Why Magnetic Flux Can Be Zero in a Strong Magnetic Field
A strong magnetic field does not always produce magnetic flux through a coil. Learn how the area vector, coil orientation, and angle determine magnetic flux, with worked calculations and HSC Physics practice questions.
A coil can sit inside a strong magnetic field of 2.0 T and still have exactly zero magnetic flux through it. That sounds contradictory. If the magnetic field is strong, surely plenty of magnetic field must pass through the coil?
The key is orientation. Rotate the coil until the magnetic field runs parallel to its flat surface. The field is still there, the coil has not changed size, and the magnetic field strength has not decreased. However, no component of the field passes through the coil's enclosed surface.
**Magnetic flux can be zero in a strong magnetic field because flux depends on the direction of the field relative to the coil, not just the field's strength.**
This is one of the most useful applications of the area vector in HSC Physics. Once you understand what that vector represents, zero flux stops looking like a strange exception and becomes a predictable result.
[IMAGE: A rectangular wire coil shown in two orientations within the same uniform magnetic field. In the first, the magnetic field passes perpendicular through the coil's flat surface, producing maximum flux. In the second, the coil is rotated so the field runs parallel to its surface, producing zero flux. Show identical field-line spacing in both cases to emphasise that field strength has not changed. Include the coil's area vector as a perpendicular arrow in each orientation.]
## 1. A strong magnetic field does not guarantee magnetic flux
Imagine holding a sheet of cardboard in falling rain.
When the cardboard is horizontal, it catches plenty of rain. Tilt it, and less rain lands on its surface. Hold it vertically, with the rain falling parallel to the cardboard, and almost none lands on it.
The amount of rain falling has not changed. Only the orientation of the cardboard has changed.
Magnetic flux behaves in a similar way. The magnetic field is represented by the falling rain, while the coil's enclosed surface is represented by the cardboard.
The analogy has a limit: magnetic field lines are not actual streams of particles. They represent the direction and strength of a magnetic field. Magnetic flux is a mathematical measure of the field passing through a chosen surface.
The important idea is that a magnetic field can be strong around a coil without passing through the area enclosed by it.
### What actually determines magnetic flux?
Three quantities matter:
- Magnetic field strength, \(B\), measured in tesla (T).
- Area enclosed by the coil, \(A\), measured in square metres (\(\mathrm{m^2}\)).
- Orientation of the magnetic field relative to the coil's surface.
For a flat surface in a uniform magnetic field, magnetic flux is:
\[
\Phi=BA\cos\theta
\]
Here:
- \(\Phi\) is magnetic flux through one turn, measured in webers (Wb).
- \(B\) is magnetic flux density, measured in tesla (T).
- \(A\) is the area enclosed by the coil, measured in square metres (\(\mathrm{m^2}\)).
- \(\theta\) is the angle between the magnetic field and the coil's area vector.
Notice that the angle is measured from the area vector, not from the flat surface of the coil. This is the detail that explains why a strong magnetic field can produce zero flux.
## 2. The area vector: the missing piece
A flat coil has a surface, but calculating magnetic flux requires us to represent that surface using a direction.
We do this with an area vector.
An area vector is an imaginary arrow with two properties:
1. It points perpendicular (at \(90^\circ\)) to the plane of the coil.
2. Its magnitude is equal to the area enclosed by the coil.
It is also called the surface normal.
Think of a rectangular coil lying flat on a desk. Its area vector would point vertically upwards or vertically downwards. It does not point along the wire, around the perimeter, or across the surface.
[IMAGE: Annotated three-dimensional rectangular wire coil. Show the flat coil as a clearly visible rectangular loop, with a perpendicular area vector labelled A pointing upwards from its centre. Include a second possible normal pointing downwards, the coil's plane, and a magnetic field vector B approaching at an angle theta to the upward normal. Clearly distinguish the surface from its perpendicular area vector.]
Why bother introducing an arrow that isn't physically part of the coil?
Because magnetic flux depends on how much of the magnetic field points in the same direction as the area vector.
We can split the magnetic field into two components:
- A component perpendicular to the coil's plane, which contributes to magnetic flux.
- A component parallel to the coil's plane, which does not contribute to magnetic flux.
Only the perpendicular component matters.
Mathematically, the component of the magnetic field along the area vector is \(B\cos\theta\). Multiplying this by the coil's area gives:
\[
\Phi=(B\cos\theta)A
\]
Therefore:
\[
\boxed{\Phi=BA\cos\theta}
\]
The cosine is not an arbitrary feature to memorise. It selects the component of the magnetic field that passes through the surface.
### The three orientations you must recognise
There are three particularly important arrangements.
[IMAGE: Three side-by-side three-dimensional views of identical rectangular coils in the same uniform magnetic field. First: B points along A, with theta = 0 degrees and maximum positive flux. Second: B is at 60 degrees to A, with theta = 60 degrees and half the maximum flux. Third: B runs parallel to the coil's plane and is perpendicular to A, with theta = 90 degrees and zero flux. Draw the area vector and magnetic field clearly in every panel.]
| Coil orientation | Angle to area vector | Magnetic flux |
|---|---:|---:|
| Field perpendicular to the coil's plane | \(0^\circ\) | \(+BA\), maximum positive |
| Field at an angle to the area vector | \(0^\circ<\theta<90^\circ\) | Between zero and \(BA\) |
| Field parallel to the coil's plane | \(90^\circ\) | Zero |
| Field opposite to the area vector | \(180^\circ\) | \(-BA\), maximum negative |
The zero-flux arrangement is the third row.
