How to Find Maximum Magnetic Flux Through a Rectangular Coil

A rectangular coil is placed in a uniform magnetic field. You rotate it through 90°, keeping its dimensions and the field strength unchanged. Surprisingly, the magnetic flux through the coil can fall from its maximum value to zero.

Before calculating anything, predict which orientation produces the greatest flux:

• The magnetic field passes straight through the broad face of the coil.
• The magnetic field runs along the plane of the coil.

The first orientation gives maximum flux. The magnetic field must be perpendicular to the plane of the coil, not parallel to it.

The calculation is straightforward once you identify the correct angle. The main challenge in HSC Physics is recognising whether an angle is measured from the coil’s plane or from its perpendicular direction, called the area vector.

01The maximum magnetic flux formula

For a rectangular coil in a uniform magnetic field, magnetic flux through one turn is calculated using:

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

The symbols represent:

For a rectangular coil, calculate its enclosed area using:

\[ A=\text{length}\times\text{width} \]

Therefore:

\[ \Phi_B= B(\text{length}\times\text{width})\cos\theta \]

To find the maximum positive magnetic flux, set \(\theta=0^\circ\).

Since \(\cos 0^\circ=1\):

\[ \boxed{\Phi_{B,\max}=BA} \]

Substituting the rectangular area:

\[ \boxed{\Phi_{B,\max}=B\ell w} \]

Here, \(\ell\) is the length and \(w\) is the width of the rectangular coil, both measured in metres.

The key result: Maximum magnetic flux occurs when the magnetic field is parallel to the area vector and perpendicular to the coil’s plane.

What about a coil with multiple turns?

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

If a coil has \(N\) identical turns, each turn experiences the same magnetic flux (assuming the field and orientation are uniform across the coil).

The total flux linkage is:

\[ \lambda=N\Phi_B \]

Therefore, maximum flux linkage is:

\[ \boxed{\lambda_{\max}=NBA} \]

Be careful with the wording in an exam:

• Magnetic flux through one turn: \(\Phi_B\), measured in webers (Wb).
• Magnetic flux linkage: \(N\Phi_B\), commonly expressed as Wb-turn.

Multiplying by the number of turns gives flux linkage, not the flux through an individual turn.

02Understanding the area vector and the angle

This is where many otherwise capable HSC Physics students lose marks.

The angle in the magnetic flux equation is measured between the magnetic field and the area vector, not between the magnetic field and the coil’s surface.

The area vector is an imaginary vector that:

• Has a magnitude equal to the area enclosed by the coil.
• Points perpendicular (normal) to the plane of the coil.

We represent it as:

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

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

The flux equation can also be written as a dot product:

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

This explains why the cosine appears. Only the component of the magnetic field perpendicular to the coil’s plane contributes to the flux.

What if the question gives the angle between the coil’s plane and the magnetic field?

Suppose an exam question states:

A rectangular coil is positioned so that its plane makes an angle of 35° with a uniform magnetic field.

You might be tempted to substitute \(35^\circ\) directly into:

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

Don’t.

The given angle is measured from the plane, whereas the equation requires the angle from the area vector.

Since the area vector is perpendicular to the plane:

\[ \theta=90^\circ-35^\circ \]

\[ \theta=55^\circ \]

Now the correct equation is:

\[ \Phi_B=BA\cos55^\circ \]

There is also a useful alternative.

Using the trigonometric identity:

\[ \cos(90^\circ-\alpha)=\sin\alpha \]

where \(\alpha\) is the acute angle between the magnetic field and the coil’s plane, we obtain:

\[ \boxed{|\Phi_B|=BA\sin\alpha} \]

Both approaches give the same flux magnitude.

How orientation changes magnetic flux

Consider what happens as the rectangular coil rotates in a constant magnetic field.

Notice the difference between maximum flux and maximum flux magnitude.

If we choose the area vector to point in the same direction as the magnetic field, the maximum positive flux is \(+BA\).

If the area vector points in the opposite direction, the flux is \(-BA\). Its magnitude is still maximum.

Therefore:

\[ |\Phi_B|_{\max}=BA \]

The sign tells us about the direction of the flux relative to our chosen area vector.

03Worked examples

Let’s put the method into practice with two typical HSC-style calculations.

