How to Convert Coil Diameter into Area for Flux Calculations

Suppose a circular coil has a diameter of 8.0 cm. A student substitutes 0.080 m directly into \(A=\pi r^2\) and calculates an area of \(2.01\times10^{-2}\,\mathrm{m^2}\). The answer is four times too large.

The mistake? They used the diameter as the radius.

For a circular coil, you must halve the diameter before calculating its area:

\[
\boxed{A=\pi r^2=\frac{\pi d^2}{4}}
\]

For an 8.0 cm diameter coil:

\[
r=\frac{0.080}{2}=0.040\,\mathrm{m}
\]

\[
A=\pi(0.040)^2
=5.03\times10^{-3}\,\mathrm{m^2}
\]

That’s the area you use in a magnetic flux calculation. However, there’s another detail that catches students out: the angle in the flux equation is measured relative to the coil’s area vector, not its surface.

Let’s work through both ideas so you can handle questions involving coil diameter, orientation, magnetic flux, and flux linkage without mixing up the geometry.

01Converting coil diameter into area

A circular coil has two measurements worth keeping separate:

• Diameter (\(d\)): the distance across the circle through its centre.
• Radius (\(r\)): the distance from the centre to the circumference.

The radius is always half the diameter:

\[
r=\frac{d}{2}
\]

The area enclosed by a circle is:

\[
A=\pi r^2
\]

Substituting \(r=d/2\):

\[
A=\pi\left(\frac{d}{2}\right)^2
\]

Therefore:

\[
\boxed{A=\frac{\pi d^2}{4}}
\]

Getting the units right

Magnetic flux calculations use area in square metres (\(\mathrm{m^2}\)).

If a question gives you a diameter in centimetres or millimetres, convert it into metres before squaring.

Be particularly careful with the conversion because you’re squaring the radius.

For example, \(5.0\,\mathrm{cm}=0.050\,\mathrm{m}\), but:

\[
(0.050)^2=0.0025\,\mathrm{m^2}
\]

You cannot simply square 5.0 and then label the answer in square metres.

02Understanding the area vector and its angle

You’ve calculated the area, but you cannot always put \(BA\) straight into a magnetic flux calculation.

The orientation of the coil matters.

Imagine holding a circular hoop in front of a stream of water. If you hold it facing directly into the stream, lots of water passes through its opening. If you turn it sideways, much less passes through.

A magnetic field behaves similarly when we consider the flux through a coil.

The useful quantity is the area projected perpendicular to the magnetic field.

What is an area vector?

An area vector is a vector perpendicular to the plane of a surface. Its magnitude is equal to the surface area.

For a circular coil:

\[
|\vec A|=A=\pi r^2
\]

Its direction is along the normal to the coil’s plane.

A normal is simply a line perpendicular to a surface.

The area vector isn’t a physical arrow attached to the coil. It’s a mathematical way of describing both the area and its orientation.

This is important because magnetic flux depends on the angle between the magnetic field and this vector.

The magnetic flux equation

Magnetic flux through one turn of a coil is:

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

Where:

• \(\Phi\) is magnetic flux, measured in webers (Wb).
• \(B\) is magnetic field strength, measured in teslas (T).
• \(A\) is the area enclosed by one turn, measured in \(\mathrm{m^2}\).
• \(\theta\) is the angle between the magnetic field and the area vector.

The angle is measured from the normal, not from the face of the coil.

This is one of the most important details to remember.

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

Exam questions don’t always give you the angle relative to the normal.

Sometimes they describe the angle between the magnetic field and the plane of the coil.

Call this angle \(\alpha\).

Because the area vector is perpendicular to the plane:

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

Substituting into the flux equation:

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

Using the trigonometric identity:

\[
\boxed{\Phi=BA\sin\alpha}
\]

This version gives the magnitude of the flux when \(\alpha\) is the acute angle between the field and the plane.

Here’s a useful comparison:

The two extreme cases are particularly useful:

• When the magnetic field is perpendicular to the coil’s plane, flux is maximum.
• When the magnetic field is parallel to the coil’s plane, flux is zero.

The area of the coil hasn’t changed in either situation. Only its orientation has changed.

A common angle trap

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

That wording alone isn’t sufficiently precise to choose the trigonometric function. You need to establish whether the angle is measured from the plane or the normal.

