangular coils, A and B, are placed in the same uniform magnetic field of strength \(0.30\ \text{T}\).

Both coils have a total area of \(0.020\ \text{m}^2\).

However:

• Coil A has 25% of its area inside the magnetic field. Its area vector is parallel to the field.
• Coil B has 60% of its area inside the magnetic field. Its area vector makes an angle of \(70^\circ\) with the field.

A student claims that coil B must have greater flux linkage because more of its area is inside the magnetic field.

Calculate the flux linkage of each coil and determine whether the student is correct.

Solution 3

Coil A has the greater flux linkage, despite having less of its area inside the magnetic field.

The reason is that flux depends on both overlap area and the angle between the magnetic field and the area vector.

Step 1: Calculate the overlap area of coil A.

\[ A_A=0.25(0.020) \]

\[ A_A=0.0050\ \text{m}^2 \]

Its area vector is parallel to the magnetic field:

\[ \theta_A=0^\circ \]

Step 2: Calculate the flux linkage of coil A.

\[ \begin{aligned} N\Phi_A &=NBA_A\cos\theta_A\\ &=20(0.30)(0.0050)\cos0^\circ\\ &=0.0300\ \text{Wb-turn} \end{aligned} \]

Therefore:

\[ \boxed{N\Phi_A=0.0300\ \text{Wb-turn}} \]

Step 3: Calculate the overlap area of coil B.

\[ A_B=0.60(0.020) \]

\[ A_B=0.012\ \text{m}^2 \]

Its area vector makes an angle of \(70^\circ\) with the field.

Step 4: Calculate the flux linkage of coil B.

\[ \begin{aligned} N\Phi_B &=NBA_B\cos\theta_B\\ &=20(0.30)(0.012)\cos70^\circ\\ &=0.0246\ \text{Wb-turn} \end{aligned} \]

Therefore:

\[ \boxed{N\Phi_B=0.0246\ \text{Wb-turn}} \]

Step 5: Compare the results.

\[ 0.0300>0.0246 \]

Coil A has the greater flux linkage.

The student focused on the overlap area but ignored the orientation of the magnetic field.

Coil B has more area inside the field, but its larger angle reduces the component of the field passing through its surface.

The quantity that matters is the product:

\[ A_{\text{in}}\cos\theta \]

Neither overlap area nor orientation can be considered independently.

Question 4

A 40-turn rectangular coil is 0.15 m wide and 0.08 m high.

It moves horizontally at a constant speed of \(0.12\ \text{m s}^{-1}\) into a uniform magnetic field of strength \(0.50\ \text{T}\).

The magnetic field is perpendicular to the plane of the coil. Initially, the coil is completely outside the field, and the field region is large enough to contain the entire coil.

Assume a sharp field boundary and negligible field outside the region.

(a) Calculate the time taken for the entire coil to enter the field.

(b) Calculate the magnitude of the average induced emf while the coil is entering.

(c) What happens to the induced emf once the coil is completely inside the field and continues moving at the same speed?

(d) Describe what happens to the direction of the induced emf when the coil leaves the field.

Solution 4

The coil takes \(1.25\ \text{s}\) to enter the field, producing an induced emf of \(0.192\ \text{V}\) during entry. Once fully inside and moving through the uniform field, the induced emf becomes zero. When it leaves, the induced emf reverses direction.

Step 1: Calculate the time taken to enter the field.

The coil must move a distance equal to its full width.

\[ t=\frac{d}{v} \]

Substituting:

\[ \begin{aligned} t&=\frac{0.15}{0.12}\\ &=1.25\ \text{s} \end{aligned} \]

Therefore:

\[ \boxed{t=1.25\ \text{s}} \]

Step 2: Determine the rate of change of overlap area.

As the coil moves into the field, its overlap area increases.

For a rectangular coil:

\[ \frac{\Delta A_{\text{in}}}{\Delta t}=hv \]

Substituting:

\[ \begin{aligned} \frac{\Delta A_{\text{in}}}{\Delta t} &=0.08(0.12)\\ &=0.0096\ \text{m}^2\text{s}^{-1} \end{aligned} \]

Step 3: Calculate the induced emf.

Faraday’s law gives:

\[ |\varepsilon|=N\frac{|\Delta\Phi|}{\Delta t} \]

Since the field is perpendicular to the coil’s plane and uniform:

\[ |\varepsilon|=NBhv \]

Substituting:

\[ \begin{aligned} |\varepsilon| &=40(0.50)(0.08)(0.12)\\ &=0.192\ \text{V} \end{aligned} \]

Therefore:

\[ \boxed{|\varepsilon|=0.192\ \text{V}} \]

Step 4: Consider the coil when it is fully inside the field.

Once the entire coil is inside the uniform field, its overlap area remains constant.

Its orientation and the magnetic field strength also remain constant.

Therefore, the magnetic flux linkage does not change.

\[ \boxed{\varepsilon=0\ \text{V}} \]

This is a particularly important distinction: movement through a magnetic field does not automatically produce an induced emf. The flux linkage must change.

Step 5: Consider the coil leaving the field.

As the coil leaves, its overlap area decreases.

The magnetic flux linkage therefore decreases, producing an emf with the opposite sign to the entry emf.

This follows from Faraday’s law and the reversal in the direction of the flux change.

Question 5 – Extension challenge

A 50-turn rectangular coil has a total enclosed area of \(0.012\ \text{m}^2\).

