How to Sketch Magnetic Flux During One Coil Rotation

A rectangular coil completes one full rotation in a uniform magnetic field. At the halfway point, its plane is back in the same geometric orientation it had at the start. Does that mean the magnetic flux has returned to its original value?

No. The coil’s area vector has reversed direction, so the signed magnetic flux has changed from maximum positive to maximum negative.

This is the key to sketching magnetic flux during one coil rotation. You need to track the angle between the magnetic field and the coil’s area vector, not simply watch the coil’s plane turn.

If the area vector starts parallel to the magnetic field, the flux follows this sequence:

Maximum positive → zero → maximum negative → zero → maximum positive.

The resulting graph is one complete cosine cycle.

01Start by tracking the coil’s area vector

Imagine a rectangular coil rotating between the north and south poles of a magnet. The magnetic field points from north to south.

As the coil rotates, its orientation changes relative to the field. To describe this accurately, we use an area vector.

The area vector is an imaginary arrow perpendicular to the plane of the coil. Its length represents the coil’s area, and its direction identifies which side of the coil is treated as positive.

The important angle, \(\theta\), is measured between:

• The magnetic field direction, \(\vec B\).
• The coil’s area vector, \(\vec A\).

For a coil rotating about an axis perpendicular to the magnetic field, this angle changes as the coil turns.

Consider what happens at four important orientations.

At 0°: Maximum positive flux

The area vector points in the same direction as the magnetic field.

The coil’s plane is perpendicular to the field, so the maximum possible amount of magnetic field passes through its area.

The flux is maximum and positive.

At 90°: Zero flux

The area vector is perpendicular to the magnetic field.

The coil’s plane is now parallel to the field. The magnetic field runs along the plane rather than passing through it.

The flux is zero.

At 180°: Maximum negative flux

The area vector now points directly opposite to the magnetic field.

The coil’s plane is perpendicular to the field again, but the direction of the area vector has reversed.

The flux is maximum in magnitude but negative.

At 270°: Zero flux again

The area vector is perpendicular to the field once more.

The flux returns to zero.

At 360°, the coil returns to its original orientation, and the flux is maximum and positive again.

Notice the distinction between 0° and 180°. The coil occupies the same geometric plane at both positions, but its directed area vector points in opposite directions.

That is why signed flux changes sign halfway through the rotation.

02Use the five key angles to sketch the graph

Once you understand the physical orientations, the mathematics becomes straightforward.

Magnetic flux through one turn is:

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

Where:

• \(\Phi\) = magnetic flux through one turn, measured in webers (Wb).
• \(B\) = magnetic field strength, measured in tesla (T).
• \(A\) = area of the coil, measured in square metres (m²).
• \(\theta\) = angle between the magnetic field and the area vector.

For a coil containing \(N\) turns, we also calculate flux linkage:

\[
\lambda=N\Phi
\]

Therefore:

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

Here, \(\lambda\) is measured in Wb turns.

The distinction is important:

• Magnetic flux describes the flux through one turn.
• Flux linkage accounts for all \(N\) turns.

Both produce the same cosine-shaped graph. Flux linkage simply has an amplitude \(N\) times larger.

For a complete rotation starting with the area vector parallel to the magnetic field, use these five positions:

These are the five anchor points for your graph.

The graph should have:

• A positive maximum at the beginning.
• A smooth decrease to zero.
• A negative minimum halfway through.
• A smooth increase back to zero.
• A return to the original positive maximum.

The curve must be smooth. Magnetic flux does not suddenly jump between positive and negative values as the coil rotates continuously.

Signed flux versus flux magnitude

There is a common mistake worth addressing here.

Some students sketch two positive peaks because they think magnetic flux cannot be negative.

That would be correct only if the question specifically asks for the magnitude of magnetic flux.

Signed flux can be negative because the magnetic field passes through the coil in the direction opposite to the chosen positive area vector.

For induction questions, retaining the sign is particularly important because the direction of the change in flux determines the induced emf.

03Convert the rotation angle into time

Sometimes an HSC question asks you to sketch magnetic flux against rotation angle.

Other questions ask for magnetic flux against time.

The physical relationship is the same, but you need one extra piece of information to construct the time graph: the coil’s angular speed.

If the coil rotates at constant angular velocity:

\[
\theta=\theta_0+\omega t
\]

Where:

• \(\theta_0\) = initial angle between the area vector and magnetic field.
• \(\omega\) = angular velocity, measured in rad s\(^{-1}\).
• \(t\) = time, measured in seconds.