04Testing the maximum-flux prediction experimentally

So far, we’ve used geometry to predict the orientation that produces maximum flux.

But how could we test that prediction experimentally?

A useful investigation involves rotating a rectangular coil inside a uniform magnetic field and measuring the induced EMF.

Setting up the experiment

Consider the following apparatus:

• A rectangular multi-turn coil is positioned between the north and south pole pieces of an electromagnet.
• The magnetic field runs across the gap between the poles.
• The coil is attached to an axle passing through its centre. The axle lies in the plane of the coil and is perpendicular to the magnetic field.
• A motor rotates the coil at a constant angular speed.
• Slip rings and brushes connect the rotating coil to a voltage sensor and data logger.
• A Hall probe measures the magnetic field strength near the coil.

The coil’s angle can be tracked as it rotates.

What should the experiment show?

Start with the coil’s area vector parallel to the magnetic field.

At this position:

\[ \theta=0^\circ \]

Therefore:

\[ \Phi_B=BA \]

The flux is at its maximum positive value.

Now rotate the coil at a constant angular speed.

As the area vector moves away from the magnetic field direction, the flux decreases. When the area vector becomes perpendicular to the magnetic field:

\[ \theta=90^\circ \]

The magnetic flux becomes zero.

Keep rotating, and the flux becomes negative before reaching its maximum negative value at \(180^\circ\).

The predicted relationship is:

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

This is a cosine relationship between magnetic flux and the angle of rotation.

The surprising result: Maximum flux does not mean maximum EMF

Here’s where the experiment becomes particularly useful.

The voltage sensor measures induced EMF, not magnetic flux directly.

According to Faraday’s law:

\[ \varepsilon=-N\frac{d\Phi_B}{dt} \]

The induced EMF depends on how quickly magnetic flux changes.

For a coil rotating at constant angular speed, let:

\[ \theta=\omega t \]

where \(\omega\) is angular speed in radians per second and \(t\) is time in seconds.

The flux varies as:

\[ \Phi_B=BA\cos(\omega t) \]

The resulting induced EMF has the form:

\[ \varepsilon=NBA\omega\sin(\omega t) \]

The important result is that the flux and induced EMF reach their maximum magnitudes at different orientations.

A student might see the largest voltage reading when the coil’s plane is parallel to the magnetic field and conclude that this must be the position of maximum flux.

That conclusion is incorrect.

When the plane is parallel to the field, the area vector is perpendicular to the field. The flux is zero, but it is changing most rapidly as the coil rotates.

At maximum flux, the flux is momentarily changing at its slowest rate. Consequently, the induced EMF is zero at that instant.

A limitation of the experimental design

There is an important flaw in a tempting first approach.

Suppose you rotate the coil to different angles, stop it, and measure the induced voltage at each position.

You will obtain zero induced EMF at every position because the coil is stationary and its magnetic flux is no longer changing.

That experiment cannot identify the maximum-flux orientation using voltage readings alone.

A better design is to rotate the coil continuously at a constant angular speed while recording both its angular position and induced EMF.

The predicted zero-EMF positions can then be compared with the known coil orientations. The direction of rotation and the initial orientation distinguish the positive and negative flux maxima.

This is an important experimental distinction: the Hall probe measures the local magnetic field, while the voltage sensor measures induced EMF. Neither instrument directly gives the total magnetic flux through the coil.

Worked scale estimate: What voltage should the rotating coil produce?

Consider an illustrative rotating coil with:

• 40 turns.
• Area of \(0.0060\,\mathrm{m^2}\).
• Magnetic field strength of \(0.20\,\mathrm T\).
• Rotation frequency of \(1.0\,\mathrm{Hz}\).

First, calculate maximum flux linkage:

\[ \begin{aligned} \lambda_{\max}&=NBA\\ &=40\times0.20\times0.0060\\ &=0.048\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Next, calculate angular speed:

\[ \begin{aligned} \omega&=2\pi f\\ &=2\pi\times1.0\\ &=6.28\,\mathrm{rad\,s^{-1}} \end{aligned} \]

For uniform rotation in a uniform magnetic field, maximum induced EMF is:

\[ \varepsilon_{\max}=NBA\omega \]

Therefore:

\[ \begin{aligned} \varepsilon_{\max}&=\lambda_{\max}\omega\\ &=0.048\times6.28\\ &=0.30\,\mathrm V \end{aligned} \]

The predicted peak induced EMF is approximately \(0.30\,\mathrm V\).

