Why a Moving Coil Can Produce No EMF in a Uniform Field

A coil can move through a magnetic field without producing an induced EMF. That sounds odd: the wire is moving through a region filled with magnetic field lines, so surely something should happen?

Not necessarily. If the entire coil translates through a **uniform, steady magnetic field**, without rotating or changing shape, the magnetic flux through it stays constant. No net EMF is induced around the coil.

The important distinction is between moving through a magnetic field and changing the magnetic flux through a coil. These are not the same thing. For induction, it is the change in flux linkage that matters, not simply whether the coil is moving.

## Why moving through a magnetic field does not guarantee an EMF

Picture a rectangular coil moving horizontally through a large region of uniform magnetic field. The field points vertically upwards, perpendicular to the plane of the coil. The coil is completely inside the field before, during, and after its movement.

Before it moves, the field passes through a certain amount of area inside the coil. After it moves, exactly the same amount of field passes through exactly the same area, in exactly the same direction.

The coil has changed position, but nothing about its relationship with the magnetic field has changed.

[IMAGE: A rectangular wire coil translating horizontally to the right through a large uniform magnetic field. Show three ghosted positions of the same coil along its path, with identical parallel upward magnetic field arrows passing perpendicularly through each coil. Label the constant translation velocity v, magnetic field B, coil area A, and perpendicular area vector A. Make clear that the coil remains entirely inside the field throughout its motion.]

This is the central idea: **induction depends on a change in magnetic flux linkage, not on displacement alone.**

### Start with the area vector

To understand why, we need to be precise about what the coil's orientation means.

The area vector, written as \(\vec A\), is an imaginary vector that:

- Has a magnitude equal to the area enclosed by one turn of the coil.
- Points perpendicular to the plane of the coil.
- Represents the orientation of the coil's surface.

There are two possible directions for this perpendicular vector. We choose one as the positive direction and keep that choice consistent.

Magnetic flux through one turn is calculated using:

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

Or, in terms of magnitudes:

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

where:

- \(\Phi_B\) is magnetic flux, measured in webers (Wb).
- \(B\) is magnetic flux density, measured in tesla (T).
- \(A\) is the area enclosed by one turn, measured in square metres (\(\mathrm{m^2}\)).
- \(\theta\) is the angle between the magnetic field and the area vector.

Notice that the angle is measured from the perpendicular to the coil, not from the plane of the coil.

For a coil with \(N\) identical turns, the flux linkage is:

\[
\boxed{N\Phi_B=NBA\cos\theta}
\]

The number of turns matters because every turn contributes its own magnetic flux.

### What happens when the coil translates?

Suppose the coil moves from position 1 to position 2 without rotating or changing shape.

Consider each quantity in the flux linkage equation:

| Quantity | Before translation | After translation |
|---|---|---|
| Magnetic field strength, \(B\) | Constant | Constant |
| Coil area, \(A\) | Constant | Constant |
| Angle, \(\theta\) | Constant | Constant |
| Number of turns, \(N\) | Constant | Constant |

Every factor remains unchanged.

Therefore:

\[
(N\Phi_B)_{\mathrm{before}}
=
(N\Phi_B)_{\mathrm{after}}
\]

So:

\[
\boxed{\Delta(N\Phi_B)=0}
\]

Faraday's law gives the induced EMF:

\[
\boxed{\mathcal E=-\frac{\Delta(N\Phi_B)}{\Delta t}}
\]

Here, \(\mathcal E\) is the induced EMF in volts (V), and \(\Delta t\) is the time interval in seconds (s).

Since the flux linkage does not change:

\[
\boxed{\mathcal E=0\ \mathrm V}
\]

This is why a coil can travel a considerable distance through a magnetic field and still have no net induced EMF.

The result does not depend on how far the coil travels. It depends on whether its flux linkage changes during that movement.

### A useful way to picture it

Imagine sliding a rectangular picture frame across a wall covered in evenly spaced parallel lines. The frame moves, but the number of lines passing through its enclosed area stays the same.

The analogy captures the important idea: changing position does not automatically change what passes through the enclosed area.

There is one limitation. Magnetic field lines are not physical strings passing through a frame. The analogy is simply a way to picture the geometry represented by the flux equation.

## The experiment: Does a moving coil actually produce an EMF?

To test the prediction, we need an arrangement that measures an induced EMF while allowing the coil to move.

Imagine a rectangular coil mounted on an insulating carriage. It is connected to a sensitive galvanometer, an instrument that detects small currents and indicates their direction. The coil moves horizontally through the central region of a large electromagnet, where the magnetic field is approximately uniform.

The connecting leads must be arranged carefully. They should follow a fixed, controlled path so that movement of the leads does not introduce an additional motional EMF into the measurement.

[IMAGE: Experimental apparatus showing a rectangular coil on an insulating horizontal sliding carriage inside the uniform central region between the broad pole faces of an electromagnet. Show the coil connected to a sensitive galvanometer by carefully routed flexible leads, with the uniform magnetic field perpendicular to the coil plane. Include a direction arrow for the carriage motion and identify the galvanometer, electromagnet, coil, and field region.]

### What should the original experiment show?

There are two possible predictions.

**Prediction 1:** The coil moves through the magnetic field, so an EMF must be induced.

This prediction treats movement itself as the cause of induction.