When the field runs parallel to the coil's plane, it is perpendicular to the area vector. Therefore:
\[
\Phi=BA\cos90^\circ=0
\]
This remains true regardless of how large \(B\) becomes, provided the idealised geometry remains unchanged.
For example, increasing the field from \(0.5\,\mathrm{T}\) to \(5.0\,\mathrm{T}\) will not change the flux if the field remains exactly parallel to the coil's plane.
That is the central answer to the title question.
A coil can sit inside a strong magnetic field of 2.0 T and still have exactly zero magnetic flux through it. That sounds contradictory. If the magnetic field is strong, surely plenty of magnetic field must pass through the coil?
The key is orientation. Rotate the coil until the magnetic field runs parallel to its flat surface. The field is still there, the coil has not changed size, and the magnetic field strength has not decreased. However, no component of the field passes through the coil's enclosed surface.
**Magnetic flux can be zero in a strong magnetic field because flux depends on the direction of the field relative to the coil, not just the field's strength.**
This is one of the most useful applications of the area vector in HSC Physics. Once you understand what that vector represents, zero flux stops looking like a strange exception and becomes a predictable result.
[IMAGE: A rectangular wire coil shown in two orientations within the same uniform magnetic field. In the first, the magnetic field passes perpendicular through the coil's flat surface, producing maximum flux. In the second, the coil is rotated so the field runs parallel to its surface, producing zero flux. Show identical magnetic field-line spacing in both cases to emphasise that field strength has not changed. Include the perpendicular area vector in each orientation.]
## 1. A strong magnetic field does not guarantee magnetic flux
Imagine holding a sheet of cardboard in falling rain.
When the cardboard is horizontal, it catches plenty of rain. Tilt it, and less rain lands on its surface. Hold it vertically, with the rain falling parallel to the cardboard, and almost none lands on it.
The amount of rain falling has not changed. Only the orientation of the cardboard has changed.
Magnetic flux behaves in a similar way. The magnetic field is represented by the falling rain, while the coil's enclosed surface is represented by the cardboard.
The analogy has a limit: magnetic field lines are not actual streams of particles. They represent the direction and strength of a magnetic field. Magnetic flux is a mathematical measure of the field passing through a chosen surface.
The important idea is that a magnetic field can be strong around a coil without passing through the area enclosed by it.
### What actually determines magnetic flux?
Three quantities matter:
- Magnetic field strength, \(B\), measured in tesla (T).
- Area enclosed by the coil, \(A\), measured in square metres (\(\mathrm{m^2}\)).
- Orientation of the magnetic field relative to the coil's surface.
For a flat surface in a uniform magnetic field, magnetic flux is:
\[
\Phi=BA\cos\theta
\]
Here:
- \(\Phi\) is magnetic flux through one turn, measured in webers (Wb).
- \(B\) is magnetic flux density, measured in tesla (T).
- \(A\) is the area enclosed by the coil, measured in square metres (\(\mathrm{m^2}\)).
- \(\theta\) is the angle between the magnetic field and the coil's area vector.
Notice that the angle is measured from the area vector, not from the flat surface of the coil. This is the detail that explains why a strong magnetic field can produce zero flux.
## 2. The area vector: the missing piece
A flat coil has a surface, but calculating magnetic flux requires us to represent that surface using a direction.
We do this with an area vector.
An area vector is an imaginary arrow with two properties:
1. It points perpendicular (at \(90^\circ\)) to the plane of the coil.
2. Its magnitude is equal to the area enclosed by the coil.
It is also called the surface normal.
Think of a rectangular coil lying flat on a desk. Its area vector would point vertically upwards or vertically downwards. It does not point along the wire, around the perimeter, or across the surface.
[IMAGE: Annotated three-dimensional rectangular wire coil. Show the flat coil as a clearly visible rectangular loop, with a perpendicular area vector labelled A pointing upwards from its centre. Include a second possible normal pointing downwards, the coil's plane, and a magnetic field vector B approaching at an angle theta to the upward normal. Clearly distinguish the surface from its perpendicular area vector.]
Why bother introducing an arrow that isn't physically part of the coil?
Because magnetic flux depends on how much of the magnetic field points in the same direction as the area vector.
We can split the magnetic field into two components:
- A component perpendicular to the coil's plane, which contributes to magnetic flux.
- A component parallel to the coil's plane, which does not contribute to magnetic flux.
Only the perpendicular component matters.
Mathematically, the component of the magnetic field along the area vector is \(B\cos\theta\). Multiplying this by the coil's area gives:
\[
\Phi=(B\cos\theta)A
\]
Therefore:
\[
\boxed{\Phi=BA\cos\theta}
\]
The cosine is not an arbitrary feature to memorise. It selects the component of the magnetic field that passes through the surface.
### The three orientations you must recognise
There are three particularly important arrangements.
[IMAGE: Three side-by-side three-dimensional views of identical rectangular coils in the same uniform magnetic field. First: B points along A, with theta = 0 degrees and maximum positive flux. Second: B is at 60 degrees to A, with theta = 60 degrees and half the maximum flux. Third: B runs parallel to the coil's plane and is perpendicular to A, with theta = 90 degrees and zero flux. Draw the area vector and magnetic field clearly in every panel.]
| Coil orientation | Angle to area vector | Magnetic flux |
|---|---:|---:|
| Field perpendicular to the coil's plane | \(0^\circ\) | \(+BA\), maximum positive |
| Field at an angle to the area vector | \(0^\circ<\theta<90^\circ\) | Between zero and \(BA\) |
| Field parallel to the coil's plane | \(90^\circ\) | Zero flux |
| Field opposite to the area vector | \(180^\circ\) | \(-BA\), maximum negative |
The zero-flux arrangement is the third row.