Notice that this voltage peak occurs when the magnetic flux is zero, not when it is maximum.

This calculation assumes a rigid, flat coil, a uniform magnetic field, constant rotational speed, and an ideal measurement of induced EMF.

05A reliable method for HSC exam questions

When an exam question asks you to find maximum magnetic flux through a rectangular coil, use this sequence.

Pay particular attention to the wording.

Two final reminders:

• Convert all dimensions into metres before calculating area.
• Only multiply by the number of turns when calculating flux linkage, rather than flux through one turn.

06Questions and solutions

The following questions progress from standard calculations to more demanding applications. Pay attention to the information each question gives you about the coil’s orientation.

Question 1

A rectangular coil has 60 turns and dimensions of 12.0 cm by 5.0 cm. It is placed in a uniform magnetic field of strength \(0.25\,\mathrm T\), with the magnetic field perpendicular to the plane of the coil.

Calculate:

(a) The maximum magnetic flux through one turn.

(b) The maximum magnetic flux linkage.

Assume all turns enclose the same area and have the same orientation.

Solution 1

The maximum flux is \(1.5\times10^{-3}\,\mathrm{Wb}\) per turn, and the maximum flux linkage is \(0.090\,\mathrm{Wb\text{-}turn}\).

Step 1: Calculate the area.

Convert the dimensions into metres:

\[ \ell=0.120\,\mathrm m \]

\[ w=0.0500\,\mathrm m \]

Therefore:

\[ \begin{aligned} A&=\ell w\\ &=0.120\times0.0500\\ &=0.00600\,\mathrm{m^2} \end{aligned} \]

Step 2: Identify the angle.

The magnetic field is perpendicular to the coil’s plane.

Therefore, the area vector is parallel to the magnetic field:

\[ \theta=0^\circ \]

Step 3: Calculate maximum flux per turn.

\[ \begin{aligned} \Phi_{B,\max}&=BA\\ &=0.25\times0.00600\\ &=0.00150\,\mathrm{Wb} \end{aligned} \]

Step 4: Calculate maximum flux linkage.

\[ \begin{aligned} \lambda_{\max}&=N\Phi_{B,\max}\\ &=60\times0.00150\\ &=0.090\,\mathrm{Wb\text{-}turn} \end{aligned} \]

The two answers are:

• (a) \(1.50\times10^{-3}\,\mathrm{Wb}\).
• (b) \(0.090\,\mathrm{Wb\text{-}turn}\).

The important distinction is that the first answer describes one turn, while the second includes all 60 turns.

Question 2

A rectangular coil has 150 turns and dimensions of 18.0 cm by 6.0 cm. It is placed in a uniform magnetic field of strength \(0.50\,\mathrm T\).

The plane of the coil makes an angle of \(25^\circ\) with the magnetic field.

Calculate:

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

(b) The magnetic flux linkage.

(c) The percentage of maximum possible flux represented by the actual flux.

Solution 2

The flux per turn is \(2.28\times10^{-3}\,\mathrm{Wb}\), the flux linkage is \(0.342\,\mathrm{Wb\text{-}turn}\), and the actual flux is 42.3% of its maximum possible value.

Step 1: Calculate the area.

\[ \begin{aligned} A&=0.180\times0.0600\\ &=0.0108\,\mathrm{m^2} \end{aligned} \]

Step 2: Convert the given angle.

The \(25^\circ\) angle is measured from the coil’s plane.

Therefore:

\[ \theta=90^\circ-25^\circ \]

\[ \theta=65^\circ \]

Step 3: Calculate the magnetic flux per turn.

\[ \begin{aligned} \Phi_B&=BA\cos\theta\\ &=0.50\times0.0108\times\cos65^\circ\\ &=0.00228\,\mathrm{Wb} \end{aligned} \]

Step 4: Calculate flux linkage.

\[ \begin{aligned} \lambda&=N\Phi_B\\ &=150\times0.00228\\ &=0.342\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Step 5: Compare the actual flux with maximum flux.

