How to Compare Magnetic Flux Through Circular and Square Coils

Two coils sit in the same magnetic field. One is circular, and the other is square. Which experiences more magnetic flux?

Before calculating, consider two possibilities. If both coils enclose the same area and have the same orientation, does their different shape matter? No. They experience exactly the same magnetic flux per turn.

But if the coils are made from the same length of wire, the answer changes. A circular loop encloses more area than a square loop with the same perimeter, so the circular coil experiences more flux when both have the same orientation.

The key is that magnetic flux depends on three things: magnetic field strength, enclosed area, and the angle between the field and the coil’s area vector. The shape matters only through the area it encloses.

01The equation that compares circular and square coils

Magnetic flux measures how much magnetic field passes through a surface.

For a flat coil in a uniform magnetic field:

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

The symbols represent:

One weber (Wb) is equivalent to one tesla multiplied by one square metre.

The important detail is that the equation uses the area enclosed by the coil, not the length of wire around its edge.

Calculating the area of each coil

For a circular coil, use the area of a circle:

\[ A_{\mathrm{circle}}=\pi r^2 \]

where \(r\) is the radius.

For a square coil, use:

\[ A_{\mathrm{square}}=s^2 \]

where \(s\) is the side length.

Substituting these into the magnetic flux equation gives:

Circular coil:

\[ \Phi_c=B\pi r^2\cos\theta \]

Square coil:

\[ \Phi_s=Bs^2\cos\theta \]

If both coils have the same orientation and sit in the same uniform magnetic field, their flux ratio is:

\[ \boxed{\frac{\Phi_c}{\Phi_s} =\frac{\pi r^2}{s^2}} \]

This is the first useful shortcut.

You do not need to calculate the magnetic flux of both coils separately if the field strength and orientation are identical. You can compare their areas directly.

What happens if their areas are equal?

Suppose a circular coil and a square coil each enclose an area of \(0.020\,\mathrm{m^2}\).

Both are placed in a uniform magnetic field of \(0.40\,\mathrm T\), with their area normals pointing in the same direction.

Their fluxes are identical:

\[ \Phi_c=\Phi_s=BA\cos\theta \]

The circle does not receive extra flux simply because it is circular.

The magnetic field responds to the surface area and orientation, not whether the boundary is curved or straight.

02The area vector: getting the angle right

There is one detail that regularly causes mistakes in HSC Physics questions: the angle in the flux equation is measured from the normal to the coil, not from the plane of the coil.

The area vector is an imaginary vector that:

• Has a magnitude equal to the coil’s enclosed area.
• Points perpendicular to the plane of the coil.

We write it as:

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

where \(\hat n\) is a unit vector perpendicular to the coil.

The magnetic flux equation can therefore be written as a dot product:

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

This explains why the cosine appears.

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

Suppose a question states that the plane of a coil makes an angle of \(30^\circ\) with the magnetic field.

You cannot substitute \(30^\circ\) directly into \(BA\cos\theta\).

You first need to find the angle between the magnetic field and the normal.

For an acute angle between the field and the plane:

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

Therefore:

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

\[ \theta=60^\circ \]

The flux is then:

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

Alternatively, because \(\cos(90^\circ-\alpha)=\sin\alpha\), you can write:

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

Both methods give the same magnitude.

How orientation changes flux

The following table shows what happens as the coil rotates.

Notice something important.

When the magnetic field is perpendicular to the plane of the coil, the flux is greatest. When the magnetic field runs parallel to the plane, the flux is zero.

This is because the field has no component passing through the surface when it runs parallel to it.

A common HSC trap: confusing flux and torque

The angle convention also appears in questions about the torque on a current-carrying coil.

However, the equations are different:

Magnetic flux:

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

Magnetic torque:

\[ \tau=NIAB\sin\theta \]

In both equations, θ can represent the angle between the magnetic field and the coil’s normal.

Do not swap the sine and cosine functions. They describe different physical quantities.