For a coil completing one rotation in period \(T\):

\[
\omega=\frac{2\pi}{T}
\]

Substituting the changing angle into the flux-linkage equation gives:

\[
\lambda=NBA\cos(\theta_0+\omega t)
\]

If the coil starts with its area vector parallel to the field, then \(\theta_0=0\).

The equation becomes:

\[
\boxed{\lambda=NBA\cos(\omega t)}
\]

This produces a cosine graph against time.

The five important time positions

For constant angular velocity, each quarter-turn takes one-quarter of the rotation period.

There is one important assumption here.

A cosine graph against time requires constant angular velocity.

If the coil rotates faster during the first half of its journey and slower during the second half, the flux-versus-angle graph remains a cosine curve, but the flux-versus-time graph will be stretched and compressed.

The physical relationship has not changed. Only the timing of the rotation has changed.

04How an experiment would check the flux graph

There is a useful experimental question behind this mathematical model:

How could you test whether the flux through a rotating coil actually follows a cosine relationship?

A rotating-coil investigation can answer this by measuring the induced emf and tracking the coil’s angular position.

Setting up the apparatus

Imagine a rectangular coil mounted on a rotating axle between the pole pieces of an electromagnet.

The apparatus contains:

• An electromagnet producing a uniform magnetic field.
• A rectangular coil with a known number of turns and area.
• A rotating axle connected to an angular-position sensor.
• Slip rings and brushes connecting the rotating coil to a voltage sensor.
• An oscilloscope or data logger recording the induced emf.
• A Hall probe used to measure the magnetic field strength.

The coil starts with its area vector parallel to the magnetic field.

As the axle turns, the angular sensor records the coil’s orientation. At the same time, the voltage sensor records the induced emf.

What should the experiment show?

We can predict the flux-linkage pattern from the coil’s geometry.

However, the oscilloscope does not directly measure magnetic flux. It measures induced emf.

Faraday’s law connects the two:

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

Here:

• \(\mathcal E_{\text{avg}}\) = average induced emf.
• \(\Delta(N\Phi)\) = change in flux linkage.
• \(\Delta t\) = time interval.

This gives us an indirect way to reconstruct the flux-linkage graph.

For example:

• At maximum positive flux linkage, the flux is momentarily not changing, so the induced emf is zero.
• As the flux linkage decreases towards zero, the induced emf develops.
• When flux linkage passes through zero, its rate of change is greatest for constant angular speed.
• At maximum negative flux linkage, the induced emf returns to zero.

The induced emf and flux-linkage graphs are therefore shifted by one-quarter of a cycle.

This is a particularly important experimental observation to understand: the voltage graph is not the flux graph.

Why measuring only the average voltage fails

Suppose you connect the rotating coil to a meter that reports only the signed average induced voltage over one complete rotation.

The result is zero for an ideal coil completing a full rotation at constant speed.

Why?

The coil finishes with the same flux linkage it started with. Therefore, its net change in flux linkage over the complete rotation is zero.

That does not mean there was no induced emf during the rotation. Positive and negative contributions cancel when averaged over the full cycle.

A better design records the voltage continuously and measures the angular position at the same time.

Using Faraday’s law over short time intervals, the changes in flux linkage can be reconstructed. The known starting value, \(NBA\), provides the initial reference.

A small measurement calculation

Suppose an experimental coil has:

• 100 turns.
• Area of \(0.010\text{ m}^2\).
• Magnetic field strength of \(0.20\text{ T}\).
• Rotation period of \(0.50\text{ s}\).

Its maximum flux linkage is:

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

\[
=100(0.20)(0.010)
\]

\[
=0.20\text{ Wb turns}
\]

The first quarter-turn takes:

\[
\Delta t=\frac{0.50}{4}=0.125\text{ s}
\]

During this interval, flux linkage decreases from \(+0.20\) to zero.

The average induced emf magnitude is:

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

\[
=\frac{0.20}{0.125}
\]

\[
\boxed{|\mathcal E_{\text{avg}}|=1.6\text{ V}}
\]

This is an average value over the quarter-turn, not the instantaneous maximum emf.

The important experimental result is that the measured voltage changes as the coil rotates, allowing us to reconstruct the changing flux linkage.

This investigation assumes a uniform, constant magnetic field, a coil that remains completely within the field, constant coil area, and a fixed rotation axis. Real apparatus may produce deviations due to non-uniform fields, friction, or variations in angular speed.

05Worked examples

06Questions and solutions

The following questions build from standard flux calculations to more demanding situations involving rotation direction, angular velocity, and the meaning of a signed flux graph.

Question 1

An 80-turn coil has an area of \(0.012\text{ m}^2\) and rotates in a uniform magnetic field of strength \(0.50\text{ T}\).