First, calculate maximum flux:

\[ \begin{aligned} \Phi_{B,\max}&=BA\\ &=0.50\times0.0108\\ &=0.00540\,\mathrm{Wb} \end{aligned} \]

Now calculate the percentage:

\[ \begin{aligned} \text{Percentage}&=\frac{\Phi_B}{\Phi_{B,\max}}\times100\\ &=\frac{0.00228}{0.00540}\times100\\ &=42.3\% \end{aligned} \]

The actual flux is 42.3% of the maximum possible flux.

The common mistake is using \(\cos25^\circ\), which incorrectly treats the given angle as the angle between B and the area vector. Always identify what the angle is measured from before substituting.

Question 3

A 100-turn rectangular coil has dimensions of 20.0 cm by 8.0 cm. It is positioned in a uniform magnetic field of strength \(0.30\,\mathrm T\).

Initially, the angle between the magnetic field and the coil’s area vector is \(40^\circ\).

The coil is then rotated about an axis in its plane until the area vector makes an angle of \(140^\circ\) with the magnetic field.

Taking the initial area vector as the positive reference direction, calculate:

(a) The initial signed magnetic flux through one turn.

(b) The final signed magnetic flux through one turn.

(c) The initial and final flux linkages.

(d) The change in flux linkage.

Solution 3

The flux linkage changes from \(+0.368\) to \(-0.368\,\mathrm{Wb\text{-}turn}\). Therefore, the change in flux linkage is \(-0.736\,\mathrm{Wb\text{-}turn}\).

The negative sign indicates that the final flux linkage is opposite to the chosen positive reference direction.

Step 1: Calculate the coil’s area.

\[ \begin{aligned} A&=0.200\times0.0800\\ &=0.0160\,\mathrm{m^2} \end{aligned} \]

Step 2: Calculate the maximum flux magnitude per turn.

\[ \begin{aligned} BA&=0.30\times0.0160\\ &=0.00480\,\mathrm{Wb} \end{aligned} \]

This is the maximum possible flux magnitude for the given field and area.

Step 3: Calculate the initial flux.

The initial angle is \(40^\circ\).

\[ \begin{aligned} \Phi_{B,i}&=BA\cos40^\circ\\ &=0.00480\times0.7660\\ &=+0.00368\,\mathrm{Wb} \end{aligned} \]

Step 4: Calculate the final flux.

The final angle is \(140^\circ\).

\[ \begin{aligned} \Phi_{B,f}&=BA\cos140^\circ\\ &=0.00480\times(-0.7660)\\ &=-0.00368\,\mathrm{Wb} \end{aligned} \]

Notice that the magnitudes are identical, but the signs are opposite.

Step 5: Calculate the initial and final flux linkages.

Initially:

\[ \begin{aligned} \lambda_i&=N\Phi_{B,i}\\ &=100\times0.00368\\ &=+0.368\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Finally:

\[ \begin{aligned} \lambda_f&=N\Phi_{B,f}\\ &=100\times(-0.00368)\\ &=-0.368\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Step 6: Calculate the change in flux linkage.

Always subtract the initial value from the final value:

\[ \begin{aligned} \Delta\lambda&=\lambda_f-\lambda_i\\ &=-0.368-(+0.368)\\ &=-0.736\,\mathrm{Wb\text{-}turn} \end{aligned} \]

The change has twice the magnitude of the initial flux linkage because the coil moves from an initially positive flux to an equally large negative flux.

A tempting mistake is to compare only the magnitudes and conclude that the change is zero. That ignores the reversal of the area vector relative to the magnetic field.

Question 4

An 80-turn rectangular coil has dimensions of 15.0 cm by 10.0 cm. It rotates at a constant frequency of \(2.0\,\mathrm{Hz}\) in a uniform magnetic field of strength \(0.20\,\mathrm T\).

The rotation axis passes through the centre of the coil, lies in its plane, and is perpendicular to the magnetic field.

A voltage sensor connected through slip rings records its largest induced EMF when the plane of the coil is parallel to the magnetic field.

A student claims that this must also be the position of maximum magnetic flux.

Evaluate the student’s claim.

Calculate:

(a) The maximum magnetic flux through one turn.

(b) The maximum flux linkage.

(c) The maximum induced EMF.

Assume uniform magnetic field strength, constant angular speed, and ideal measurement conditions.