Consider a coil with:

• Diameter \(=6.0\,\mathrm{cm}\)
• Magnetic field strength \(B=0.40\,\mathrm{T}\)

Its area is:

\[
A=\pi(0.030)^2
=2.83\times10^{-3}\,\mathrm{m^2}
\]

Now compare two different situations.

Situation A: The field is \(30^\circ\) to the plane.

The angle to the normal is \(60^\circ\).

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

\[
\Phi=5.65\times10^{-4}\,\mathrm{Wb}
\]

Situation B: The field is \(30^\circ\) to the normal.

Now the angle in the original flux equation really is \(30^\circ\).

\[
\Phi=BA\cos30^\circ
\]

\[
\Phi=9.80\times10^{-4}\,\mathrm{Wb}
\]

The second flux is substantially larger, despite both questions containing the number \(30^\circ\).

Before choosing sine or cosine, identify exactly where the angle is measured.

03From coil area to flux linkage

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

Most coils contain multiple turns of wire. If a coil has 40 turns, and each turn experiences the same magnetic flux, the total flux linkage is 40 times the flux through one turn.

Flux linkage is represented by \(\lambda\):

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

Where:

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

Combining this with the magnetic flux equation:

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

Or, when the angle is measured from the coil’s plane:

\[
\boxed{\lambda=NBA\sin\alpha}
\]

This combined equation is particularly useful for HSC calculations.

The key distinction is that \(A\) is still the area enclosed by one turn. You multiply the flux by \(N\) afterwards.

You can also substitute the diameter formula directly:

\[
\boxed{\lambda=NB\frac{\pi d^2}{4}\cos\theta}
\]

This gives you a direct route from coil diameter to flux linkage.

04How an experiment checks the angle convention

It’s useful to connect these calculations to an actual experimental arrangement.

Suppose you want to investigate how the orientation of a circular coil affects its magnetic flux linkage.

A suitable setup uses an electromagnet with two broad pole pieces facing each other. A circular coil sits centrally in the gap, mounted on a rotatable axle.

The coil’s terminals connect to a voltage sensor. A Hall probe measures the magnetic field strength, and a protractor scale at the axle allows the angle of the coil to be adjusted.

What should the experiment show?

Keep the following quantities constant:

• The number of turns, \(N\).
• The coil’s diameter and area, \(A\).
• The magnetic field strength, \(B\).

Now rotate the coil and measure its orientation relative to the field.

The predicted flux linkage is:

\[
\lambda=NBA\sin\alpha
\]

Here, \(\alpha\) is the angle between the field and the plane of the coil.

Therefore, the predicted relationship is:

These are theoretical relative values, not claimed experimental measurements. The maximum flux linkage is used as the reference value of 1.

The prediction is that flux linkage increases as the coil turns towards a position where its plane is perpendicular to the field.

At that position, the area vector is parallel to the magnetic field, giving maximum flux linkage.

What does the voltage sensor actually measure?

There’s an important experimental distinction here.

A voltage sensor connected to a stationary coil does not directly measure its magnetic flux. It measures induced potential difference.

When you rotate the coil, its flux linkage changes, producing an induced voltage.

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

\[
\mathcal{E}=-\frac{\Delta\lambda}{\Delta t}
\]

For a changing emf, the area under an emf-versus-time graph gives the negative change in flux linkage:

\[
\Delta\lambda=-\int\mathcal{E}\,dt
\]

This is why a voltage-time graph can provide information about the change in flux linkage, rather than simply displaying the static flux through the coil.

For more on interpreting the area under a physics graph, see HSC Physics Graphs: How to Use Gradient and Area.

Why use an electromagnet rather than a bar magnet?

A bar magnet is useful for demonstrating electromagnetic induction, but it introduces a problem when you’re trying to isolate the effect of coil orientation.

Its magnetic field varies considerably with position. A single measurement of magnetic field strength at the centre of the coil may not represent the field across the entire enclosed area.

Broad electromagnet pole pieces provide a more uniform field in the central region.

Keeping the coil centred and smaller than the pole faces helps reduce the effects of non-uniformity and fringing fields.

This doesn’t make the field perfectly uniform. It makes the approximation \( \Phi=BA\cos\theta \) more reasonable.

The experiment therefore tests the geometry of the flux calculation under controlled conditions, rather than assuming every magnetic field is uniform.

05Measurement details that can change your answer

Before moving to exam questions, there are a few practical details worth understanding.