**Prediction 2:** The coil moves, but its area, orientation, and the magnetic field through it remain unchanged. Therefore, no net EMF is induced around the coil.

The second prediction is correct for the ideal arrangement.

If the coil is already entirely inside the uniform field, moving it steadily should produce no sustained galvanometer deflection attributable to the coil's translation.

This observation supports the flux-linkage model. It also shows why the phrase *relative motion between a conductor and a magnetic field* is not, by itself, a complete test for electromagnetic induction.

### Why the measurement needs care

A galvanometer does not measure magnetic flux directly. It detects current caused by an EMF in the connected circuit.

If the connecting wires move through the magnetic field, they can experience their own motional EMFs. These could affect the reading even when the coil itself has no net induced EMF.

A poorly arranged experiment could therefore appear to contradict the prediction.

The improved arrangement keeps the leads outside the changing region where possible, fixes their geometry, and accounts for any EMF they might experience. It also uses a uniform central field rather than the less predictable field near the magnet's edges.

The conclusion is specific: a rigid coil undergoing pure translation through a static, uniform magnetic field has no net induced EMF around its closed circuit. It is not a claim that moving conductors can never experience magnetic forces or local potential differences.

A coil can move through a magnetic field without producing an induced EMF. That sounds odd: the wire is moving through a region filled with magnetic field lines, so surely something should happen?

Not necessarily. If the entire coil translates through a **uniform, steady magnetic field**, without rotating or changing shape, the magnetic flux through it stays constant. No net EMF is induced around the coil.

The important distinction is between moving through a magnetic field and changing the magnetic flux through a coil. These are not the same thing. For induction, it is the change in flux linkage that matters, not simply whether the coil is moving.

## Why moving through a magnetic field does not guarantee an EMF

Picture a rectangular coil moving horizontally through a large region of uniform magnetic field. The field points vertically upwards, perpendicular to the plane of the coil. The coil is completely inside the field before, during, and after its movement.

Before it moves, the field passes through a certain amount of area inside the coil. After it moves, exactly the same amount of field passes through exactly the same area, in exactly the same direction.

The coil has changed position, but nothing about its relationship with the magnetic field has changed.

[IMAGE: A rectangular wire coil translating horizontally to the right through a large uniform magnetic field. Show three ghosted positions of the same coil along its path, with identical parallel upward magnetic field arrows passing perpendicularly through each coil. Label the constant translation velocity v, magnetic field B, coil area A, and perpendicular area vector A. Make clear that the coil remains entirely inside the field throughout its motion.]

This is the central idea: **induction depends on a change in magnetic flux linkage, not on displacement alone.**

### Start with the area vector

To understand why, we need to be precise about what the coil's orientation means.

The area vector, written as \(\vec A\), is an imaginary vector that:

- Has a magnitude equal to the area enclosed by one turn of the coil.
- Points perpendicular to the plane of the coil.
- Represents the orientation of the coil's surface.

There are two possible directions for this perpendicular vector. We choose one as the positive direction and keep that choice consistent.

Magnetic flux through one turn is calculated using:

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

Or, in terms of magnitudes:

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

where:

- \(\Phi_B\) is magnetic flux, measured in webers (Wb).
- \(B\) is magnetic flux density, measured in tesla (T).
- \(A\) is the area enclosed by one turn, measured in square metres (\(\mathrm{m^2}\)).
- \(\theta\) is the angle between the magnetic field and the area vector.

Notice that the angle is measured from the perpendicular to the coil, not from the plane of the coil.

[IMAGE: An explanatory three-dimensional diagram of a tilted rectangular coil showing its flat surface, perpendicular area vector A, magnetic field vector B, and angle theta between B and A. Include a second orientation with the field parallel to the coil plane to show why the flux becomes zero. Clearly distinguish the coil's plane from its normal.]

For a coil with \(N\) identical turns, the flux linkage is:

\[
N\Phi_B=NBA\cos\theta
\]

The number of turns matters because every turn contributes its own magnetic flux.

### What happens when the coil translates?

Suppose the coil moves from position 1 to position 2 without rotating or changing shape.

Consider each quantity in the flux linkage equation:

| Quantity | Before translation | After translation |
|---|---|---|
| Magnetic field strength, \(B\) | Constant | Constant |
| Coil area, \(A\) | Constant | Constant |
| Angle, \(\theta\) | Constant | Constant |
| Number of turns, \(N\) | Constant | Constant |

Every factor remains unchanged.

Therefore:

\[
(N\Phi_B)_{\mathrm{before}}
=
(N\Phi_B)_{\mathrm{after}}
\]

So:

\[
\Delta(N\Phi_B)=0
\]

Faraday's law gives the induced EMF:

\[
\mathcal E=-\frac{\Delta(N\Phi_B)}{\Delta t}
\]

Here, \(\mathcal E\) is the induced EMF in volts (V), and \(\Delta t\) is the time interval in seconds (s).

Since the flux linkage does not change:

\[
\mathcal E=0\ \mathrm V
\]

This is why a coil can travel a considerable distance through a magnetic field and still have no net induced EMF.

The result does not depend on how far the coil travels. It depends on whether its flux linkage changes during that movement.