When the field runs parallel to the coil's plane, it is perpendicular to the area vector. Therefore:
\[
\Phi=BA\cos90^\circ=0
\]
This remains true regardless of how large \(B\) becomes, provided the idealised geometry remains unchanged.
For example, increasing the field from \(0.5\,\mathrm{T}\) to \(5.0\,\mathrm{T}\) will not change the flux if the field remains exactly parallel to the coil's plane.
That is the central answer to the title question.
## 3. Why the angle matters more than you might expect
Let's look more closely at what happens to the magnetic field as the coil rotates.
Suppose a coil is initially positioned so that its area vector points directly along the magnetic field.
The angle is:
\[
\theta=0^\circ
\]
The perpendicular component of the magnetic field is:
\[
B_\perp=B\cos0^\circ=B
\]
The entire field contributes to magnetic flux.
Now rotate the coil until its area vector is at \(60^\circ\) to the field.
\[
B_\perp=B\cos60^\circ=0.5B
\]
Only half the magnetic field strength contributes in the direction perpendicular to the coil's surface.
Finally, rotate the coil until its area vector is perpendicular to the magnetic field.
\[
B_\perp=B\cos90^\circ=0
\]
There is no perpendicular component.
The field can still be extremely strong, but its contribution to magnetic flux is zero.
### A nearly parallel field is not quite zero flux
There is another distinction worth understanding.
If the magnetic field is almost parallel to the coil's plane, the flux will be small, but not necessarily zero.
Consider a coil in a \(2.0\,\mathrm{T}\) field, with an area of \(0.015\,\mathrm{m^2}\). Its area vector makes an angle of \(89^\circ\) with the magnetic field.
\[
\Phi=BA\cos\theta
\]
\[
\Phi=(2.0)(0.015)\cos89^\circ
\]
\[
\boxed{\Phi=5.2\times10^{-4}\,\mathrm{Wb}}
\]
The flux is small because \(\cos89^\circ\) is close to zero.
If the angle increases to exactly \(90^\circ\), the flux becomes zero.
This is useful when interpreting diagrams. A coil that looks almost edge-on to the magnetic field might have a very small flux, but you should not assume it is exactly zero unless the geometry establishes a \(90^\circ\) angle to the area vector.
### What if the angle is measured from the coil's plane?
Some questions give the angle between the magnetic field and the flat surface of the coil rather than its normal.
Call this angle \(\alpha\).
Because the area vector is perpendicular to the plane:
\[
\theta=90^\circ-\alpha
\]
Substituting into the magnetic flux equation:
\[
\Phi=BA\cos(90^\circ-\alpha)
\]
Using the complementary-angle identity:
\[
\boxed{|\Phi|=BA\sin\alpha}
\]
This gives a useful alternative way to identify zero flux.
If the magnetic field is parallel to the coil's plane, then:
\[
\alpha=0^\circ
\]
Therefore:
\[
|\Phi|=BA\sin0^\circ=0
\]
The field has no component perpendicular to the coil's surface.
If you want more practice distinguishing angles measured from the plane and angles measured from the normal, see [How to Calculate Magnetic Flux When the Angle Is to the Coil's Plane](https://tutorgum.com/learn/resources/hsc-physics-how-to-find-flux-when-the-angle-is-given-to-the-coil-s-plane/).
### Does negative flux mean the magnetic field is weaker?
No.
The sign of magnetic flux depends on the direction of the chosen area vector.
Suppose we choose the area vector to point to the right. A magnetic field pointing to the right produces positive flux. A field pointing to the left produces negative flux.
At \(180^\circ\):
\[
\Phi=BA\cos180^\circ=-BA
\]
The magnitude of the flux is still at its maximum. Only its direction relative to the chosen normal has changed.
This matters when calculating changes in flux. A coil rotating from \(0^\circ\) to \(180^\circ\) does not have zero change in flux just because the initial and final magnitudes are equal.
The flux has reversed direction.
## 4. What happens when the coil has many turns?
So far, we've considered a single loop.
Real coils often contain many turns of wire. Each turn can experience approximately the same magnetic flux.
This introduces magnetic flux linkage.
Flux linkage is the magnetic flux through one turn multiplied by the number of turns:
\[
\boxed{\lambda=N\Phi}
\]
where:
- \(\lambda\) is magnetic flux linkage, measured in Wb-turn.
- \(N\) is the number of turns.
- \(\Phi\) is magnetic flux through one turn, measured in Wb.
Substituting the magnetic flux equation:
\[
\boxed{\lambda=NBA\cos\theta}
\]
This equation assumes that all turns have the same area and orientation and experience approximately the same magnetic field.
### Can a coil with 1,000 turns still have zero flux linkage?
Yes.
Suppose every turn is positioned so that the magnetic field runs parallel to its plane.
The flux through each turn is:
\[
\Phi=BA\cos90^\circ=0
\]
For 1,000 turns:
\[
\lambda=1000\times0
\]
\[
\boxed{\lambda=0\,\mathrm{Wb\text{-}turn}}
\]
Adding more turns cannot overcome the orientation.
This is an important distinction:
- Magnetic flux describes the field passing through one turn.
- Flux linkage accounts for the number of turns experiencing that flux.
- Neither quantity is the same as magnetic field strength.
A larger number of turns increases flux linkage when the flux per turn is non-zero. It does not create flux when every turn has zero flux.