Solution 4

The student’s claim is incorrect. The coil’s magnetic flux is zero when its plane is parallel to the magnetic field, but its induced EMF is maximum because the flux is changing most rapidly.

Step 1: Calculate the rectangular area.

\[ \begin{aligned} A&=0.150\times0.100\\ &=0.0150\,\mathrm{m^2} \end{aligned} \]

Step 2: Calculate maximum flux per turn.

The maximum flux occurs when the area vector is parallel to the magnetic field.

\[ \begin{aligned} \Phi_{B,\max}&=BA\\ &=0.20\times0.0150\\ &=0.00300\,\mathrm{Wb} \end{aligned} \]

Step 3: Calculate maximum flux linkage.

\[ \begin{aligned} \lambda_{\max}&=N\Phi_{B,\max}\\ &=80\times0.00300\\ &=0.240\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Step 4: Calculate angular speed.

The coil rotates at \(2.0\) revolutions per second.

Use:

\[ \omega=2\pi f \]

Therefore:

\[ \begin{aligned} \omega&=2\pi\times2.0\\ &=12.57\,\mathrm{rad\,s^{-1}} \end{aligned} \]

Step 5: Calculate maximum induced EMF.

For a coil rotating uniformly in a uniform magnetic field:

\[ \varepsilon_{\max}=NBA\omega \]

Using the maximum flux linkage:

\[ \begin{aligned} \varepsilon_{\max}&=\lambda_{\max}\omega\\ &=0.240\times12.57\\ &=3.02\,\mathrm V \end{aligned} \]

Therefore, the maximum induced EMF is approximately:

\[ \boxed{\varepsilon_{\max}=3.0\,\mathrm V} \]

The student’s mistake is confusing the value of magnetic flux with its rate of change.

When the coil’s plane is parallel to the magnetic field, the area vector is perpendicular to B. Therefore:

\[ \Phi_B=0 \]

However, as the coil continues rotating, the flux changes most rapidly at this position, producing maximum induced EMF.

At the opposite position, where flux reaches its maximum magnitude, the induced EMF momentarily becomes zero.

Question 5

Two 25-turn rectangular coils are made from equal lengths of wire. Each turn has a perimeter of \(0.80\,\mathrm m\).

Both coils are placed in the same uniform magnetic field of strength \(0.40\,\mathrm T\).

Coil A is a rigid rectangle measuring \(0.30\,\mathrm m\) by \(0.10\,\mathrm m\). It can be positioned with its plane perpendicular to the magnetic field.

Coil B is an adjustable rectangular frame. Its side lengths can change while its perimeter remains fixed at \(0.80\,\mathrm m\). However, its plane is constrained to make an angle of \(60^\circ\) with the magnetic field.

Assume that:

• Both coils have 25 turns.
• Every turn has the same area and orientation within its respective coil.
• The adjustable frame can change its aspect ratio without changing its perimeter.
• Both coils experience the same uniform magnetic field.
• Wire thickness and any mechanical limitations are negligible.

Answer the following:

(a) Calculate the maximum possible flux linkage of coil A.

(b) Determine the largest area coil B can enclose while remaining rectangular.

(c) Calculate the maximum possible flux linkage of coil B under its \(60^\circ\) orientation constraint.

(d) Which coil produces the greater flux linkage, and by what percentage?

(e) At what acute angle between coil B’s plane and the magnetic field would its maximum flux linkage exactly equal that of coil A? For what range of plane angles would coil B produce greater flux linkage?

Solution 5

Coil B can achieve the greater flux linkage, despite its less favourable orientation. Its maximum flux linkage is \(0.346\,\mathrm{Wb\text{-}turn}\), compared with \(0.300\,\mathrm{Wb\text{-}turn}\) for coil A.

The key is that magnetic flux depends on both area and orientation. Coil B can enclose a larger area by changing its rectangular shape.

Step 1: Calculate the maximum flux linkage of coil A.

Coil A has dimensions:

\[ \ell=0.30\,\mathrm m \]

\[ w=0.10\,\mathrm m \]

Its area is:

\[ \begin{aligned} A_A&=0.30\times0.10\\ &=0.030\,\mathrm{m^2} \end{aligned} \]

The plane of coil A is perpendicular to the magnetic field, so its area vector is parallel to B.