For a separate explanation of the motor-effect relationship, see Torque on a Current-Carrying Coil for HSC Physics.

03Comparing flux linkage, not just flux

So far, we have considered the magnetic flux through one turn of a coil.

Real coils often contain many turns. This introduces another quantity: magnetic flux linkage.

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

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

where:

• \(\lambda\) is the magnetic flux linkage.
• \(N\) is the number of turns.
• \(\Phi_B\) is the magnetic flux through one turn.

Flux linkage has the SI unit Wb. It is also commonly written as Wb-turn to emphasise the number of turns.

This distinction matters when comparing circular and square coils.

Imagine a circular coil with 20 turns and a square coil with 40 turns. Even if the circular coil has greater flux through each turn, the square coil could have greater total flux linkage.

For two coils in the same uniform magnetic field:

\[ \lambda_c= N_cB\pi r^2\cos\theta_c \]

\[ \lambda_s=N_sBs^2\cos\theta_s \]

Therefore, to compare their flux linkages, you must consider:

1. The area enclosed by each turn.
2. The orientation of each coil.
3. The number of turns.

Do not multiply by the number of turns when a question asks only for the magnetic flux through one turn.

This distinction becomes particularly important when calculating induced EMF using Faraday’s law.

For an average induced EMF:

\[ \mathcal E_{\mathrm{avg}} =-\frac{\Delta(N\Phi_B)}{\Delta t} \]

The relevant quantity is now the change in flux linkage over time.

For an application of this relationship, see How Wireless Charging Transfers Energy Through Changing Magnetic Flux.

04What if the circular and square coils have the same perimeter?

This is where the comparison becomes interesting.

Suppose you have two coils made from exactly the same length of wire.

One is circular, and the other is square.

Would they enclose the same area?

No. A circle encloses more area than a square with the same perimeter.

Let’s establish why.

For a circular coil with perimeter \(P\):

\[ P=2\pi r \]

Rearranging:

\[ r=\frac{P}{2\pi} \]

Its area is therefore:

\[ A_c=\pi\left(\frac{P}{2\pi}\right)^2 \]

\[ \boxed{A_c=\frac{P^2}{4\pi}} \]

For a square coil:

\[ P=4s \]

Therefore:

\[ s=\frac{P}{4} \]

Its area is:

\[ A_s=\left(\frac{P}{4}\right)^2 \]

\[ \boxed{A_s=\frac{P^2}{16}} \]

We can now compare the areas:

\[ \frac{A_c}{A_s} = \frac{P^2/(4\pi)}{P^2/16} \]

Simplifying:

\[ \boxed{\frac{A_c}{A_s}=\frac{4}{\pi}} \]

\[ \frac{A_c}{A_s}\approx1.273 \]

The circular coil encloses approximately 27.3% more area.

If both coils have the same orientation, magnetic field strength, and number of turns, the circular coil also has approximately 27.3% greater flux linkage.

05Testing whether shape actually matters: an experimental investigation

So far, we have used mathematical relationships to compare the coils.

How could we test the prediction experimentally?

The central question is whether a difference in flux comes from the shape itself or simply from the different areas enclosed by the coils.

A fair experiment needs to separate these two possibilities.

Setting up the apparatus

Imagine an electromagnet with two broad pole pieces facing each other.

The gap between the pole pieces contains an approximately uniform magnetic field.

A flat test coil is positioned in the middle of this gap. Its plane is perpendicular to the magnetic field, so its area normal is parallel to the field.

Two flexible wires connect the coil to a fluxmeter or a voltage sensor capable of integrating the induced voltage.

The procedure is:

1. Place the circular coil inside the uniform magnetic field.
2. Record its initial flux linkage using the fluxmeter.
3. Withdraw the coil completely from the field.
4. Record the change in flux linkage.
5. Repeat using the square coil under the same conditions.