### A useful way to picture it

Imagine sliding a rectangular picture frame across a wall covered in evenly spaced parallel lines. The frame moves, but the number of lines passing through its enclosed area stays the same.

The analogy captures the important idea: changing position does not automatically change what passes through the enclosed area.

There is one limitation. Magnetic field lines are not physical strings passing through a frame. The analogy is simply a way to picture the geometry represented by the flux equation.

## The experiment: Does a moving coil actually produce an EMF?

To test the prediction, we need an arrangement that measures an induced EMF while allowing the coil to move.

Consider an illustrative experiment using a rectangular coil mounted on an insulating carriage. The coil is connected to a sensitive galvanometer, an instrument that detects small currents and indicates their direction.

The carriage moves horizontally through the central region of a large electromagnet, where the magnetic field is approximately uniform. The coil remains perpendicular to the field throughout the first part of the investigation.

The connecting leads must be arranged carefully. They should follow a controlled path so that movement of the leads does not introduce an additional motional EMF into the measurement.

[IMAGE: Experimental apparatus showing a rectangular coil on an insulating horizontal sliding carriage inside the uniform central region between the broad pole faces of an electromagnet. Show the coil connected to a sensitive galvanometer by carefully routed flexible leads, with the uniform magnetic field perpendicular to the coil plane. Include a direction arrow for the carriage motion and identify the galvanometer, electromagnet, coil, and field region.]

### What should the experiment show?

There are two possible predictions.

**Prediction 1:** The coil moves through the magnetic field, so an EMF must be induced.

This prediction treats movement itself as the cause of induction.

**Prediction 2:** The coil moves, but its area, orientation, and the magnetic field through it remain unchanged. Therefore, no net EMF is induced around the coil.

The second prediction is correct for the ideal arrangement.

If the coil is already entirely inside the uniform field, moving it steadily should produce no sustained galvanometer deflection attributable to its translation.

Now consider what happens if the same coil enters or leaves the field region.

The expected qualitative pattern is:

| Stage of movement | Expected galvanometer reading |
|---|---|
| Coil stationary outside the field | Zero |
| Coil entering the field | Brief deflection |
| Coil moving entirely inside the field | Returns to zero |
| Coil leaving the field | Brief deflection in the opposite direction |
| Coil stationary outside again | Zero |

These are predicted results for the idealised arrangement, not claimed measurements from an actual experiment.

The pattern supports the flux-linkage model. It also shows why the phrase *relative motion between a conductor and a magnetic field* is not, by itself, a complete test for electromagnetic induction.

### Why the measurement needs care

A galvanometer does not measure magnetic flux directly. It detects current caused by an EMF in the connected circuit.

If the connecting wires move through the magnetic field, they can experience their own motional EMFs. These could affect the reading even when the coil itself has no net induced EMF.

A poorly arranged experiment could therefore appear to contradict the prediction.

The improved arrangement keeps the leads outside the strong-field region where possible, controls their geometry, and uses the central uniform region rather than the less predictable field near the magnet's edges.

The conclusion is specific: a rigid coil undergoing pure translation through a static, uniform magnetic field has no net induced EMF around its closed circuit. It is not a claim that moving conductors can never experience magnetic forces or local potential differences.

## But don't moving charges experience a magnetic force?

Yes. This is an important part of the explanation.

When the coil moves through a magnetic field, its free charge carriers move with the wire. These moving charges experience a magnetic force.

The magnetic force on a moving charge is:

\[
\vec F_B=q(\vec v\times\vec B)
\]

where:

- \(q\) is the charge in coulombs (C).
- \(\vec v\) is the charge's velocity in metres per second (\(\mathrm{m\,s^{-1}}\)).
- \(\vec B\) is the magnetic field in tesla (T).

The direction of this force is perpendicular to both the velocity and the magnetic field.

So why doesn't this force produce a current around the entire coil?

The answer comes down to what happens on opposite sides of the rectangular coil.

Imagine the coil lying flat in the plane of the page. The magnetic field points into the page, and the coil moves to the right.

Using the right-hand rule, the magnetic force on positive charges points upwards.

Both vertical sides of the coil experience this same upward force.

[IMAGE: A rectangular conducting coil viewed face-on, translating to the right through a uniform magnetic field directed into the page, represented by crosses. Show positive charge magnetic-force arrows pointing upwards on both vertical sides. Indicate that the two vertical sides have equal upward motional effects, while the horizontal sides have no force component along their length. Do not draw a net circulating current.]

Here is the important detail:

- The left vertical side experiences an upward motional effect.
- The right vertical side experiences an upward motional effect of equal magnitude.
- However, when we travel around the complete loop, these contributions act in opposite directions.

For the pictured arrangement, the magnitude of the motional EMF along each vertical side is:

\[
\mathcal E_{\mathrm{side}}=B\ell v
\]

where \(\ell\) is the length of the vertical side.

The two contributions cancel:

\[
\mathcal E_{\mathrm{net}}
=
B\ell v-B\ell v
\]

\[
\mathcal E_{\mathrm{net}}=0
\]

The magnetic force can redistribute charges locally, but it does not create a net circulating EMF around the complete coil.

This is another way to understand the result predicted by Faraday's law.

**The tempting mistake is to see a magnetic force on moving charges and immediately assume there must be an induced current.** You must consider the complete circuit, not just one section of wire.