Initially, the area vector is parallel to the magnetic field. The coil completes one full rotation.

Calculate the flux linkage at:

• 0°.
• 90°.
• 180°.
• 270°.
• 360°.

Describe the shape of the resulting graph.

Solution 1

The flux-linkage values are \(+0.48\), \(0\), \(-0.48\), \(0\), and \(+0.48\text{ Wb turns}\), respectively. The graph is one complete cosine cycle.

Step 1: Calculate maximum flux linkage.

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

\[
=80(0.50)(0.012)
\]

\[
\boxed{\lambda_{\max}=0.48\text{ Wb turns}}
\]

Step 2: Calculate the values at each angle.

Use:

\[
\lambda=0.48\cos\theta
\]

Step 3: Describe the graph.

The graph starts at maximum positive flux linkage, crosses zero after one-quarter rotation, reaches maximum negative flux linkage halfway through, and returns to its original value after one complete rotation.

The important point is that the 180° value is negative, not positive.

Question 2

A 100-turn coil has an area of \(0.020\text{ m}^2\) and rotates in a uniform magnetic field of strength \(0.25\text{ T}\).

At \(t=0\), its area vector makes an angle of 120° with the magnetic field.

The coil rotates at a constant angular velocity of \(600^\circ\text{ s}^{-1}\), with the angle increasing.

Calculate:

a) The maximum flux linkage.

b) The first time at which the flux linkage reaches its negative maximum.

c) The first zero crossing after this negative maximum.

d) The flux linkage at \(t=0\), \(0.10\text{ s}\), \(0.25\text{ s}\), \(0.40\text{ s}\), \(0.55\text{ s}\), and \(0.60\text{ s}\).

Solution 2

The first negative maximum occurs at \(0.10\text{ s}\), and the first zero crossing after it occurs at \(0.25\text{ s}\).

Step 1: Calculate maximum flux linkage.

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

\[
=100(0.25)(0.020)
\]

\[
\boxed{\lambda_{\max}=0.50\text{ Wb turns}}
\]

Step 2: Write the angle equation.

The initial angle is 120°, and the angular velocity is \(600^\circ\text{ s}^{-1}\).

\[
\theta=120^\circ+600^\circ t
\]

Therefore:

\[
\lambda=0.50\cos(120^\circ+600^\circ t)
\]

Step 3: Find the first negative maximum.

Negative maximum occurs when the area vector is directly opposite to the magnetic field.

Therefore:

\[
\theta=180^\circ
\]

Substitute into the angle equation:

\[
180=120+600t
\]

\[
600t=60
\]

\[
\boxed{t=0.10\text{ s}}
\]

Step 4: Find the first zero crossing after the negative maximum.

The next zero occurs at 270°.

\[
270=120+600t
\]

\[
600t=150
\]

\[
\boxed{t=0.25\text{ s}}
\]

Step 5: Calculate the requested flux-linkage values.

The graph completes one full cycle in \(0.60\text{ s}\), but it starts at a negative value because the coil’s initial angle is 120°.

A common mistake is to automatically start every flux graph at its positive maximum. That only works when the initial area vector is parallel to the magnetic field.

Question 3

An 80-turn coil has an area of \(0.015\text{ m}^2\) and rotates in a uniform magnetic field of strength \(0.40\text{ T}\).

Its area vector initially points parallel to the magnetic field.

The coil’s rotation is controlled as follows:

• During the first \(0.20\text{ s}\), it rotates from 0° to 180° at constant angular velocity.
• During the next \(0.60\text{ s}\), it rotates from 180° to 360° at a different constant angular velocity.

Calculate:

a) The maximum flux linkage.

b) The two times at which the flux linkage becomes zero.

c) The flux linkage at \(t=0.35\text{ s}\).

d) Explain why the flux-linkage graph against time is not an ordinary cosine curve, even though the flux-linkage graph against angle is still a cosine curve.

Solution 3

The flux linkage reaches zero at \(0.10\text{ s}\) and \(0.50\text{ s}\). At \(0.35\text{ s}\), it is approximately \(-0.34\text{ Wb turns}\).

Step 1: Calculate maximum flux linkage.

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

\[
=80(0.40)(0.015)
\]

\[
\boxed{\lambda_{\max}=0.48\text{ Wb turns}}
\]

Step 2: Calculate the angular velocity during each part of the rotation.

During the first half-turn:

\[
\omega_1=\frac{180^\circ}{0.20}
\]

\[
\boxed{\omega_1=900^\circ\text{ s}^{-1}}
\]

During the second half-turn:

\[
\omega_2=\frac{180^\circ}{0.60}
\]

\[
\boxed{\omega_2=300^\circ\text{ s}^{-1}}
\]

The coil rotates three times faster during the first half-turn.