For a broader treatment of the relationship between field strength, area, angle, and flux linkage, see [Magnetic Flux for HSC Physics: Using Φ = BA cos θ](https://tutorgum.com/learn/resources/hsc-physics-magnetic-flux/).
## 5. An experiment that separates magnetic flux from induced voltage
There is a useful experimental problem here.
Suppose you connect a coil to a voltmeter and place it inside a strong magnetic field. The voltmeter reads zero.
Does that prove the magnetic flux through the coil is zero?
No. This is a measurement problem, and understanding it helps prevent another common HSC mistake.
### The first design: a stationary coil
Imagine a rectangular search coil placed between the poles of an electromagnet.
The electromagnet produces a uniform magnetic field across the coil. The coil is connected to a sensitive voltmeter.
The coil is initially positioned with its area vector parallel to the field.
There is substantial magnetic flux through it.
Now rotate the coil until its area vector is perpendicular to the field. The magnetic flux becomes zero.
If you stop the coil in either position, the voltmeter reads zero.
Why?
Because the voltmeter measures induced electromotive force (emf), which depends on the rate of change of magnetic flux linkage. It does not directly measure the amount of magnetic flux.
A stationary coil in a constant magnetic field has no changing flux, regardless of whether its flux is large or zero.
The original design cannot distinguish these two situations using the voltmeter alone.
### The revised design: rotate the coil continuously
We can improve the experiment by mounting the coil on a rotating axle.
The apparatus consists of:
1. An electromagnet with two opposing poles, producing a uniform magnetic field across the gap.
2. A rectangular search coil mounted on an axle within that gap.
3. An axle positioned in the plane of the coil and perpendicular to the magnetic field, allowing the coil's area vector to rotate relative to the field.
4. Slip rings connecting the rotating coil to an oscilloscope.
5. A Hall probe placed beside the coil to check the magnetic field strength.
6. An angle scale attached to the axle to track the angle between the magnetic field and the area vector.
The coil starts with its area vector aligned with the magnetic field. It then rotates at a constant angular speed.
[IMAGE: Detailed experimental apparatus showing an electromagnet with north and south poles, a rectangular multi-turn search coil rotating in the uniform magnetic field between them, and a pivot axis perpendicular to the field. Show the coil's area vector rotating from parallel to the field towards perpendicular to it. Include slip rings and connecting wires leading to an oscilloscope, a Hall probe beside the coil, and an angle scale at the axle. Make the physical connections and spatial arrangement clear.]
### What should the experiment show?
As the coil rotates, its magnetic flux linkage changes.
For a coil rotating from an initial angle of \(0^\circ\), the expected pattern is:
| Angle to area vector | Magnetic flux linkage | Induced emf at constant rotation speed |
|---|---|---|
| \(0^\circ\) | Maximum positive | Zero at that instant |
| \(60^\circ\) | Half maximum | Large |
| \(90^\circ\) | Zero | Maximum magnitude |
| \(120^\circ\) | Half maximum, negative | Large |
| \(180^\circ\) | Maximum negative magnitude | Zero at that instant |
The surprising result is at \(90^\circ\).
At the exact instant the magnetic flux becomes zero, the induced emf reaches its maximum magnitude, provided the coil is rotating at constant angular speed.
This happens because magnetic flux is changing most rapidly around that orientation.
The relationship is described by Faraday's law:
\[
\varepsilon_{\mathrm{avg}}=-\frac{\Delta\lambda}{\Delta t}
\]
Here:
- \(\varepsilon_{\mathrm{avg}}\) is the average induced emf, measured in volts (V).
- \(\Delta\lambda\) is the change in magnetic flux linkage, measured in Wb-turn.
- \(\Delta t\) is the time interval, measured in seconds (s).
The negative sign represents the direction of the induced emf according to Lenz's law. The sign of the measured voltage also depends on the chosen lead connections.
### A small experimental calculation
Suppose the search coil has:
- 100 turns
- An area of \(0.010\,\mathrm{m^2}\) per turn
- A magnetic field strength of \(0.40\,\mathrm{T}\)
The coil rotates from \(80^\circ\) to \(90^\circ\) in \(0.050\,\mathrm{s}\).
All angles are measured from the area vector.
Initially, the flux linkage is:
\[
\lambda_i=NBA\cos80^\circ
\]
\[
\lambda_i=(100)(0.40)(0.010)(0.17365)
\]
\[
\lambda_i=0.0695\,\mathrm{Wb\text{-}turn}
\]
At \(90^\circ\):
\[
\lambda_f=NBA\cos90^\circ=0
\]
Therefore:
\[
\Delta\lambda=\lambda_f-\lambda_i
\]
\[
\Delta\lambda=0-0.0695
\]
\[
\Delta\lambda=-0.0695\,\mathrm{Wb\text{-}turn}
\]
The magnitude of the average induced emf is:
\[
|\varepsilon_{\mathrm{avg}}|
=\frac{|\Delta\lambda|}{\Delta t}
\]
\[
=\frac{0.0695}{0.050}
\]
\[
\boxed{|\varepsilon_{\mathrm{avg}}|=1.39\,\mathrm{V}}
\]
The coil reaches zero magnetic flux at the end of this interval, but it has experienced a substantial induced emf while rotating.
These are illustrative calculated values, not reported experimental measurements.
### What does the experiment establish?
It establishes two different ideas:
- Magnetic flux describes the field passing through the coil's surface.
- Induced emf depends on how quickly the flux linkage changes.
A coil can have zero flux and a non-zero induced emf while rotating. The same coil can have zero induced emf while stationary, even when its magnetic flux is at its maximum.