## When does translating a coil actually produce an EMF?

Translation can produce an EMF when it changes the magnetic flux linkage.

There are three particularly useful cases to recognise.

### Case 1: The coil crosses the boundary of a magnetic field

Suppose the coil starts outside a magnetic field and moves into it.

Initially, very little of the coil's enclosed area is inside the field. As the coil enters, more of its area is exposed to the magnetic field.

The area contributing to the magnetic flux increases, so the flux linkage changes.

[IMAGE: Three sequential positions of a rectangular coil entering a finite uniform magnetic field from the left. Show the first coil completely outside, the second partly inside, and the third fully inside. The magnetic field is perpendicular to the coil plane. Highlight the changing overlap area in the middle position. Include a horizontal velocity arrow.]

Once the entire coil is inside the uniform field, the overlap area stops changing. The flux linkage becomes constant, so the induced EMF returns to zero.

This explains why a coil can produce an EMF while entering a field but not while translating through the field's uniform interior.

For more detail on calculating the changing overlap area, see [How to Calculate Magnetic Flux Through a Coil Partly in a Field](https://tutorgum.com/learn/resources/hsc-physics-how-to-find-flux-through-a-coil-partly-inside-a-magnetic-field/).

### Case 2: The coil moves through a non-uniform magnetic field

A uniform field has the same magnetic field strength and direction throughout the relevant region.

A non-uniform field does not.

Imagine a coil moving from a region where the magnetic field strength is \(0.10\ \mathrm T\) into a region where it is \(0.30\ \mathrm T\).

Even if the coil maintains exactly the same area and orientation, the magnetic flux through it changes because \(B\) changes.

Therefore, an induced EMF can occur.

The key distinction is that the coil is no longer experiencing the same magnetic field at every position.

### Case 3: The coil rotates while translating

Translation and rotation are different types of motion.

A coil can translate without changing its orientation. But if it rotates, its area vector changes direction.

Remember:

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

Even if \(B\) and \(A\) remain constant, changing \(\theta\) changes the magnetic flux.

For example, a coil initially has its area vector parallel to the magnetic field, so \(\theta=0^\circ\). It then rotates until \(\theta=60^\circ\).

Its flux per turn changes from:

\[
\Phi_{\mathrm{initial}}=BA
\]

to:

\[
\Phi_{\mathrm{final}}=BA\cos60^\circ
\]

\[
\Phi_{\mathrm{final}}=0.5BA
\]

The flux has halved.

The rotation has changed the flux linkage, even if the coil's centre has not moved at all.

If you need to distinguish an angle measured from the coil's surface from one measured from its 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/).

### Comparing the different situations

| Situation | What happens to flux linkage? | Net induced EMF? |
|---|---|---|
| Translation inside a uniform field | Remains constant | No |
| Entering a field region | Increases or decreases | Yes |
| Leaving a field region | Increases or decreases | Yes |
| Translation through a field gradient | May change | Potentially |
| Rotation in a uniform field | Changes if orientation changes relative to the field | Potentially |
| Deformation of the coil | May change because area changes | Potentially |

The most reliable decision process is to compare the flux linkage at different positions.

Do not begin by asking how fast the coil moves. First identify what happens to \(B\), \(A\), \(\theta\), and \(N\).

## Worked examples

### Worked example: A coil translating entirely inside a uniform field

A rectangular coil has 60 turns and an area of \(3.0\times10^{-3}\ \mathrm{m^2}\). It is placed in a uniform magnetic field of \(0.25\ \mathrm T\), with its area vector parallel to the field.

The coil translates \(0.60\ \mathrm m\) in \(0.40\ \mathrm s\). It remains entirely inside the uniform field and does not rotate or change shape.

Calculate its initial flux linkage and the average induced EMF during its movement.

**Step 1**: Calculate the magnetic flux through one turn.

The area vector is parallel to the magnetic field, so:

\[
\theta=0^\circ
\]

Using:

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

Substitute:

\[
\Phi_B=0.25(3.0\times10^{-3})\cos0^\circ
\]

\[
\Phi_B=7.5\times10^{-4}\ \mathrm{Wb}
\]

**Step 2**: Calculate the initial flux linkage.

There are 60 turns, so:

\[
N\Phi_B=60(7.5\times10^{-4})
\]

\[
N\Phi_B=0.045\ \mathrm{Wb\,turn}
\]

**Step 3**: Consider what happens after translation.

The coil remains entirely inside the same uniform field.

Its area, orientation, and number of turns are unchanged. Therefore, the final flux linkage is also:

\[
(N\Phi_B)_{\mathrm{final}}
=0.045\ \mathrm{Wb\,turn}
\]

**Step 4**: Calculate the average induced EMF.

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

Substitute:

\[
\mathcal E_{\mathrm{avg}}
=
-\frac{0.045-0.045}{0.40}
\]

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

The coil moves at \(1.5\ \mathrm{m\,s^{-1}}\), but its speed does not change the result.

The important point is that its flux linkage remains constant throughout the translation.

### Worked example: A coil entering a uniform field at an angle

A rectangular coil has 60 turns. Its width along the direction of travel is \(0.12\ \mathrm m\), and its height is \(0.10\ \mathrm m\).