Step 3: Find the first zero crossing.

The first zero occurs at 90°.

\[
t=\frac{90}{900}
\]

\[
\boxed{t=0.10\text{ s}}
\]

Step 4: Find the second zero crossing.

The coil reaches 180° after \(0.20\text{ s}\).

It then rotates another 90° at \(300^\circ\text{ s}^{-1}\).

\[
\Delta t=\frac{90}{300}
\]

\[
\Delta t=0.30\text{ s}
\]

Therefore:

\[
t=0.20+0.30
\]

\[
\boxed{t=0.50\text{ s}}
\]

Step 5: Calculate flux linkage at \(t=0.35\text{ s}\).

This time occurs during the second half of the rotation.

The coil has been rotating at \(300^\circ\text{ s}^{-1}\) for:

\[
0.35-0.20=0.15\text{ s}
\]

Its angular position is therefore:

\[
\theta=180^\circ+300(0.15)
\]

\[
\theta=225^\circ
\]

Now calculate flux linkage:

\[
\lambda=0.48\cos225^\circ
\]

\[
=0.48(-0.707)
\]

\[
\boxed{\lambda=-0.34\text{ Wb turns}}
\]

Step 6: Explain the shape of the two graphs.

Against rotation angle, the graph remains a standard cosine curve because:

\[
\lambda=NBA\cos\theta
\]

The relationship between flux linkage and angle has not changed.

However, the time graph is different because the coil does not rotate at a constant angular velocity.

The first half of the cosine curve is compressed into \(0.20\text{ s}\), while the second half is stretched across \(0.60\text{ s}\).

The tempting mistake is to assume that one complete mechanical rotation must always produce a standard cosine graph against time.

It only does so when the angular velocity is constant.

Question 4

Two identical coils are placed in the same uniform magnetic field of strength \(0.40\text{ T}\).

Each coil has 80 turns and an area of \(0.015\text{ m}^2\).

Both coils initially have their area vectors at 60° to the magnetic field.

However, their rotation axes are different:

• Coil A rotates about an axis perpendicular to the magnetic field and lying in the plane of the coil.
• Coil B rotates about an axis parallel to the magnetic field.

Both coils complete one mechanical revolution every \(0.50\text{ s}\).

Calculate the flux linkage of each coil at \(t=0\) and \(t=0.125\text{ s}\).

Explain why their flux-linkage graphs are different, even though both coils complete one full mechanical rotation in the same time.

Solution 4

Coil A has a changing flux linkage, while coil B has a constant flux linkage of \(+0.24\text{ Wb turns}\).

The difference comes from the direction of each coil’s rotation axis.

Step 1: Calculate maximum flux linkage.

Both coils have identical physical properties.

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

\[
=80(0.40)(0.015)
\]

\[
\boxed{\lambda_{\max}=0.48\text{ Wb turns}}
\]

Step 2: Calculate the initial flux linkage.

Both coils start at 60°.

\[
\lambda_0=0.48\cos60^\circ
\]

\[
\boxed{\lambda_0=+0.24\text{ Wb turns}}
\]

Step 3: Calculate the angular velocity of coil A.

The coil completes one rotation in \(0.50\text{ s}\).

\[
\omega=\frac{360^\circ}{0.50}
\]

\[
\boxed{\omega=720^\circ\text{ s}^{-1}}
\]

Step 4: Calculate the flux linkage of coil A at \(t=0.125\text{ s}\).

Its area vector rotates through:

\[
\Delta\theta=720(0.125)
\]

\[
\Delta\theta=90^\circ
\]

The new angle is:

\[
\theta=60^\circ+90^\circ
\]

\[
\theta=150^\circ
\]

Therefore:

\[
\lambda_A=0.48\cos150^\circ
\]

\[
\boxed{\lambda_A=-0.416\text{ Wb turns}}
\]

Coil A’s area vector changes its angle relative to the magnetic field as it rotates.

Step 5: Calculate the flux linkage of coil B.

This is where the question becomes interesting.

Coil B rotates about an axis parallel to the magnetic field.

Its area vector rotates around the magnetic field direction, but the angle between the two vectors remains constant.

It therefore stays at 60° throughout the rotation.

\[
\lambda_B=0.48\cos60^\circ
\]

\[
\boxed{\lambda_B=+0.24\text{ Wb turns}}
\]

This value does not change with time.

The key lesson is that mechanical rotation alone does not guarantee changing magnetic flux.

What matters is whether the angle between the magnetic field and the coil’s area vector actually changes.