What if the coil wire has noticeable thickness?

In most HSC problems, the coil is treated as a thin wire, and the stated diameter is used directly.

However, suppose a question gives you both an inner and an outer diameter.

The effective area is estimated using the diameter measured along the centreline of the wire.

For a uniformly thick circular winding:

\[
d_{\mathrm{mean}}=\frac{d_{\mathrm{inner}}+d_{\mathrm{outer}}}{2}
\]

You then use this mean diameter to calculate the enclosed area:

\[
A=\frac{\pi d_{\mathrm{mean}}^2}{4}
\]

Why not use the outer diameter?

Because the magnetic flux surface is bounded by the centreline of the conducting loop in the usual thin-wire model. Using the outer diameter would overestimate the enclosed area.

Does rotating a coil change its area?

No.

A rigid circular coil retains exactly the same physical area when rotated.

What changes is its projected area perpendicular to the magnetic field:

\[
A_{\perp}=A\cos\theta
\]

This is the area that determines magnetic flux in a uniform field.

If the coil is stretched or reshaped, however, its actual enclosed area can change. In that situation, you must recalculate \(A\) as well as considering any change in orientation.

Does flux always have to be positive?

No.

Magnetic flux is a signed quantity when an area-vector direction has been chosen.

A positive area normal can be defined using the right-hand rule: curl your fingers in the chosen positive direction around the coil, and your thumb points along the positive area vector.

If the magnetic field has a component in the opposite direction to this normal, the flux is negative.

For questions asking only for the magnitude of flux, use the absolute value. For questions involving changes in flux linkage or induced emf, keep the sign consistent.

Assumptions behind the standard calculation

The equations in this article assume:

• The coil is planar and circular.
• The magnetic field is uniform across the enclosed area.
• Each turn has approximately the same area and orientation.
• The field direction and coil orientation are clearly defined.

If the magnetic field varies significantly across the coil, the simple \(BA\cos\theta\) calculation is no longer sufficient.

The same area-vector idea also appears in torque calculations for current-carrying coils. You can explore that connection in Torque on a Current-Carrying Coil for HSC Physics.

06Questions and solutions

Try each question before reading its solution. The questions progress from a standard flux calculation to situations where diameter, orientation, and changes in the field must be considered together.

Question 1

A circular coil has 25 turns and a diameter of 8.0 cm. It is placed in a uniform magnetic field of strength \(0.20\,\mathrm{T}\), with the magnetic field perpendicular to the plane of the coil.

Calculate:

(a) The area enclosed by one turn.

(b) The magnetic flux through one turn.

(c) The flux linkage of the coil.

Solution 1

The area is \(5.03\times10^{-3}\,\mathrm{m^2}\), the flux per turn is \(1.01\times10^{-3}\,\mathrm{Wb}\), and the flux linkage is \(2.51\times10^{-2}\,\mathrm{Wb\text{-}turn}\).

Step 1: Convert the diameter and calculate the radius.

\[
d=0.080\,\mathrm{m}
\]

\[
r=\frac{0.080}{2}=0.040\,\mathrm{m}
\]

Step 2: Calculate the area.

\[
\begin{aligned}
A&=\pi r^2\\
&=\pi(0.040)^2\\
&=5.03\times10^{-3}\,\mathrm{m^2}
\end{aligned}
\]

Step 3: Identify the angle.

The field is perpendicular to the plane of the coil, so it is parallel to the area vector.

Therefore:

\[
\theta=0^\circ
\]

Step 4: Calculate the flux through one turn.

\[
\begin{aligned}
\Phi&=BA\cos\theta\\
&=0.20(5.03\times10^{-3})\cos0^\circ\\
&=1.01\times10^{-3}\,\mathrm{Wb}
\end{aligned}
\]

Step 5: Calculate flux linkage.

\[
\begin{aligned}
\lambda&=N\Phi\\
&=25(1.01\times10^{-3})\\
&=2.51\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

The field is aligned with the area vector, so the coil experiences maximum flux for this field strength and area.

Question 2

Two circular coils, A and B, are placed in the same uniform magnetic field of strength \(0.30\,\mathrm{T}\).

Their properties are:

Calculate the flux linkage of each coil and determine which has the greater flux linkage. Express your answer as a ratio.

Solution 2

Coil B has the greater flux linkage, approximately 3.09 times that of coil A.