It moves at \(0.80\ \mathrm{m\,s^{-1}}\) into a uniform magnetic field of \(0.40\ \mathrm T\). The magnetic field makes an angle of \(60^\circ\) with the coil's area vector.

Initially, the coil is completely outside the field. Calculate the average induced EMF while the entire coil enters the field.

Assume the field outside the region is negligible and the coil maintains a fixed orientation.

[IMAGE: Rectangular 60-turn coil entering a uniform magnetic field across a sharp vertical boundary. Show the coil partly inside the field, with its width along the horizontal direction of travel and height perpendicular to that direction. Include the magnetic field vector at 60 degrees to the coil's area vector, the area vector normal to the coil, the velocity arrow, and dimensions 0.12 m and 0.10 m. Do not show the calculated EMF.]

**Step 1**: Calculate the time taken for the coil to enter completely.

The coil must move a horizontal distance of \(0.12\ \mathrm m\).

Using:

\[
v=\frac{\Delta x}{\Delta t}
\]

Rearrange:

\[
\Delta t=\frac{\Delta x}{v}
\]

Substitute:

\[
\Delta t=\frac{0.12}{0.80}
\]

\[
\Delta t=0.15\ \mathrm s
\]

**Step 2**: Calculate the increase in area inside the magnetic field.

The coil enters completely, so the increase in area is its full rectangular area:

\[
\Delta A=0.12\times0.10
\]

\[
\Delta A=0.012\ \mathrm{m^2}
\]

**Step 3**: Calculate the change in flux linkage.

The angle is measured from the area vector, so we use:

\[
\Delta(N\Phi_B)=NB\Delta A\cos\theta
\]

Substitute:

\[
\Delta(N\Phi_B)
=
60(0.40)(0.012)\cos60^\circ
\]

\[
\Delta(N\Phi_B)=0.144\ \mathrm{Wb\,turn}
\]

The flux linkage increases because more of the coil's area enters the field.

**Step 4**: Calculate the average induced EMF.

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

Substitute:

\[
\mathcal E_{\mathrm{avg}}
=
-\frac{0.144}{0.15}
\]

\[
\boxed{\mathcal E_{\mathrm{avg}}=-0.96\ \mathrm V}
\]

The magnitude of the induced EMF is \(0.96\ \mathrm V\).

The negative sign indicates that the induced EMF acts opposite to the chosen positive loop direction, defined using the area vector.

Notice what changed between this example and the previous one.

In the first example, the coil translated without changing its flux linkage. Here, the coil translates across a field boundary, so the area exposed to the field increases.

Once the coil is completely inside the uniform field, the EMF returns to zero. When it eventually leaves the field, the flux linkage decreases and the induced EMF reverses direction.

## Questions and solutions

The following questions build from straightforward translation to situations where several changes occur simultaneously. Pay particular attention to the difference between magnetic flux and the rate of change of magnetic flux.

### Question 1

A 40-turn rectangular coil has an area of \(2.5\times10^{-3}\ \mathrm{m^2}\). It is positioned in a uniform magnetic field of \(0.32\ \mathrm T\), perpendicular to the plane of the coil.

The coil translates \(0.45\ \mathrm m\) in \(0.30\ \mathrm s\), remaining entirely within the uniform field. Its orientation does not change.

Calculate:

a) The initial magnetic flux through one turn.

b) The initial flux linkage.

c) The average induced EMF during the translation.

### Solution 1

**The average induced EMF is zero volts.** Although the coil moves, its magnetic flux linkage remains constant.

**Part a: Magnetic flux through one turn**

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

Therefore:

\[
\theta=0^\circ
\]

Using:

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

Substitute:

\[
\Phi_B=0.32(2.5\times10^{-3})\cos0^\circ
\]

\[
\boxed{\Phi_B=8.0\times10^{-4}\ \mathrm{Wb}}
\]

**Part b: Initial flux linkage**

\[
N\Phi_B=40(8.0\times10^{-4})
\]

\[
\boxed{N\Phi_B=0.032\ \mathrm{Wb\,turn}}
\]

**Part c: Average induced EMF**

Because the coil remains inside the same uniform field and does not rotate, its final flux linkage is also \(0.032\ \mathrm{Wb\,turn}\).

Therefore:

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

\[
\mathcal E_{\mathrm{avg}}
=
-\frac{0.032-0.032}{0.30}
\]

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

The tempting approach is to use the coil's velocity to calculate an EMF. However, the velocity alone is insufficient. Both sides of the coil experience equal motional contributions that cancel around the complete circuit.

### Question 2

Two identical 100-turn rectangular coils, A and B, each have an area of \(0.010\ \mathrm{m^2}\). Both are entirely inside the same static, uniform magnetic field of \(0.20\ \mathrm T\).

- Coil A has its area vector at \(60^\circ\) to the magnetic field. It translates \(0.20\ \mathrm m\) in \(0.10\ \mathrm s\).
- Coil B has its area vector parallel to the magnetic field. It translates \(0.40\ \mathrm m\) in the same time.

Neither coil rotates or changes shape.

Calculate the flux linkage of each coil and their average induced EMFs. Explain why the coil moving faster does not necessarily experience a greater induced EMF.