Question 5

A 100-turn coil has an area of \(0.015\text{ m}^2\) and rotates in a uniform magnetic field of strength \(0.40\text{ T}\).

A fixed positive winding and terminal convention has been selected.

At a particular instant:

• The measured signed flux linkage is \(+0.30\text{ Wb turns}\).
• The measured induced emf is \(+2.6\text{ V}\).
• The coil’s angular speed is \(5.0\text{ rad s}^{-1}\).

The coil rotates at constant angular speed, but the direction of rotation is unknown.

Determine:

a) The possible angles between the area vector and the magnetic field.

b) Whether the measured emf identifies one unique angle.

c) The possible combinations of angle and rotation direction that produce the given measurements.

d) What additional measurement would allow the coil’s orientation to be identified uniquely.

Solution 5

There are two possible configurations: the area vector can be at 60° while rotating towards increasing angles, or at 300° while rotating towards decreasing angles. The measurements alone cannot distinguish between them.

Step 1: Calculate maximum flux linkage.

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

\[
=100(0.40)(0.015)
\]

\[
\boxed{\lambda_{\max}=0.60\text{ Wb turns}}
\]

Step 2: Use the measured flux linkage to find the possible angles.

The flux-linkage equation is:

\[
\lambda=NBA\cos\theta
\]

Substitute the measured value:

\[
0.30=0.60\cos\theta
\]

Rearrange:

\[
\cos\theta=\frac{0.30}{0.60}
\]

\[
\cos\theta=0.50
\]

There are two possible angles between 0° and 360°:

\[
\boxed{\theta=60^\circ\text{ or }300^\circ}
\]

This is because cosine has the same positive value in the first and fourth quadrants.

The flux-linkage measurement alone cannot tell us which orientation the coil occupies.

Step 3: Check the magnitude of the induced emf.

For constant angular velocity, the magnitude of the induced emf is:

\[
|\mathcal E|
=
NBA\omega|\sin\theta|
\]

Substitute the known values:

\[
|\mathcal E|
=
0.60(5.0)|\sin60^\circ|
\]

\[
=3.0(0.866)
\]

\[
\boxed{|\mathcal E|=2.60\text{ V}}
\]

This agrees with the measured emf.

However, notice that:

\[
|\sin60^\circ|=|\sin300^\circ|
\]

Both possible angles produce the same emf magnitude.

Step 4: Use the emf sign to examine the rotation direction.

The measured emf is positive.

According to Faraday’s law, its sign depends on the direction in which the flux linkage is changing, as well as the chosen terminal convention.

Consider the two possible orientations.

Case 1: The area vector is at 60°.

If the angle increases, the flux linkage decreases from its positive value towards zero.

The induced emf is positive with the chosen terminal convention.

Therefore, this configuration is consistent with the measurements.

Case 2: The area vector is at 300°.

If the angle increases, the flux linkage increases towards its positive maximum.

The induced emf would be negative.

However, if the angle decreases, the flux linkage decreases towards zero, producing a positive induced emf.

This configuration is also consistent with the measurements.

The two possible combinations are therefore:

Step 5: Identify the additional measurement required.

An angular-position sensor that also records the direction of rotation would resolve the ambiguity.

If the angle is increasing, the coil must be at 60°.

If the angle is decreasing, the coil must be at 300°.

The important point is that a flux measurement tells you the coil’s orientation is consistent with certain angles, while the emf measurement provides information about how that flux is changing.

Neither measurement, without knowing the rotation direction, uniquely identifies the coil’s orientation.

This is why interpreting a rotating-coil problem requires more than simply substituting numbers into the cosine equation.

06From magnetic flux to induced emf

Once you can sketch magnetic flux or flux linkage during one complete rotation, the next step is understanding how that changing flux produces an induced voltage.

Faraday’s law states:

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

The negative sign represents the direction described by Lenz’s law.

For a coil rotating at constant angular velocity:

• Flux linkage is maximum at 0° and 180°.
• Induced emf is zero at these positions.
• Flux linkage is zero at 90° and 270°.
• The magnitude of induced emf is greatest at these zero-crossing positions.

This relationship is useful when sketching the induced-emf graph from a flux-linkage graph.

The key decision is to identify what the question actually asks you to plot: magnetic flux, flux linkage, or induced emf. They are connected, but their graphs are not interchangeable.

For a broader understanding of the relationship between changing magnetic flux and induced voltage, continue with How Transformers Work: From Magnetic Flux to Voltage.

If you need to revisit the starting geometry, Magnetic Flux for HSC Physics: Using Φ = BA cos θ covers the underlying flux relationship.