The fluxmeter can determine the change in flux linkage from the integrated induced voltage:

\[ \left|\int\mathcal E\,dt\right| =N|\Delta\Phi_B| \]

If the coil starts fully inside the field and ends completely outside it, the final flux is zero. The magnitude of the integrated reading therefore gives the initial flux linkage.

The withdrawal speed affects the size of the induced voltage at a particular instant, but not the total integrated voltage for the same initial and final fluxes.

The first comparison: equal perimeter

Suppose each turn of the circular and square coils has a perimeter of \(0.400\,\mathrm m\).

Both coils have 50 turns and are placed in a uniform magnetic field of \(0.30\,\mathrm T\).

Their normals are parallel to the magnetic field.

From our earlier calculation:

• Circular coil area: \(0.0127\,\mathrm{m^2}\).
• Square coil area: \(0.0100\,\mathrm{m^2}\).

The predicted flux linkages are:

\[ \lambda_c=50(0.30)(0.0127) \]

\[ \lambda_c\approx0.191\,\mathrm{Wb} \]

For the square coil:

\[ \lambda_s=50(0.30)(0.0100) \]

\[ \lambda_s=0.150\,\mathrm{Wb} \]

The circular coil should produce a larger fluxmeter reading.

These are illustrative theoretical predictions, not reported experimental measurements.

However, there is a problem with this experiment.

The coils have different enclosed areas.

We cannot conclude that the shape itself caused the difference. The area is a variable we have not controlled.

Revising the experiment: equal enclosed areas

To isolate the effect of shape, we need to make the enclosed areas equal.

Keep the circular coil unchanged, but increase the square coil’s side length until its area matches the circular coil.

The required square side length is:

\[ s=\sqrt{A_c} \]

\[ s=\sqrt{0.0127} \]

\[ s\approx0.1128\,\mathrm m \]

Now both coils have:

• The same enclosed area.
• The same number of turns.
• The same magnetic field strength.
• The same orientation.

The predicted flux linkages become:

\[ \lambda_c=\lambda_s \]

\[ \boxed{\lambda\approx0.191\,\mathrm{Wb}} \]

If the measured integrated readings agree within experimental uncertainty, this supports the prediction that shape itself does not affect flux in a uniform magnetic field.

The revised experiment has addressed the original design’s main weakness: the areas are now controlled.

When does this model stop working?

There is one important limitation.

The equation \(\Phi_B=BA\cos\theta\) assumes that the magnetic field is uniform across the coil’s entire enclosed area.

If the magnetic field varies significantly across the surface, equal-area circular and square coils can experience different fluxes.

For example, one coil might enclose a region where the field is stronger near the centre, while another extends further into a weaker-field region.

In that situation, the field strength must be considered across different parts of the surface. The simple uniform-field equation is no longer sufficient.

For standard uniform-field comparisons, however, the area-vector method is the appropriate starting point.

06A reliable method for HSC exam questions

When comparing circular and square coils, use the following sequence.

Before moving on, remember these four comparisons:

• Equal area and equal orientation mean equal flux per turn.
• Equal perimeter does not mean equal area.
• Different orientations mean you must compare projected areas, not just actual areas.
• Different numbers of turns can change which coil has the greater flux linkage.

07Questions and solutions

The following questions progress from straightforward area comparisons to problems involving orientation, experimental measurements, and changing magnetic fields.

Question 1

A one-turn circular coil has a radius of \(6.0\,\mathrm{cm}\). A one-turn square coil has a side length of \(10.0\,\mathrm{cm}\).

Both coils are positioned perpendicular to a uniform magnetic field of \(0.25\,\mathrm T\).

Calculate the magnetic flux through each coil and determine which experiences greater flux.

Solution 1

The circular coil experiences greater magnetic flux, approximately 13% more than the square coil.

Both coils have their area normals parallel to the magnetic field, so \(\theta=0^\circ\).

Step 1: Calculate the area of the circular coil.