The tempting prediction is that coil A will have greater flux linkage because it has twice as many turns. However, coil B has a larger area and a more favourable orientation. We need to consider all three factors.

Step 1: Calculate the area of coil A.

\[
r_A=\frac{0.040}{2}=0.020\,\mathrm{m}
\]

\[
A_A=\pi(0.020)^2
=1.257\times10^{-3}\,\mathrm{m^2}
\]

Step 2: Calculate the flux linkage of coil A.

The angle is measured from the plane, so:

\[
\lambda_A=NB A\sin\alpha
\]

Substitute:

\[
\begin{aligned}
\lambda_A
&=80(0.30)(1.257\times10^{-3})\sin20^\circ\\
&=1.03\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

Step 3: Calculate the area of coil B.

\[
r_B=\frac{0.060}{2}=0.030\,\mathrm{m}
\]

\[
A_B=\pi(0.030)^2
=2.827\times10^{-3}\,\mathrm{m^2}
\]

Step 4: Calculate the flux linkage of coil B.

\[
\begin{aligned}
\lambda_B
&=40(0.30)(2.827\times10^{-3})\sin70^\circ\\
&=3.19\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

Step 5: Calculate the ratio.

\[
\begin{aligned}
\frac{\lambda_B}{\lambda_A}
&=\frac{3.19\times10^{-2}}{1.03\times10^{-2}}\\
&=3.09
\end{aligned}
\]

Therefore:

\[
\boxed{\lambda_B:\lambda_A=3.09:1}
\]

Coil B has approximately 3.09 times the flux linkage of coil A.

The important idea is that turn count alone does not determine flux linkage. Diameter affects area quadratically, while orientation affects the perpendicular component of the area.

Question 3

A circular coil is made from relatively thick wire. Its inner diameter is 9.6 cm, and its outer diameter is 10.0 cm.

The coil contains 60 turns and is placed in a uniform magnetic field of strength \(0.18\,\mathrm{T}\). The field makes an angle of \(40^\circ\) with the plane of the coil.

Assuming the wire thickness is uniform and the field is uniform across the coil, calculate:

(a) The effective diameter of the coil.

(b) The effective area enclosed by one turn.

(c) The flux linkage.

Explain why using the outer diameter alone would give an incorrect result.

Solution 3

The effective diameter is \(9.8\,\mathrm{cm}\), the area is \(7.54\times10^{-3}\,\mathrm{m^2}\), and the flux linkage is approximately \(5.24\times10^{-2}\,\mathrm{Wb\text{-}turn}\).

Step 1: Calculate the effective diameter.

For a uniformly thick wire, use the mean of the inner and outer diameters.

\[
\begin{aligned}
d_{\mathrm{mean}}
&=\frac{9.6+10.0}{2}\\
&=9.8\,\mathrm{cm}
\end{aligned}
\]

Convert to metres:

\[
d=0.098\,\mathrm{m}
\]

Step 2: Calculate the radius.

\[
r=\frac{0.098}{2}=0.049\,\mathrm{m}
\]

Step 3: Calculate the enclosed area.

\[
\begin{aligned}
A&=\pi r^2\\
&=\pi(0.049)^2\\
&=7.54\times10^{-3}\,\mathrm{m^2}
\end{aligned}
\]

Step 4: Choose the correct angle.

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

Therefore:

\[
\Phi=BA\sin40^\circ
\]

Step 5: Calculate flux through one turn.

\[
\begin{aligned}
\Phi
&=0.18(7.54\times10^{-3})\sin40^\circ\\
&=8.73\times10^{-4}\,\mathrm{Wb}
\end{aligned}
\]

Step 6: Calculate flux linkage.

\[
\begin{aligned}
\lambda&=N\Phi\\
&=60(8.73\times10^{-4})\\
&=5.24\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

The outer diameter would overestimate the enclosed area because it measures across the outside of the wire rather than along the centreline of the conducting loop.

This is a small correction for thin wire, but it becomes more significant when the wire thickness is a noticeable fraction of the coil diameter.

Question 4

A 50-turn circular coil is initially held in a uniform magnetic field of \(0.30\,\mathrm{T}\).

Its initial diameter is 8.0 cm, and the angle between the magnetic field and the plane of the coil is \(30^\circ\).