The Hall probe helps check the field strength, but it does not measure flux directly. The flux is determined by combining the field measurement with the coil's area and orientation.
This distinction is particularly useful when answering questions that provide a voltmeter reading and ask you to infer something about magnetic flux.
## 6. Worked examples
### Worked example: A strong magnetic field produces zero flux
A rectangular coil has 150 turns and dimensions \(20\,\mathrm{cm}\times10\,\mathrm{cm}\). It is placed in a uniform magnetic field of \(1.8\,\mathrm{T}\). The magnetic field runs parallel to the plane of the coil.
Calculate:
a) The magnetic flux through one turn.
b) The magnetic flux linkage.
[IMAGE: Rectangular 150-turn coil positioned edge-on to a uniform magnetic field of 1.8 T. Show the field arrows running parallel to the coil's plane and the area vector perpendicular to the field. Label the coil dimensions as 20 cm by 10 cm.]
**Step 1**: Calculate the area of the coil.
Convert the dimensions into metres:
\[
20\,\mathrm{cm}=0.20\,\mathrm{m}
\]
\[
10\,\mathrm{cm}=0.10\,\mathrm{m}
\]
Therefore:
\[
A=0.20\times0.10
\]
\[
A=0.020\,\mathrm{m^2}
\]
**Step 2**: Identify the angle.
The magnetic field runs parallel to the coil's plane.
Therefore, it is perpendicular to the area vector:
\[
\theta=90^\circ
\]
**Step 3**: Calculate the magnetic flux through one turn.
Use:
\[
\Phi=BA\cos\theta
\]
Substitute:
\[
\Phi=(1.8)(0.020)\cos90^\circ
\]
\[
\boxed{\Phi=0\,\mathrm{Wb}}
\]
**Step 4**: Calculate the flux linkage.
Use:
\[
\lambda=N\Phi
\]
Substitute:
\[
\lambda=150\times0
\]
\[
\boxed{\lambda=0\,\mathrm{Wb\text{-}turn}}
\]
The magnetic field is strong, but the flux and flux linkage are both zero because the field has no component perpendicular to the coil's surface.
### Worked example: A coil nearly parallel to the magnetic field
A circular coil has 120 turns and a radius of \(5.0\,\mathrm{cm}\). It is placed in a uniform magnetic field of \(0.65\,\mathrm{T}\). The magnetic field makes an angle of \(20^\circ\) with the plane of the coil.
Calculate:
a) The magnetic flux through one turn.
b) The magnetic flux linkage.
c) The percentage of the maximum possible flux linkage.
[IMAGE: Circular multi-turn coil viewed in three dimensions, tilted so that its plane makes a 20-degree angle with the uniform magnetic field. Show the perpendicular area vector making a 70-degree angle with the field. Label the radius and distinguish the coil's plane from its normal.]
**Step 1**: Calculate the area of the circular coil.
Convert the radius into metres:
\[
r=0.050\,\mathrm{m}
\]
For a circle:
\[
A=\pi r^2
\]
\[
A=\pi(0.050)^2
\]
\[
A=0.00785\,\mathrm{m^2}
\]
**Step 2**: Find the angle to the area vector.
The question gives the angle to the plane, not the normal.
Therefore:
\[
\theta=90^\circ-20^\circ
\]
\[
\theta=70^\circ
\]
**Step 3**: Calculate the magnetic flux through one turn.
\[
\Phi=BA\cos\theta
\]
Substitute:
\[
\Phi=(0.65)(0.00785)\cos70^\circ
\]
\[
\Phi=(0.65)(0.00785)(0.3420)
\]
\[
\boxed{\Phi=1.75\times10^{-3}\,\mathrm{Wb}}
\]
**Step 4**: Calculate the magnetic flux linkage.
There are 120 turns.
\[
\lambda=N\Phi
\]
\[
\lambda=120(1.75\times10^{-3})
\]
\[
\boxed{\lambda=0.210\,\mathrm{Wb\text{-}turn}}
\]
**Step 5**: Compare this with the maximum possible flux linkage.
Maximum flux occurs when the magnetic field is parallel to the area vector.
Therefore:
\[
\lambda_{\max}=NBA
\]
\[
\lambda_{\max}=120(0.65)(0.00785)
\]
\[
\lambda_{\max}=0.612\,\mathrm{Wb\text{-}turn}
\]
The percentage of maximum flux linkage is:
\[
\frac{\lambda}{\lambda_{\max}}\times100
\]
\[
=\frac{0.210}{0.612}\times100
\]
\[
\boxed{34.2\%}
\]
Although the coil is inside a relatively strong magnetic field, its orientation means it experiences only about 34.2% of its maximum possible flux linkage.
## 7. A reliable decision process for HSC questions
When a question asks why magnetic flux is zero, or asks you to calculate flux from a diagram, use this process.
**1. Identify the coil's plane.**
Work out which surface encloses the area.
**2. Draw the area vector.**
It must be perpendicular to the coil's plane.
**3. Compare the magnetic field with the area vector.**
Do not automatically measure the angle from the wire or the flat surface.
**4. Identify the angle.**
If the field is parallel to the plane, the angle to the area vector is \(90^\circ\).
**5. Calculate the flux.**
Use:
\[
\Phi=BA\cos\theta
\]
**6. Check whether flux linkage is required.**
Only multiply by the number of turns if the question asks for flux linkage:
\[
\lambda=N\Phi
\]
A useful final check is to predict the result before using your calculator:
- Field perpendicular to the coil's plane: maximum flux magnitude.
- Field at an angle: partial flux.
- Field parallel to the coil's plane: zero flux.