Convert the radius into metres:

\[ r=0.060\,\mathrm m \]

\[ A_c=\pi r^2 \]

\[ A_c=\pi(0.060)^2 \]

\[ A_c=0.01131\,\mathrm{m^2} \]

Step 2: Calculate the area of the square coil.

\[ A_s=s^2 \]

\[ A_s=(0.100)^2 \]

\[ A_s=0.0100\,\mathrm{m^2} \]

Step 3: Calculate the flux through each coil.

For the circular coil:

\[ \begin{aligned} \Phi_c&=BA_c\cos0^\circ\\ &=0.25(0.01131)\\ &=2.83\times10^{-3}\,\mathrm{Wb} \end{aligned} \]

For the square coil:

\[ \begin{aligned} \Phi_s&=BA_s\cos0^\circ\\ &=0.25(0.0100)\\ &=2.50\times10^{-3}\,\mathrm{Wb} \end{aligned} \]

Step 4: Compare the fluxes.

\[ \frac{\Phi_c}{\Phi_s} =\frac{2.83}{2.50} \]

\[ \frac{\Phi_c}{\Phi_s}=1.13 \]

The circular coil has approximately 1.13 times the flux of the square coil.

This is because its enclosed area is larger, even though both coils experience the same magnetic field and orientation.

Question 2

A circular coil has 20 turns and a radius of \(5.0\,\mathrm{cm}\). A square coil has 50 turns and a side length of \(8.0\,\mathrm{cm}\).

Both are placed in a uniform magnetic field of \(0.40\,\mathrm T\).

The normal to the circular coil makes an angle of \(30^\circ\) with the magnetic field.

The plane of the square coil makes an angle of \(30^\circ\) with the magnetic field.

Calculate:

(a) The magnetic flux through one turn of each coil.

(b) The flux linkage of each coil.

(c) Which coil has the greater flux per turn and which has the greater flux linkage?

Solution 2

The circular coil has greater flux per turn, but the square coil has greater flux linkage because it contains more turns.

This is a useful example of why flux and flux linkage must be treated separately.

Step 1: Identify the angles.

For the circular coil, the normal-field angle is already given:

\[ \theta_c=30^\circ \]

For the square coil, the given angle is between its plane and the field.

Therefore:

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

\[ \theta_s=60^\circ \]

Step 2: Calculate the area of each coil.

For the circular coil:

\[ A_c=\pi(0.050)^2 \]

\[ A_c=0.007854\,\mathrm{m^2} \]

For the square coil:

\[ A_s=(0.080)^2 \]

\[ A_s=0.00640\,\mathrm{m^2} \]

Step 3: Calculate the magnetic flux per turn.

For the circular coil:

\[ \begin{aligned} \Phi_c&=BA_c\cos30^\circ\\ &=0.40(0.007854)(0.866)\\ &=2.72\times10^{-3}\,\mathrm{Wb} \end{aligned} \]

For the square coil:

\[ \begin{aligned} \Phi_s&=BA_s\cos60^\circ\\ &=0.40(0.00640)(0.5)\\ &=1.28\times10^{-3}\,\mathrm{Wb} \end{aligned} \]

The circular coil has greater flux through each individual turn.

Step 4: Calculate the flux linkage.

For the circular coil:

\[ \lambda_c=N_c\Phi_c \]

\[ \lambda_c=20(2.72\times10^{-3}) \]

\[ \boxed{\lambda_c=0.0544\,\mathrm{Wb}} \]

For the square coil:

\[ \lambda_s=N_s\Phi_s \]

\[ \lambda_s=50(1.28\times10^{-3}) \]

\[ \boxed{\lambda_s=0.0640\,\mathrm{Wb}} \]

Step 5: Compare the results.

The circular coil has greater flux per turn:

\[ 0.00272>0.00128 \]

However, the square coil has greater flux linkage:

\[ 0.0640>0.0544 \]

The square coil’s greater number of turns compensates for its smaller flux per turn.

The tempting mistake is to assume that the coil with greater magnetic flux must also have greater flux linkage. That is only true if the number of turns is equal.