The coil is then reshaped and reoriented. In its final state:

• Its diameter has increased to 8.8 cm.
• The magnetic field strength has decreased to \(0.15\,\mathrm{T}\).
• The angle between the field and the plane of the coil has increased to \(60^\circ\).

The number of turns remains constant.

Calculate:

(a) The initial flux linkage.

(b) The final flux linkage.

(c) The percentage change in flux linkage.

Explain why halving the magnetic field strength does not halve the flux linkage.

Solution 4

The initial flux linkage is \(3.77\times10^{-2}\,\mathrm{Wb\text{-}turn}\), and the final flux linkage is \(3.95\times10^{-2}\,\mathrm{Wb\text{-}turn}\). The flux linkage increases by approximately 4.8%.

The field strength halves, but the coil’s area and orientation both change. These effects must be considered together.

Step 1: Calculate the initial area.

\[
r_i=\frac{0.080}{2}=0.040\,\mathrm{m}
\]

\[
A_i=\pi(0.040)^2
=5.027\times10^{-3}\,\mathrm{m^2}
\]

Step 2: Calculate the initial flux linkage.

The angle is measured from the plane, so use:

\[
\lambda_i=NB A_i\sin30^\circ
\]

Substitute:

\[
\begin{aligned}
\lambda_i
&=50(0.30)(5.027\times10^{-3})(0.5)\\
&=3.77\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

Step 3: Calculate the final area.

\[
r_f=\frac{0.088}{2}=0.044\,\mathrm{m}
\]

\[
A_f=\pi(0.044)^2
=6.082\times10^{-3}\,\mathrm{m^2}
\]

Step 4: Calculate the final flux linkage.

\[
\lambda_f=NB A_f\sin60^\circ
\]

Substitute:

\[
\begin{aligned}
\lambda_f
&=50(0.15)(6.082\times10^{-3})(0.866)\\
&=3.95\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

Step 5: Calculate the percentage change.

\[
\%\text{ change}
=\frac{\lambda_f-\lambda_i}{\lambda_i}\times100
\]

\[
\begin{aligned}
&=\frac{0.0395-0.0377}{0.0377}\times100\\
&\approx4.8\%
\end{aligned}
\]

Therefore, flux linkage increases by approximately 4.8%.

Why does this happen?

There are three competing effects:

• The magnetic field halves, reducing flux linkage.
• The diameter increases by 10%, increasing the area by 21%.
• The plane angle changes from \(30^\circ\) to \(60^\circ\), increasing the orientation factor from \(0.5\) to approximately \(0.866\).

The combined effect is a small increase in flux linkage.

A useful shortcut for checking this result is to compare the initial and final equations directly:

\[
\frac{\lambda_f}{\lambda_i}
=
\left(\frac{d_f}{d_i}\right)^2
\left(\frac{B_f}{B_i}\right)
\left(\frac{\sin60^\circ}{\sin30^\circ}\right)
\]

\[
=1.10^2\times0.50\times1.732
\]

\[
=1.048
\]

This confirms that the final linkage is approximately 1.048 times the initial linkage.

Question 5

A 20-turn circular coil has a diameter of 10.0 cm and is placed in a uniform magnetic field.

Initially:

• The magnetic field strength is \(0.30\,\mathrm{T}\).
• The plane of the coil makes an angle of \(30^\circ\) with the magnetic field.
• The positive area vector has a component in the same direction as the magnetic field.

The coil is then rotated through \(180^\circ\) about one of its diameters. The rotation axis is parallel to the component of the magnetic field lying in the original plane of the coil.

During the rotation, the magnetic field decreases to \(0.20\,\mathrm{T}\), but its direction remains unchanged.

At the end of the rotation, the coil occupies the same geometric plane as it did initially, but it has been turned over.

Assume the positive area-vector direction is defined by a marked positive direction around the winding and rotates with the coil.

Answer the following:

(a) Calculate the initial signed flux linkage.

(b) Calculate the final signed flux linkage.

(c) Calculate the change in flux linkage.

(d) Determine the flux linkage when the coil has completed \(90^\circ\) of its rotation.

(e) A student argues that the final flux linkage must be positive because the coil has returned to its original geometric plane. Explain precisely why this reasoning is incorrect.