- Field pointing opposite to the chosen area vector: maximum negative flux.
If your calculated result disagrees with the physical arrangement, check your angle.
## Questions and solutions
### Question 1
A single-turn rectangular coil has an area of \(0.025\,\mathrm{m^2}\). It is placed in a uniform magnetic field of \(1.6\,\mathrm{T}\).
The magnetic field runs parallel to the plane of the coil.
Calculate the magnetic flux through the coil. Explain why the result does not depend on the strength of the magnetic field in this particular arrangement.
[IMAGE: A single rectangular wire loop positioned with its flat plane parallel to uniform horizontal magnetic field arrows. Show the area vector perpendicular to the field. Label B = 1.6 T and A = 0.025 m².]
### Solution 1
The magnetic flux is exactly \(0\,\mathrm{Wb}\), regardless of the field strength, provided the field remains parallel to the coil's plane.
The area vector is perpendicular to the plane of the coil. Therefore, the magnetic field is at \(90^\circ\) to the area vector.
Using:
\[
\Phi=BA\cos\theta
\]
Substitute:
\[
\Phi=(1.6)(0.025)\cos90^\circ
\]
\[
\boxed{\Phi=0\,\mathrm{Wb}}
\]
The magnetic field contributes no component in the direction of the area vector.
Increasing \(B\) changes the strength of the magnetic field but does not change its orientation. Since the cosine term remains zero, the magnetic flux remains zero.
The important distinction is that zero magnetic flux does not mean zero magnetic field.
### Question 2
A rectangular coil has 40 turns and dimensions \(18\,\mathrm{cm}\times12\,\mathrm{cm}\). It is placed in a uniform magnetic field of \(0.75\,\mathrm{T}\).
The magnetic field makes an angle of \(10^\circ\) with the plane of the coil.
Calculate:
a) The magnetic flux through one turn.
b) The magnetic flux linkage.
c) The percentage of maximum possible flux linkage.
[IMAGE: A 40-turn rectangular coil tilted at a shallow angle to uniform magnetic field arrows. Show the 10-degree angle between the field and the coil's plane, and the perpendicular area vector at 80 degrees to the field. Label both rectangular dimensions.]
### Solution 2
The magnetic flux through one turn is \(2.81\times10^{-3}\,\mathrm{Wb}\), and the flux linkage is approximately \(0.113\,\mathrm{Wb\text{-}turn}\). This represents 17.4% of the maximum possible flux linkage.
The key is to convert the given angle to the angle between the field and the area vector.
**Step 1**: Calculate the area.
Convert the dimensions:
\[
18\,\mathrm{cm}=0.18\,\mathrm{m}
\]
\[
12\,\mathrm{cm}=0.12\,\mathrm{m}
\]
Therefore:
\[
A=0.18\times0.12
\]
\[
A=0.0216\,\mathrm{m^2}
\]
**Step 2**: Find the angle to the area vector.
The given angle is measured from the plane.
\[
\theta=90^\circ-10^\circ
\]
\[
\theta=80^\circ
\]
**Step 3**: Calculate flux through one turn.
\[
\Phi=BA\cos\theta
\]
\[
\Phi=(0.75)(0.0216)\cos80^\circ
\]
\[
\Phi=(0.75)(0.0216)(0.17365)
\]
\[
\boxed{\Phi=2.81\times10^{-3}\,\mathrm{Wb}}
\]
**Step 4**: Calculate flux linkage.
\[
\lambda=N\Phi
\]
\[
\lambda=40(2.81\times10^{-3})
\]
\[
\boxed{\lambda=0.113\,\mathrm{Wb\text{-}turn}}
\]
**Step 5**: Calculate the maximum possible flux linkage.
The maximum occurs when \(\theta=0^\circ\).
\[
\lambda_{\max}=NBA
\]
\[
\lambda_{\max}=40(0.75)(0.0216)
\]
\[
\lambda_{\max}=0.648\,\mathrm{Wb\text{-}turn}
\]
Therefore:
\[
\text{Percentage}
=\frac{0.113}{0.648}\times100
\]
\[
\boxed{\text{Percentage}=17.4\%}
\]
The result makes physical sense. The magnetic field is almost parallel to the coil's plane, so only a small component passes through its enclosed area.
A common mistake would be to substitute \(10^\circ\) directly into the cosine equation. That would incorrectly treat the angle as being measured from the area vector.
### Question 3
A 60-turn rectangular coil has an area of \(0.025\,\mathrm{m^2}\). It is placed in a uniform magnetic field of \(0.40\,\mathrm{T}\).
Its area vector initially makes an angle of \(65^\circ\) with the magnetic field. The coil is then rotated until the area vector makes an angle of \(115^\circ\) with the field.
Calculate:
a) The initial magnetic flux through one turn.
b) The final magnetic flux through one turn.
c) The change in magnetic flux through one turn.
d) The change in magnetic flux linkage.
A student argues that the initial and final fluxes have equal magnitudes, so the change in flux must be zero. Explain the error.
[IMAGE: Two positions of the same rectangular coil in a fixed uniform magnetic field. Show the initial area vector at 65 degrees to B and the final area vector at 115 degrees to B. Make the reversal of the area vector's component along B clear. Label the initial and final orientations without showing calculated answers.]
### Solution 3
The flux changes from \(+4.23\times10^{-3}\,\mathrm{Wb}\) to \(-4.23\times10^{-3}\,\mathrm{Wb}\). The change in flux is \(-8.45\times10^{-3}\,\mathrm{Wb}\) per turn, giving a change in flux linkage of approximately \(-0.507\,\mathrm{Wb\text{-}turn}\).