[IMAGE: Side-by-side comparison of two identical rectangular coils translating through the same uniform magnetic field. Coil A has its area vector at 60 degrees to B, while coil B has its area vector parallel to B. Show horizontal velocity arrows of different lengths to represent their different speeds. Both coils must remain completely within the uniform field.]

### Solution 2

**Both coils have zero average induced EMF, despite having different flux linkages and different speeds.**

The distinction is that flux measures the magnetic field passing through a coil, while induced EMF depends on how quickly that flux linkage changes.

**Coil A**

Its flux linkage is:

\[
N\Phi_A=NBA\cos\theta
\]

Substitute:

\[
N\Phi_A
=
100(0.20)(0.010)\cos60^\circ
\]

\[
\boxed{N\Phi_A=0.100\ \mathrm{Wb\,turn}}
\]

The coil remains entirely inside the uniform field, and its orientation is unchanged.

Therefore:

\[
\Delta(N\Phi_A)=0
\]

Its average induced EMF is:

\[
\boxed{\mathcal E_A=0\ \mathrm V}
\]

**Coil B**

Its area vector is parallel to the magnetic field, so:

\[
\theta=0^\circ
\]

Its flux linkage is:

\[
N\Phi_B=100(0.20)(0.010)\cos0^\circ
\]

\[
\boxed{N\Phi_B=0.200\ \mathrm{Wb\,turn}}
\]

Again, its flux linkage remains constant during translation.

Therefore:

\[
\boxed{\mathcal E_B=0\ \mathrm V}
\]

**Comparison**

Coil B has twice the flux linkage of coil A, and it travels twice as far in the same time.

However, neither coil experiences a change in flux linkage.

This exposes two common mistakes:

- A larger magnetic flux does not automatically mean a larger induced EMF.
- A greater speed does not automatically mean a greater induced EMF.

The rate of change of flux linkage is what determines the induced EMF.

### Question 3

A 30-turn rectangular coil has an area of \(0.015\ \mathrm{m^2}\). Its area vector remains parallel to the magnetic field as it translates through a region where the magnetic field strength increases.

The average magnetic field over the coil changes from \(0.18\ \mathrm T\) to \(0.30\ \mathrm T\) during a translation of \(0.50\ \mathrm m\), which takes \(0.25\ \mathrm s\).

Assume the field is approximately uniform over the coil at each measured position.

a) Calculate the change in flux linkage.

b) Calculate the average induced EMF.

c) Explain why the result differs from that of a coil translating through a uniform field.

[IMAGE: Rectangular coil moving horizontally through a magnetic field gradient. Show the field arrows becoming progressively denser from left to right, indicating increasing magnetic field strength. The coil maintains a fixed orientation, with its area vector parallel to the field. Include the translation direction and the two field-strength values, 0.18 T and 0.30 T.]

### Solution 3

**The coil experiences an average induced EMF of \(0.216\ \mathrm V\) in magnitude because the magnetic field strength increases as it moves.**

Its area and orientation are constant, but its magnetic flux changes.

**Part a: Change in flux linkage**

The area vector is parallel to the field, so:

\[
\theta=0^\circ
\]

Initial flux linkage:

\[
(N\Phi_B)_i=NBA_i\cos0^\circ
\]

\[
=30(0.18)(0.015)
\]

\[
=0.081\ \mathrm{Wb\,turn}
\]

Final flux linkage:

\[
(N\Phi_B)_f=30(0.30)(0.015)
\]

\[
=0.135\ \mathrm{Wb\,turn}
\]

Therefore:

\[
\Delta(N\Phi_B)=0.135-0.081
\]

\[
\boxed{\Delta(N\Phi_B)=0.054\ \mathrm{Wb\,turn}}
\]

**Part b: Average induced EMF**

Using Faraday's law:

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

Substitute:

\[
\mathcal E_{\mathrm{avg}}
=
-\frac{0.054}{0.25}
\]

\[
\boxed{\mathcal E_{\mathrm{avg}}=-0.216\ \mathrm V}
\]

The magnitude is \(0.216\ \mathrm V\).

The negative sign indicates that the induced EMF opposes the increase in positive flux linkage.

**Part c: Explanation**

In a uniform field, translating a rigid coil does not change the magnetic field strength experienced by the coil.

Here, the coil moves into a region with a stronger magnetic field. Its flux linkage therefore increases, even though its area and orientation remain fixed.

The important difference is the spatial variation in magnetic field strength.

A coil does not need to rotate or change shape to experience an induced EMF. A change in the magnetic field across its enclosed area is sufficient.

### Question 4

A 20-turn coil has a constant area of \(0.010\ \mathrm{m^2}\). It translates through a region where the magnetic field strength increases while the coil simultaneously rotates.

The following illustrative data are recorded:

| Time (s) | Magnetic field strength (T) | Angle to area vector |
|---|---:|---:|
| 0.00 | 0.20 | \(0^\circ\) |
| 0.25 | 0.30 | \(30^\circ\) |
| 0.50 | 0.40 | \(60^\circ\) |

Assume the magnetic field is approximately uniform over the coil at each position, and the area vector maintains the same general direction relative to the field.

a) Calculate the flux linkage at each time.

b) Calculate the average induced EMF during the first \(0.25\ \mathrm s\).

c) Calculate the average induced EMF during the second \(0.25\ \mathrm s\).

d) Calculate the average induced EMF over the entire \(0.50\ \mathrm s\).

e) Explain why obtaining zero average EMF over the complete interval does not mean that no induction occurred.