Solution 5

The initial flux linkage is \(+2.36\times10^{-2}\,\mathrm{Wb\text{-}turn}\), the final flux linkage is \(-1.57\times10^{-2}\,\mathrm{Wb\text{-}turn}\), and the change is \(-3.93\times10^{-2}\,\mathrm{Wb\text{-}turn}\). At the halfway orientation, the flux linkage is zero.

The crucial distinction is between the geometric plane of a coil and its oriented area vector.

Step 1: Calculate the area of one turn.

Convert the diameter:

\[
d=0.100\,\mathrm{m}
\]

Calculate the radius:

\[
r=0.050\,\mathrm{m}
\]

Therefore:

\[
\begin{aligned}
A&=\pi r^2\\
&=\pi(0.050)^2\\
&=7.854\times10^{-3}\,\mathrm{m^2}
\end{aligned}
\]

The coil’s area remains constant throughout the rotation.

Step 2: Determine the initial angle to the area vector.

The initial angle between the magnetic field and the plane is \(30^\circ\).

The area vector is perpendicular to the plane, so:

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

\[
\theta_i=60^\circ
\]

The question states that the initial area vector has a component in the same direction as the magnetic field. Therefore, the initial flux linkage is positive.

Step 3: Calculate the initial flux linkage.

\[
\lambda_i=NB_iA\cos60^\circ
\]

Substitute:

\[
\begin{aligned}
\lambda_i
&=20(0.30)(7.854\times10^{-3})(0.5)\\
&=+2.36\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

Step 4: Determine the final angle to the area vector.

The coil has rotated through \(180^\circ\) about a diameter.

Its geometric plane has returned to its original orientation, but its positive area vector now points in the opposite direction.

Therefore, its final angle to the unchanged magnetic field is:

\[
\theta_f=120^\circ
\]

Notice that this is not \(60^\circ\).

The acute angle between the field and the plane is still \(30^\circ\), but the signed angle between the field and the oriented area vector has changed.

Step 5: Calculate the final flux linkage.

The magnetic field has decreased to \(0.20\,\mathrm{T}\).

\[
\lambda_f=NB_fA\cos120^\circ
\]

Substitute:

\[
\begin{aligned}
\lambda_f
&=20(0.20)(7.854\times10^{-3})(-0.5)\\
&=-1.57\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

The negative sign indicates that the magnetic field is now directed partly opposite to the positive area vector.

Step 6: Calculate the change in flux linkage.

Change means final value minus initial value:

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

\[
\begin{aligned}
\Delta\lambda
&=-1.57\times10^{-2}-2.36\times10^{-2}\\
&=-3.93\times10^{-2}\,\mathrm{Wb\text{-}turn}
\end{aligned}
\]

The change is negative because the signed flux linkage has moved from a positive value to a negative value.

Step 7: Determine the flux linkage halfway through the rotation.

After \(90^\circ\) of rotation, the positive area vector is perpendicular to the magnetic field.

Therefore:

\[
\theta=90^\circ
\]

Using the flux linkage equation:

\[
\begin{aligned}
\lambda&=NBA\cos90^\circ\\
&=0
\end{aligned}
\]

So:

\[
\boxed{\lambda_{\mathrm{halfway}}=0}
\]

The field does not need to become zero for the flux linkage to become zero. The orientation alone makes the component of the area vector along the field equal to zero.

Why the student’s reasoning fails

The student has treated the coil’s plane as though it fully describes its orientation.

It doesn’t.

A geometric plane has no unique normal direction until an orientation is assigned. Once we define a positive area vector using the winding direction, flipping the coil reverses that vector.

The final coil therefore has:

• The same physical area.
• The same geometric plane.
• The same acute angle between the field and the plane.
• The opposite positive area-vector direction.
• A smaller magnetic field strength.

The tempting calculation would be to use \(+60^\circ\) again and conclude that the final flux linkage is simply two-thirds of the initial value.

That gives the correct magnitude but the wrong sign.

The area-vector approach resolves the problem because it tracks the orientation of the surface, not just the appearance of the coil.

07Where to go next

Once you can convert diameter into area, identify the correct angle, and calculate flux linkage, you’re ready to use these quantities in Faraday’s law.

The next step is understanding how a change in flux linkage produces an induced emf, including how to interpret the sign and magnitude of that emf.

Keep the central relationship in mind:

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

For a circular coil, the diameter determines the area, the area vector determines the angle, and the number of turns determines the total flux linkage. Get those three parts right, and the rest of the calculation becomes much more straightforward.