It is positioned across two adjacent magnetic-field regions:

• Region A has a uniform magnetic field of \(0.30\ \text{T}\) directed into the page.
• Region B has a uniform magnetic field of \(0.15\ \text{T}\) directed out of the page.

The coil’s plane is perpendicular to both magnetic fields.

Choose an area vector pointing into the page as the positive direction.

Initially, one-third of the coil’s area is in region A and two-thirds is in region B.

The coil is then moved horizontally. After \(0.40\ \text{s}\), two-thirds of its area is in region A and one-third is in region B.

Assume the two field regions are uniform, the transition region is negligible, and the coil maintains its orientation.

(a) Calculate the initial net magnetic flux through one turn.

(b) Calculate the final net magnetic flux through one turn.

(c) Calculate the change in flux linkage.

(d) Calculate the magnitude of the average induced emf during the movement.

(e) A student argues that because the initial net flux is zero, there cannot be an induced emf while the coil moves. Explain the error in this reasoning.

Solution 5

The initial net flux is zero, but the final flux is \(1.80\times10^{-3}\ \text{Wb}\). The average induced emf has a magnitude of \(0.225\ \text{V}\).

The important idea is that magnetic flux is signed. Contributions from opposite magnetic-field directions can cancel initially, but changing the overlap areas can destroy that cancellation.

Step 1: Establish the sign convention.

The chosen area vector points into the page.

Therefore:

• Region A contributes positive flux.
• Region B contributes negative flux.

The net flux is the sum of these signed contributions.

Step 2: Calculate the initial flux.

The initial area in region A is:

\[ A_A=\frac{1}{3}(0.012) \]

\[ A_A=0.0040\ \text{m}^2 \]

The initial area in region B is:

\[ A_B=\frac{2}{3}(0.012) \]

\[ A_B=0.0080\ \text{m}^2 \]

Calculate the flux contributions:

\[ \Phi_A=0.30(0.0040) \]

\[ \Phi_A=0.0012\ \text{Wb} \]

For region B, the field points opposite to the chosen normal:

\[ \Phi_B=-0.15(0.0080) \]

\[ \Phi_B=-0.0012\ \text{Wb} \]

Add the contributions:

\[ \begin{aligned} \Phi_{\text{initial}} &=\Phi_A+\Phi_B\\ &=0.0012-0.0012\\ &=0\ \text{Wb} \end{aligned} \]

Therefore:

\[ \boxed{\Phi_{\text{initial}}=0\ \text{Wb}} \]

The initial flux is zero because two non-zero contributions cancel.

Step 3: Calculate the final flux.

After the coil moves, its area in region A becomes:

\[ A_A=\frac{2}{3}(0.012) \]

\[ A_A=0.0080\ \text{m}^2 \]

Its area in region B becomes:

\[ A_B=\frac{1}{3}(0.012) \]

\[ A_B=0.0040\ \text{m}^2 \]

The flux contributions are now:

\[ \Phi_A=0.30(0.0080) \]

\[ \Phi_A=0.0024\ \text{Wb} \]

And:

\[ \Phi_B=-0.15(0.0040) \]

\[ \Phi_B=-0.0006\ \text{Wb} \]

Adding the contributions:

\[ \begin{aligned} \Phi_{\text{final}} &=0.0024-0.0006\\ &=0.0018\ \text{Wb} \end{aligned} \]

Therefore:

\[ \boxed{\Phi_{\text{final}}=1.80\times10^{-3}\ \text{Wb}} \]

Step 4: Calculate the change in flux linkage.

The change in flux per turn is:

\[ \Delta\Phi =\Phi_{\text{final}}-\Phi_{\text{initial}} \]

\[ \Delta\Phi=0.0018-0 \]

\[ \Delta\Phi=0.0018\ \text{Wb} \]

The coil has 50 turns, so:

\[ \begin{aligned} \Delta(N\Phi) &=N\Delta\Phi\\ &=50(0.0018)\\ &=0.090\ \text{Wb-turn} \end{aligned} \]

Therefore:

\[ \boxed{\Delta(N\Phi)=0.090\ \text{Wb-turn}} \]

Step 5: Calculate the average induced emf.

Using Faraday’s law:

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

Substituting:

\[ \begin{aligned} |\varepsilon| &=\frac{0.090}{0.40}\\ &=0.225\ \text{V} \end{aligned} \]

Therefore:

\[ \boxed{|\varepsilon|=0.225\ \text{V}} \]

With the chosen positive area normal, the signed emf is:

\[ \boxed{\varepsilon=-0.225\ \text{V}} \]

The negative sign indicates that the induced emf is opposite to the chosen positive loop direction.

Step 6: Explain the student’s mistake.

The student has confused zero magnetic flux with zero change in magnetic flux.

Initially, the positive contribution from region A exactly cancels the negative contribution from region B.

However, moving the coil changes the amount of area in each region. The two contributions no longer cancel.

Faraday’s law depends on the change in flux linkage, not whether the initial flux happens to be zero.

This is the deeper point: a coil can have zero net magnetic flux at one instant and still experience an induced emf if that flux changes.

Once you can calculate the flux through a partially immersed coil, the next step is determining how that flux linkage changes with time.

That is the connection to Faraday’s law:

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

The same area, angle, and flux-linkage reasoning applies when a coil rotates, a magnetic field changes, or two coils exchange energy through a changing magnetic field.

For a practical application, see How Wireless Charging Transfers Energy Through Changing Magnetic Flux.