### Solution 4

**The coil experiences opposite average EMFs during the two halves of its motion, even though its average EMF over the entire interval is zero.**

The key is that the magnetic field strength and orientation do not change in a way that keeps the flux linkage constant at every instant.

**Part a: Calculate the flux linkage at each time**

Use:

\[
N\Phi_B=NBA\cos\theta
\]

At \(t=0.00\ \mathrm s\):

\[
N\Phi_0
=
20(0.20)(0.010)\cos0^\circ
\]

\[
\boxed{N\Phi_0=0.0400\ \mathrm{Wb\,turn}}
\]

At \(t=0.25\ \mathrm s\):

\[
N\Phi_1
=
20(0.30)(0.010)\cos30^\circ
\]

\[
\boxed{N\Phi_1=0.0520\ \mathrm{Wb\,turn}}
\]

At \(t=0.50\ \mathrm s\):

\[
N\Phi_2
=
20(0.40)(0.010)\cos60^\circ
\]

\[
\boxed{N\Phi_2=0.0400\ \mathrm{Wb\,turn}}
\]

Notice something interesting.

The initial and final flux linkages are identical, but the middle value is greater.

**Part b: Average EMF during the first interval**

From \(0.00\) to \(0.25\ \mathrm s\):

\[
\mathcal E_1
=
-\frac{0.0520-0.0400}{0.25}
\]

\[
\boxed{\mathcal E_1=-0.0480\ \mathrm V}
\]

The flux linkage increases, so the induced EMF is negative relative to the chosen positive direction.

**Part c: Average EMF during the second interval**

From \(0.25\) to \(0.50\ \mathrm s\):

\[
\mathcal E_2
=
-\frac{0.0400-0.0520}{0.25}
\]

\[
\boxed{\mathcal E_2=+0.0480\ \mathrm V}
\]

The flux linkage decreases, so the induced EMF reverses direction.

**Part d: Average EMF over the entire interval**

Using the initial and final flux linkages:

\[
\mathcal E_{\mathrm{avg}}
=
-\frac{0.0400-0.0400}{0.50}
\]

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

**Part e: Explanation**

The zero average EMF is a consequence of the initial and final flux linkages being equal.

It does not mean that the flux linkage remained constant throughout the movement.

During the first half, flux linkage increased. During the second half, it decreased.

The two changes produced opposite average EMFs that cancelled when averaged over the complete interval.

This is an important distinction in exam questions: **equal initial and final flux linkages do not guarantee zero instantaneous EMF throughout the movement.**

### Question 5

A 50-turn rectangular coil moves horizontally through a finite strip of uniform magnetic field.

The following conditions apply:

- The coil is \(0.20\ \mathrm m\) wide in the direction of travel.
- Its height is \(0.10\ \mathrm m\).
- The magnetic field strip is also \(0.20\ \mathrm m\) wide.
- The magnetic field strength is \(0.30\ \mathrm T\).
- The field is perpendicular to the plane of the coil.
- The coil moves at a constant speed of \(0.50\ \mathrm{m\,s^{-1}}\).
- The coil has a total resistance of \(3.0\ \Omega\).

Initially, the coil is completely outside the magnetic field region. It moves completely through the field strip and finishes outside it on the opposite side.

Assume the magnetic field is negligible outside the strip, the field boundaries are sharp, the resistance is constant, and the coil's self-inductance is negligible.

[IMAGE: A rectangular 50-turn coil moving from left to right through a finite rectangular magnetic field region. The coil width along the direction of travel is 0.20 m, exactly matching the width of the field region. Its height is 0.10 m. Show the initial position outside the field, the instant of maximum overlap when the coil exactly coincides with the field strip, and the final position outside the field. Draw the uniform magnetic field perpendicular to the coil plane. Include the 0.50 m/s velocity arrow. Do not include a flux graph, EMF graph, current directions, or calculated answers.]

a) Describe how the magnetic flux linkage changes during the complete journey. Explain why there is no interval during which the coil remains entirely inside the field.

b) Calculate the maximum flux linkage.

c) Calculate the magnitude and duration of each induced EMF pulse.

d) Calculate the magnitude of the current and the charge transferred during the entry pulse.

e) Determine the net signed charge transferred over the complete journey. Also calculate the total thermal energy dissipated in the coil.

f) Explain why zero net signed charge does not mean that no induction occurred or that no energy was transferred.

### Solution 5

**The coil experiences two equal and opposite EMF pulses, each with a magnitude of \(0.75\ \mathrm V\). Although the net signed charge transferred is zero, the coil dissipates \(0.15\ \mathrm J\) of electrical energy as heat.**

The important detail is that the field strip and coil have exactly the same width. The coil reaches maximum overlap at one instant, then immediately begins leaving the field.

**Part a: Describe the flux linkage**

Initially, the coil is completely outside the field.

Therefore:

\[
N\Phi_{\mathrm{initial}}=0
\]

As the coil enters the field, its overlap area increases. The magnetic flux linkage increases at a constant rate because the coil moves at constant speed through a uniform field with a sharp boundary.