Therefore:

\[ \theta_A=0^\circ \]

Its maximum flux linkage is:

\[ \begin{aligned} \lambda_A&=NBA_A\\ &=25\times0.40\times0.030\\ &=0.300\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Step 2: Determine the maximum possible area of coil B.

This is the important geometric part of the problem.

Coil B has a fixed perimeter of \(0.80\,\mathrm m\).

Let its side lengths be \(x\) and \(y\).

The perimeter equation is:

\[ 2x+2y=0.80 \]

Simplifying:

\[ x+y=0.40 \]

We want to maximise the rectangular area:

\[ A_B=xy \]

For a fixed sum, the product of two positive numbers is greatest when the numbers are equal.

We can show this mathematically:

\[ (x-y)^2\geq0 \]

Expanding:

\[ x^2-2xy+y^2\geq0 \]

This gives:

\[ (x+y)^2\geq4xy \]

Therefore:

\[ xy\leq\frac{(x+y)^2}{4} \]

Substituting \(x+y=0.40\):

\[ \begin{aligned} A_{B,\max}&=\frac{0.40^2}{4}\\ &=0.040\,\mathrm{m^2} \end{aligned} \]

The maximum area occurs when:

\[ x=y=0.20\,\mathrm m \]

In other words, the adjustable rectangular frame achieves its maximum area when it becomes a square.

Step 3: Calculate the maximum flux linkage of coil B.

Coil B’s plane makes an angle of \(60^\circ\) with the magnetic field.

Remember that the area vector is perpendicular to the plane.

Therefore:

\[ \theta_B=90^\circ-60^\circ \]

\[ \theta_B=30^\circ \]

Using the maximum area:

\[ \begin{aligned} \lambda_B&=NBA_B\cos\theta_B\\ &=25\times0.40\times0.040\times\cos30^\circ\\ &=0.346\,\mathrm{Wb\text{-}turn} \end{aligned} \]

Step 4: Compare the two coils.

Coil A:

\[ \lambda_A=0.300\,\mathrm{Wb\text{-}turn} \]

Coil B:

\[ \lambda_B=0.346\,\mathrm{Wb\text{-}turn} \]

Calculate the percentage increase:

\[ \begin{aligned} \text{Increase} &=\frac{0.346-0.300}{0.300}\times100\\ &=15.5\% \end{aligned} \]

Coil B produces approximately 15.5% greater flux linkage.

Why does this happen?

Coil A has the ideal orientation, but its area is smaller.

Coil B is tilted, which reduces its flux, but its larger area more than compensates for that disadvantage.

The relevant comparison is the effective area perpendicular to the magnetic field, not simply the actual area or the orientation considered separately.

Step 5: Determine the angle at which the two coils have equal flux linkage.

Now suppose the angle of coil B changes, while its area remains at its maximum value.

Let \(\alpha\) represent the acute angle between coil B’s plane and the magnetic field.

The magnitude of flux linkage for coil B is:

\[ \lambda_B=NBA_B\sin\alpha \]

For both coils to have equal flux linkage:

\[ \lambda_B=\lambda_A \]

Substitute the known values:

\[ 25\times0.40\times0.040\times\sin\alpha = 0.300 \]

Simplifying:

\[ 0.40\sin\alpha=0.300 \]

\[ \sin\alpha=\frac{0.300}{0.40} \]

\[ \sin\alpha=0.75 \]

Therefore:

\[ \boxed{\alpha=48.6^\circ} \]

At approximately \(48.6^\circ\), the two coils have equal maximum flux linkage under the stated constraints.

For larger acute plane-to-field angles, coil B produces greater flux linkage. For smaller angles, coil A produces greater flux linkage.

This is the important conclusion: the coil with the most favourable orientation does not necessarily have the greatest magnetic flux. You must consider the area and orientation together.

Once you can calculate magnetic flux at different coil orientations, the next step is understanding how quickly that flux changes. This is what determines induced EMF under Faraday’s law.

Continue with How to Read a Magnetic-Flux-Time Graph to Calculate EMF to connect flux calculations with induced voltage.

If you want to explore how the enclosed area affects flux when the coil’s shape changes, see How to Compare Magnetic Flux Through Circular and Square Coils.