Question 3

A circular coil and a square coil are each made from \(0.48\,\mathrm m\) of wire per turn.

Both coils contain one turn and are placed in a uniform magnetic field of \(0.35\,\mathrm T\).

The circular coil is positioned with its area normal parallel to the magnetic field.

The square coil can be rotated freely.

(a) Calculate the maximum possible magnetic flux through each coil.

(b) Can the square coil be rotated to produce the same magnetic flux as the circular coil?

Justify your answer mathematically.

Solution 3

The circular coil has a maximum flux of approximately \(6.42\times10^{-3}\,\mathrm{Wb}\), while the square coil’s maximum is \(5.04\times10^{-3}\,\mathrm{Wb}\).

The square coil cannot produce the same flux, regardless of its orientation.

Step 1: Calculate the area of the circular coil.

The circumference is:

\[ P=2\pi r \]

Rearranging:

\[ r=\frac{0.48}{2\pi} \]

\[ r=0.0764\,\mathrm m \]

Therefore:

\[ A_c=\pi(0.0764)^2 \]

\[ A_c\approx0.01833\,\mathrm{m^2} \]

Step 2: Calculate the area of the square coil.

\[ s=\frac{0.48}{4} \]

\[ s=0.12\,\mathrm m \]

Therefore:

\[ A_s=(0.12)^2 \]

\[ A_s=0.0144\,\mathrm{m^2} \]

Step 3: Calculate the maximum flux through the circular coil.

The circular coil has its normal parallel to the field, so its flux is already at its maximum.

\[ \Phi_{c,\max}=BA_c \]

\[ \Phi_{c,\max}=0.35(0.01833) \]

\[ \boxed{\Phi_{c,\max}=0.00642\,\mathrm{Wb}} \]

Step 4: Calculate the maximum flux through the square coil.

The square coil also achieves maximum flux when its normal is parallel to the field.

\[ \Phi_{s,\max}=BA_s \]

\[ \Phi_{s,\max}=0.35(0.0144) \]

\[ \boxed{\Phi_{s,\max}=0.00504\,\mathrm{Wb}} \]

Step 5: Determine whether rotating the square coil can produce equal flux.

The square coil’s flux magnitude is:

\[ |\Phi_s|=BA_s|\cos\theta| \]

Because:

\[ |\cos\theta|\leq1 \]

its flux can never exceed \(BA_s\).

To match the circular coil’s maximum flux, the square coil would require:

\[ 0.35(0.0144)\cos\theta =0.00642 \]

Rearranging:

\[ \cos\theta=\frac{0.00642}{0.00504} \]

\[ \cos\theta\approx1.27 \]

This is impossible because the cosine of a real angle cannot have a magnitude greater than one.

The square coil cannot match the circular coil’s maximum flux.

This is more than a simple comparison of areas. The circular coil already has the greatest possible flux for its orientation, while the square coil has a smaller maximum available flux.

Question 4

A circular coil and a square coil each have:

• 40 turns.
• An enclosed area of \(1.50\times10^{-2}\,\mathrm{m^2}\) per turn.
• Their area normals initially parallel to a uniform magnetic field of \(0.20\,\mathrm T\).

Both coils are completely withdrawn from the magnetic field.

The circular coil takes \(0.12\,\mathrm s\) to leave the field, while the square coil takes \(0.24\,\mathrm s\).

(a) Calculate the magnitude of the average induced EMF in each coil.

(b) Compare the total integrated induced voltage for the two coils.

(c) A student concludes that the circular coil initially experienced more magnetic flux because its induced EMF was larger. Explain why this conclusion is incorrect.

Solution 4

The circular coil has an average induced EMF of \(1.0\,\mathrm V\), while the square coil has an average induced EMF of \(0.50\,\mathrm V\). However, both coils experience the same initial flux linkage.

The difference in induced EMF comes from the different withdrawal times, not the shape of the coils.

Step 1: Calculate the initial flux through one turn.