The student's mistake is ignoring the sign of magnetic flux.
**Step 1**: Calculate the initial flux.
\[
\Phi_i=BA\cos\theta_i
\]
Substitute:
\[
\Phi_i=(0.40)(0.025)\cos65^\circ
\]
\[
\Phi_i=(0.010)(0.4226)
\]
\[
\boxed{\Phi_i=+4.23\times10^{-3}\,\mathrm{Wb}}
\]
**Step 2**: Calculate the final flux.
\[
\Phi_f=BA\cos\theta_f
\]
Substitute:
\[
\Phi_f=(0.40)(0.025)\cos115^\circ
\]
\[
\Phi_f=(0.010)(-0.4226)
\]
\[
\boxed{\Phi_f=-4.23\times10^{-3}\,\mathrm{Wb}}
\]
**Step 3**: Calculate the change in flux.
Remember:
\[
\Delta\Phi=\Phi_f-\Phi_i
\]
Substitute:
\[
\Delta\Phi=(-4.23\times10^{-3})
-(4.23\times10^{-3})
\]
\[
\boxed{\Delta\Phi=-8.45\times10^{-3}\,\mathrm{Wb}}
\]
The magnitude of the change is \(8.45\times10^{-3}\,\mathrm{Wb}\).
**Step 4**: Calculate the change in flux linkage.
Since the coil has 60 turns:
\[
\Delta\lambda=N\Delta\Phi
\]
\[
\Delta\lambda=60(-8.45\times10^{-3})
\]
\[
\boxed{\Delta\lambda=-0.507\,\mathrm{Wb\text{-}turn}}
\]
The initial and final flux magnitudes are equal, but their directions relative to the chosen area vector are opposite.
The coil has rotated through the zero-flux orientation at \(90^\circ\). It then experiences flux in the opposite direction.
This is why calculating changes in magnetic flux requires signed values rather than magnitudes alone.
### Question 4
A 75-turn search coil has an area of \(0.0080\,\mathrm{m^2}\) per turn. It is placed in a uniform magnetic field of \(0.60\,\mathrm{T}\).
The coil rotates at constant angular speed from an angle of \(75^\circ\) to \(90^\circ\), measured between its area vector and the magnetic field. This rotation takes \(0.040\,\mathrm{s}\).
a) Calculate the magnitude of the average induced emf during the rotation.
b) The coil is then stopped at \(90^\circ\). Explain why the induced emf becomes zero.
c) Would the induced emf also be zero if the coil were stopped at \(0^\circ\)? Explain why this does not mean the magnetic flux is the same in both positions.
[IMAGE: A rectangular search coil rotating about an axle between the poles of an electromagnet. Show the area vector initially at 75 degrees to the field and finally at 90 degrees. Include the coil's electrical connections to a voltmeter or oscilloscope. The diagram should distinguish the rotating coil from the fixed magnetic field.]
### Solution 4
The average induced emf during rotation is approximately \(2.33\,\mathrm{V}\). Once the coil stops, the induced emf becomes zero because the magnetic flux linkage is no longer changing.
**Step 1**: Calculate the initial flux linkage.
Use:
\[
\lambda_i=NBA\cos\theta_i
\]
Substitute:
\[
\lambda_i=(75)(0.60)(0.0080)\cos75^\circ
\]
\[
\lambda_i=(0.36)(0.2588)
\]
\[
\lambda_i=0.0932\,\mathrm{Wb\text{-}turn}
\]
**Step 2**: Calculate the final flux linkage.
The final angle is \(90^\circ\).
\[
\lambda_f=NBA\cos90^\circ
\]
\[
\boxed{\lambda_f=0}
\]
**Step 3**: Calculate the change in flux linkage.
\[
\Delta\lambda=\lambda_f-\lambda_i
\]
\[
\Delta\lambda=0-0.0932
\]
\[
\Delta\lambda=-0.0932\,\mathrm{Wb\text{-}turn}
\]
**Step 4**: Calculate the average induced emf.
Using Faraday's law:
\[
|\varepsilon_{\mathrm{avg}}|
=\frac{|\Delta\lambda|}{\Delta t}
\]
Substitute:
\[
|\varepsilon_{\mathrm{avg}}|
=\frac{0.0932}{0.040}
\]
\[
\boxed{|\varepsilon_{\mathrm{avg}}|=2.33\,\mathrm{V}}
\]
**Step 5**: Explain what happens when the coil stops.
Once the coil stops at \(90^\circ\), its flux remains zero and constant.
Therefore:
\[
\Delta\lambda=0
\]
over any subsequent time interval.
There is no induced emf.
The same is true if the coil stops at \(0^\circ\). Its flux would be at its maximum positive value, but it would still be constant.
A stationary coil does not produce an induced emf simply because it has magnetic flux.
The important distinction is that the coil has zero flux at \(90^\circ\) and maximum flux at \(0^\circ\), even though the induced emf is zero at both stationary positions.
### Question 5 (Extension)
A search coil is constructed using a figure-eight arrangement. Each turn consists of two identical rectangular lobes, labelled A and B, connected in series.
Each lobe has an area of \(0.010\,\mathrm{m^2}\). The two lobes lie in the same plane but are wound in opposite senses, so their area vectors point in opposite directions.
The coil contains 20 such turns.
A magnetic field passes perpendicular to both lobes in the same physical direction.
Initially, the field strength is:
- Through lobe A: \(0.60\,\mathrm{T}\).
- Through lobe B: \(0.60\,\mathrm{T}\).