When the coil exactly coincides with the field strip, its overlap area is at a maximum.

However, the field strip and coil have equal widths.

There is no interval during which the coil can remain entirely inside the field. As soon as the coil reaches maximum overlap, it begins leaving the field on the opposite side.

The flux linkage then decreases at the same rate until the coil is completely outside.

The overall pattern is:

- Before entry: zero flux linkage.
- During entry: increasing flux linkage.
- At maximum overlap: maximum flux linkage.
- During exit: decreasing flux linkage.
- After exit: zero flux linkage.

The induced EMF is negative during entry and positive during exit, using an area vector pointing along the magnetic field.

**Part b: Maximum flux linkage**

First calculate the total coil area:

\[
A=0.20\times0.10
\]

\[
A=0.020\ \mathrm{m^2}
\]

The magnetic field is perpendicular to the coil plane, so:

\[
\theta=0^\circ
\]

The maximum flux linkage is:

\[
N\Phi_{\max}=NBA\cos0^\circ
\]

Substitute:

\[
N\Phi_{\max}
=
50(0.30)(0.020)
\]

\[
\boxed{N\Phi_{\max}=0.300\ \mathrm{Wb\,turn}}
\]

**Part c: Magnitude and duration of each EMF pulse**

The coil must travel \(0.20\ \mathrm m\) to move from its first contact with the field to maximum overlap.

Using:

\[
\Delta t=\frac{\Delta x}{v}
\]

\[
\Delta t=\frac{0.20}{0.50}
\]

\[
\boxed{\Delta t=0.40\ \mathrm s}
\]

The exit takes the same amount of time.

Therefore, each EMF pulse lasts \(0.40\ \mathrm s\).

During entry, the flux linkage increases from zero to \(0.300\ \mathrm{Wb\,turn}\).

Using Faraday's law:

\[
|\mathcal E|
=
\frac{|\Delta(N\Phi_B)|}{\Delta t}
\]

\[
|\mathcal E|
=
\frac{0.300}{0.40}
\]

\[
\boxed{|\mathcal E|=0.75\ \mathrm V}
\]

The exit produces an EMF of the same magnitude but opposite polarity.

**Part d: Current and charge during entry**

Use Ohm's law:

\[
I=\frac{|\mathcal E|}{R}
\]

Substitute:

\[
I=\frac{0.75}{3.0}
\]

\[
\boxed{I=0.25\ \mathrm A}
\]

The charge transferred during the entry pulse is:

\[
Q=I\Delta t
\]

Substitute:

\[
Q=0.25(0.40)
\]

\[
\boxed{Q_{\mathrm{entry}}=0.10\ \mathrm C}
\]

**Part e: Net charge and thermal energy**

The exit pulse has the same current magnitude and duration, but its direction is reversed.

Therefore, the signed charge transferred during exit is:

\[
Q_{\mathrm{exit}}=-0.10\ \mathrm C
\]

The net signed charge is:

\[
Q_{\mathrm{net}}
=
Q_{\mathrm{entry}}+Q_{\mathrm{exit}}
\]

\[
Q_{\mathrm{net}}=0.10-0.10
\]

\[
\boxed{Q_{\mathrm{net}}=0\ \mathrm C}
\]

However, the total magnitude of charge that passes through the coil during both pulses is:

\[
Q_{\mathrm{total}}=0.10+0.10
\]

\[
\boxed{Q_{\mathrm{total}}=0.20\ \mathrm C}
\]

Now calculate the thermal energy dissipated.

Electrical power is:

\[
P=I^2R
\]

Substitute:

\[
P=(0.25)^2(3.0)
\]

\[
P=0.1875\ \mathrm W
\]

There are two pulses, each lasting \(0.40\ \mathrm s\).

Therefore:

\[
E_{\mathrm{thermal}}=P(2\Delta t)
\]

\[
E_{\mathrm{thermal}}
=
0.1875(0.80)
\]

\[
\boxed{E_{\mathrm{thermal}}=0.15\ \mathrm J}
\]

**Part f: Why zero net charge does not mean zero energy transfer**

This is the most important conceptual result.

The current flows in one direction during entry and the opposite direction during exit. The signed charges cancel, giving zero net charge transferred.

However, electrical heating depends on the square of the current:

\[
P=I^2R
\]

Squaring the current removes its sign. Both pulses produce positive thermal energy.

The energy dissipated as heat must come from somewhere. In this idealised constant-speed situation, an external agent supplies mechanical work to keep the coil moving against the magnetic forces associated with the induced current.

Therefore, zero net flux change over the complete journey does not mean zero instantaneous EMF, zero current, or zero energy transfer.

It means only that the initial and final flux linkages are equal.

## Where to take this next

When you see a coil moving through a magnetic field in an HSC Physics question, resist the urge to calculate an EMF immediately from its speed.

First establish whether the coil's flux linkage actually changes.

Check the magnetic field strength, the enclosed area, the area-vector direction, and the number of turns. If all remain constant, translation alone produces no net EMF around the coil.

If one of them changes, calculate the initial and final flux linkages and apply Faraday's law.

That distinction is particularly useful when analysing coils entering and leaving field regions, rotating generators, and changing magnetic fields in electromagnetic induction problems.