Both coils have the same area and orientation.

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

\[ \Phi_B=0.20(1.50\times10^{-2}) \]

\[ \Phi_B=0.0030\,\mathrm{Wb} \]

Step 2: Calculate the initial flux linkage.

Both coils contain 40 turns.

\[ \lambda=N\Phi_B \]

\[ \lambda=40(0.0030) \]

\[ \lambda=0.12\,\mathrm{Wb} \]

Both coils have identical initial flux linkage.

Step 3: Calculate the average EMF for the circular coil.

Using the magnitude form of Faraday’s law:

\[ |\mathcal E_{\mathrm{avg}}| =\frac{|\Delta\lambda|}{\Delta t} \]

The coil is completely withdrawn, so its final flux linkage is zero.

\[ |\mathcal E_c| =\frac{0.12}{0.12} \]

\[ \boxed{|\mathcal E_c|=1.0\,\mathrm V} \]

Step 4: Calculate the average EMF for the square coil.

\[ |\mathcal E_s| =\frac{0.12}{0.24} \]

\[ \boxed{|\mathcal E_s|=0.50\,\mathrm V} \]

Step 5: Compare the integrated voltages.

The magnitude of the integrated voltage is:

\[ \left|\int\mathcal E\,dt\right| =|\Delta\lambda| \]

For both coils:

\[ \boxed{\left|\int\mathcal E\,dt\right|=0.12\,\mathrm{Wb}} \]

The total integrated voltage is identical.

The circular coil produces a greater average EMF because its flux linkage changes over a shorter time.

The student’s mistake is confusing the magnitude of the flux linkage with the rate at which it changes.

A larger induced EMF does not necessarily mean a larger initial magnetic flux.

Question 5

A circular coil and a square coil have identical enclosed areas of \(0.012\,\mathrm{m^2}\) and 25 turns each.

Initially, both coils have their area normals pointing in the positive \(x\)-direction.

A uniform magnetic field initially points in the positive \(x\)-direction with a magnitude of \(0.40\,\mathrm T\).

Over an interval of \(0.20\,\mathrm s\), the magnetic field reverses continuously along the \(x\)-axis, ending at \(-0.40\,\mathrm T\).

During the same interval:

• The circular coil rotates through \(180^\circ\), so its area normal ends pointing in the negative \(x\)-direction.
• The square coil remains stationary, so its area normal continues pointing in the positive \(x\)-direction.

Assume both coils remain in the uniform field throughout the process.

(a) Calculate the average signed induced EMF in each coil, taking the initial positive normal as the reference direction.

(b) A student argues that the circular coil experiences no induction because its initial and final magnetic fluxes are equal. Is this conclusion correct?

(c) Can the maximum instantaneous induced EMF in the circular coil be calculated from the information provided? Explain.

Solution 5

The circular coil has zero average signed EMF over the complete interval, while the square coil has an average signed EMF of \(+1.2\,\mathrm V\).

However, the circular coil does experience a changing magnetic flux during the interval. The student’s conclusion that no induction occurs is incorrect.

This question tests an important distinction: equal initial and final fluxes do not guarantee that flux remained constant between those two moments.

Step 1: Calculate the initial flux through one turn of each coil.

Initially, both area normals are parallel to the magnetic field.

Therefore:

\[ \Phi_i=BA \]

\[ \Phi_i=0.40(0.012) \]

\[ \Phi_i=0.0048\,\mathrm{Wb} \]

The initial flux linkage for both coils is:

\[ \lambda_i=25(0.0048) \]

\[ \lambda_i=0.120\,\mathrm{Wb} \]

Step 2: Determine the final flux through the circular coil.

At the end of the interval, the magnetic field points in the negative \(x\)-direction.

The circular coil has also rotated through \(180^\circ\), so its area normal points in the negative \(x\)-direction.

The field and area normal are still parallel.