Over a period of \(0.20\,\mathrm{s}\), the field through lobe A increases uniformly to \(0.90\,\mathrm{T}\), while the field through lobe B remains constant at \(0.60\,\mathrm{T}\).
Assume the fields are uniform over their respective lobes, and ignore resistance and other circuit effects.
a) Calculate the initial net magnetic flux linkage of the coil.
b) Calculate the final net magnetic flux linkage.
c) Calculate the magnitude of the average induced emf.
d) Suppose instead that both magnetic field strengths increased uniformly from \(0.60\,\mathrm{T}\) to \(0.90\,\mathrm{T}\) over the same time. Would an induced emf be produced? Explain your answer.
[IMAGE: A figure-eight search coil with two equal rectangular lobes, A and B, lying in the same plane and joined at a central crossover. Show the opposite winding senses using clear wire-direction arrows and draw the two area vectors pointing in opposite directions. Show uniform magnetic field arrows pointing in the same physical direction through both lobes. Label each lobe's area as 0.010 m² and distinguish the two field regions.]
### Solution 5
The initial net flux linkage is zero. When only the field through lobe A increases, the final net flux linkage becomes \(0.060\,\mathrm{Wb\text{-}turn}\), producing an average induced emf of \(0.30\,\mathrm{V}\).
If both fields increase equally, the net flux linkage remains zero and no net induced emf is produced.
This problem is more subtle than an ordinary flat coil because the two lobes have opposite area-vector directions.
**Step 1**: Calculate the initial flux through each lobe.
For lobe A:
\[
\Phi_A=BA
\]
\[
\Phi_A=(0.60)(0.010)
\]
\[
\Phi_A=0.0060\,\mathrm{Wb}
\]
Lobe B has the same flux magnitude:
\[
\Phi_B=0.0060\,\mathrm{Wb}
\]
However, the lobes are wound in opposite senses. Their signed flux contributions therefore have opposite signs.
The initial net flux per turn is:
\[
\Phi_{\mathrm{net}}
=\Phi_A-\Phi_B
\]
\[
=0.0060-0.0060
\]
\[
\Phi_{\mathrm{net}}=0
\]
**Step 2**: Calculate the initial net flux linkage.
There are 20 turns:
\[
\lambda_i=N\Phi_{\mathrm{net}}
\]
\[
\lambda_i=20(0)
\]
\[
\boxed{\lambda_i=0\,\mathrm{Wb\text{-}turn}}
\]
Notice that each lobe individually experiences non-zero magnetic flux. The total is zero because the signed contributions cancel.
**Step 3**: Calculate the final flux through each lobe.
The magnetic field through lobe A increases to \(0.90\,\mathrm{T}\).
\[
\Phi_{A,f}=(0.90)(0.010)
\]
\[
\Phi_{A,f}=0.0090\,\mathrm{Wb}
\]
The field through lobe B remains at \(0.60\,\mathrm{T}\).
\[
\Phi_{B,f}=(0.60)(0.010)
\]
\[
\Phi_{B,f}=0.0060\,\mathrm{Wb}
\]
The final net flux per turn is:
\[
\Phi_{\mathrm{net},f}
=0.0090-0.0060
\]
\[
\Phi_{\mathrm{net},f}=0.0030\,\mathrm{Wb}
\]
**Step 4**: Calculate the final net flux linkage.
\[
\lambda_f=N\Phi_{\mathrm{net},f}
\]
\[
\lambda_f=20(0.0030)
\]
\[
\boxed{\lambda_f=0.060\,\mathrm{Wb\text{-}turn}
}
\]
**Step 5**: Calculate the average induced emf.
The change in flux linkage is:
\[
\Delta\lambda=\lambda_f-\lambda_i
\]
\[
\Delta\lambda=0.060-0
\]
\[
\Delta\lambda=0.060\,\mathrm{Wb\text{-}turn}
\]
Using Faraday's law:
\[
|\varepsilon_{\mathrm{avg}}|
=\frac{|\Delta\lambda|}{\Delta t}
\]
\[
=\frac{0.060}{0.20}
\]
\[
\boxed{|\varepsilon_{\mathrm{avg}}|=0.30\,\mathrm{V}}
\]
**Step 6**: Consider what happens if both fields increase equally.
If both magnetic fields increase from \(0.60\,\mathrm{T}\) to \(0.90\,\mathrm{T}\), the final flux through each lobe is:
\[
\Phi_A=\Phi_B=0.0090\,\mathrm{Wb}
\]
The net flux per turn remains:
\[
\Phi_{\mathrm{net}}=0.0090-0.0090=0
\]
The net flux linkage therefore remains zero.
Consequently:
\[
\Delta\lambda=0
\]
and:
\[
\boxed{\varepsilon_{\mathrm{avg}}=0\,\mathrm{V}}
\]
The tempting assumption is that changing a strong magnetic field must produce an induced emf. But the two lobes experience equal changes in flux linkage with opposite signs, so their induced emfs cancel.
This is an extension of the ordinary planar-coil model. It demonstrates an important limit of treating a coil as simply \(N\) identical turns with the same flux: the orientation and signed contribution of each part must be considered.
The central lesson is that magnetic flux is not a measure of how much magnetic field exists around a coil. It measures the signed contribution of the magnetic field through the coil's chosen surface.
For an ordinary flat coil in a uniform magnetic field, the area vector makes the geometry straightforward. If the field is parallel to the coil's plane, the perpendicular component is zero, and so is the flux, however strong the field becomes.
The next useful step is to apply the same area-vector reasoning to changing flux linkage. That is where coil rotation, induced emf, and Faraday's law come together.