Therefore:

\[ \Phi_{c,f}=(-0.40)(-0.012) \]

\[ \Phi_{c,f}=+0.0048\,\mathrm{Wb} \]

The final flux linkage is:

\[ \lambda_{c,f}=25(0.0048) \]

\[ \lambda_{c,f}=+0.120\,\mathrm{Wb} \]

Step 3: Calculate the average induced EMF in the circular coil.

The change in flux linkage is:

\[ \Delta\lambda_c =\lambda_{c,f}-\lambda_{c,i} \]

\[ \Delta\lambda_c=0.120-0.120 \]

\[ \Delta\lambda_c=0 \]

Applying Faraday’s law:

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

\[ \mathcal E_{\mathrm{avg},c} =-\frac{0}{0.20} \]

\[ \boxed{\mathcal E_{\mathrm{avg},c}=0\,\mathrm V} \]

The circular coil has zero average signed EMF over the complete interval.

Step 4: Determine the final flux through the square coil.

The square coil does not rotate.

Its area normal therefore continues pointing in the positive \(x\)-direction.

The final magnetic field points in the negative \(x\)-direction, so the final flux is negative.

\[ \Phi_{s,f}=(-0.40)(0.012) \]

\[ \Phi_{s,f}=-0.0048\,\mathrm{Wb} \]

The final flux linkage is:

\[ \lambda_{s,f}=25(-0.0048) \]

\[ \lambda_{s,f}=-0.120\,\mathrm{Wb} \]

Step 5: Calculate the average induced EMF in the square coil.

The change in flux linkage is:

\[ \Delta\lambda_s=-0.120-0.120 \]

\[ \Delta\lambda_s=-0.240\,\mathrm{Wb} \]

Therefore:

\[ \mathcal E_{\mathrm{avg},s} =-\frac{-0.240}{0.20} \]

\[ \boxed{\mathcal E_{\mathrm{avg},s}=+1.2\,\mathrm V} \]

The positive sign follows the stated reference orientation. Reversing the chosen terminal convention would reverse the sign.

Step 6: Examine the student’s argument about the circular coil.

The student has correctly identified that the circular coil’s initial and final fluxes are equal.

However, this only tells us that the net change in flux over the complete interval is zero.

The magnetic field reverses continuously along the \(x\)-axis, so it must pass through zero at some point.

When the magnetic field is zero, the magnetic flux through the circular coil must also be zero, regardless of the coil’s orientation.

Therefore, the circular coil’s flux:

1. Starts at \(+0.0048\,\mathrm{Wb}\).
2. Falls to zero as the magnetic field reverses.
3. Returns to \(+0.0048\,\mathrm{Wb}\) by the end of the interval.

Its flux must have changed during the process.

Consequently, the instantaneous induced EMF cannot have been zero throughout the entire interval. Contributions of opposite signs cancel when calculating the average signed EMF.

Step 7: Determine whether the maximum instantaneous EMF can be calculated.

No. The information only gives the initial and final magnetic fields, the initial and final coil orientations, and the total time.

It does not specify exactly how quickly the magnetic field changes or how quickly the circular coil rotates at each instant.

To calculate the instantaneous EMF, we would need the rate of change of flux linkage at each moment:

\[ \mathcal E(t)=-\frac{d\lambda}{dt} \]

Without the time-dependent behaviour of the field and coil orientation, the maximum instantaneous EMF cannot be determined.

This is the crucial distinction: Faraday’s law allows us to calculate the average EMF from the change in flux linkage between two known states, but calculating the instantaneous EMF requires knowing how that linkage changes throughout the interval.

08What to understand next

Comparing circular and square coils is ultimately an exercise in identifying the effective area perpendicular to a magnetic field.

Once you can calculate the flux through each turn and account for the number of turns, you are ready to analyse what happens when that flux changes.

The next step is to connect changing flux linkage to induced EMF using Faraday’s law, then use Lenz’s law to determine the direction of the induced current.

That is the reasoning behind applications such as wireless charging, generators, and electromagnetic induction.