Why a Current Loop Can Have Zero Net Force but Nonzero Torque
Discover why equal and opposite magnetic forces can cancel without cancelling torque. Learn how current, magnetic field strength, coil orientation, and force positions determine whether a loop rotates, with worked examples and challenging HSC Physics questions.
A rectangular wire loop sits in a uniform magnetic field. You switch on the current, and the two vertical sides of the loop experience magnetic forces. One force points into the page, and the other points out of it.
You might expect the forces to cancel, leaving the loop doing nothing. But the loop starts rotating.
How can the net force be zero while the loop still experiences a turning effect?
The answer is that **equal and opposite forces can cancel as a resultant force without cancelling their torque**. The forces act at different positions on the loop, so they form a couple. The loop has no net magnetic force, but it does have a nonzero net torque.
This is the key distinction: net force determines translational acceleration, while net torque determines rotational acceleration.
## Why opposite magnetic forces can still rotate a loop
### Start with one conductor segment
Before considering the whole loop, look at just one straight section of wire inside the magnetic field.
Imagine a vertical conductor placed between the north and south poles of a magnet. Conventional current flows upwards through the wire, while the magnetic field points horizontally from left to right.
The conductor experiences a magnetic force perpendicular to both the current and the magnetic field. Using the right-hand rule, the force points into the page.
[IMAGE: Close-up annotated diagram of a straight vertical current-carrying wire between horizontal north and south magnet pole faces. Show conventional current I vertically upwards, uniform magnetic flux density B horizontally from left to right, and magnetic force F_B directed into the page using a cross symbol. Label the active conductor length L and the angle theta between current and B.]
The magnitude of this magnetic force is:
\[
F_B=BIL\sin\theta
\]
Here is what each symbol represents:
| Symbol | Meaning | SI unit |
|---|---|---|
| \(F_B\) | Magnetic force on the conductor | N |
| \(B\) | Magnetic flux density | T |
| \(I\) | Conventional current | A |
| \(L\) | Length of conductor inside the magnetic field | m |
| \(\theta\) | Angle between the current direction and the magnetic field | Degrees |
There is an important distinction between \(B\) and \(F_B\).
- \(B\) describes the magnetic field at the conductor's position. It is measured in tesla.
- \(F_B\) is the force experienced by the conductor because of its interaction with that field. It is measured in newtons.
The magnetic field is not the force. A stronger magnetic field produces a greater force if the current, conductor length, and angle remain unchanged.
For a conductor perpendicular to the field, \(\theta=90^\circ\), so:
\[
F_B=BIL
\]
For a conductor parallel to the field, \(\theta=0^\circ\), so:
\[
F_B=0
\]
Notice that \(L\) refers only to the section of conductor inside the magnetic field, not necessarily the total length of wire in the circuit.
If you want to practise identifying the correct angle and active conductor length, see [How to Find Magnetic Force on an Angled Wire](https://tutorgum.com/learn/resources/hsc-physics-how-to-find-magnetic-force-on-an-angled-wire-segment/).
### Now consider the complete rectangular loop
Place a rectangular loop in the same uniform magnetic field. The plane of the loop is aligned with the field, so the field runs horizontally through the rectangle.
Conventional current travels upwards along the left vertical side and downwards along the right vertical side. The current directions on these two sides are opposite.
Apply the right-hand rule to each side:
- The left vertical side experiences a force into the page.
- The right vertical side experiences a force out of the page.
The forces have equal magnitudes because both sides have the same length, carry the same current, and experience the same uniform magnetic field.
What happens to the other two sides? They are parallel to the magnetic field, so they experience zero magnetic force.
[IMAGE: Large, clearly annotated rectangular single-turn current loop viewed face-on. Show a uniform magnetic field B directed horizontally to the right across the entire loop. Label the left vertical side with upward current I and a magnetic force into the page (cross symbols); label the right vertical side with downward current I and a magnetic force out of the page (dot symbols). Show the top and bottom conductors parallel to B, the vertical active length L, the horizontal separation w between the two force lines, and a central vertical rotation axis. Keep the two force arrows on opposite sides of the rectangle.]
The two magnetic forces cancel when added as vectors:
\[
\vec F_{\text{net}}=
\vec F_{\text{left}}+\vec F_{\text{right}}=0
\]
There is no resultant magnetic force on the loop.
However, notice where the forces act. They are separated by the horizontal width of the rectangle. The left force pushes one side into the page, while the right force pushes the other side out of the page.
The two sides therefore try to rotate the loop in the same rotational direction.
This arrangement is called a **couple**: two equal, opposite, parallel forces acting along different lines of action.
The important point is that cancelling forces do not necessarily cancel their turning effects.
### Why the turning effects add
Torque measures the turning effect of a force about an axis.
For a force acting perpendicular to a lever arm, its torque magnitude is:
\[
\tau=Fd
\]
Here, \(F\) is the force in newtons, and \(d\) is the perpendicular distance from the axis of rotation to the force's line of action, measured in metres.
For the rectangular loop, each force acts at a distance of \(w/2\) from the central rotation axis, where \(w\) is the loop's width.
Each side produces a torque in the same rotational direction:
\[
\tau_{\text{net}}
=
F_B\frac{w}{2}
+
F_B\frac{w}{2}
\]
Therefore:
\[
\tau_{\text{net}}=F_Bw
\]
Substitute the magnetic force equation, using a conductor length \(L\) perpendicular to the field:
\[
\tau_{\text{net}}=BILw
\]
Since the rectangular loop has area \(A=Lw\):
\[
\boxed{\tau_{\text{net}}=BIA}
\]
This is the maximum torque for a single-turn rectangular loop in this arrangement.
For a coil containing \(N\) turns, each turn experiences the same torque, so:
\[
\boxed{\tau=NIAB}
\]
The loop can therefore have:
- Zero resultant magnetic force.
- A nonzero resultant torque.
- Rotational acceleration without translational acceleration.
That is the exact answer to the original puzzle.
## How an experiment demonstrates the difference
The idea can be tested using a rectangular coil placed between the poles of a strong magnet.
Consider the apparatus in the following order:
1. A magnet has two broad pole pieces facing each other, producing a magnetic field across the gap.
2. A rectangular conducting coil sits between the pole pieces, with both vertical sides inside the field.
3. The coil is mounted on a low-friction axle through its centre.
4. Slip-ring contacts connect the rotating coil to a power supply, allowing current to continue flowing as the coil turns.
5. A pointer attached to the axle indicates the coil's angular displacement.
A Hall probe can also be used to check the magnetic flux density at the positions of the two vertical sides.
[IMAGE: Experimental apparatus showing a rectangular conducting coil mounted on a central low-friction vertical axle between the north and south pole pieces of a magnet. Show both vertical sides of the coil inside the uniform field, horizontal magnetic field arrows across the gap, slip-ring electrical contacts connected to a DC power supply, and a pointer or angular scale at the axle. Include two Hall-probe measurement positions beside the opposite vertical sides.]
### What should the experiment show?
Suppose the two vertical sides experience equal and opposite magnetic forces.
A student might predict that the coil cannot move because the forces cancel. That prediction correctly identifies the zero resultant force, but it ignores the positions at which the forces act.
The actual prediction is that the coil experiences a torque and begins rotating.
When the current is switched on, the coil turns. The forces on the opposite sides still cancel as a resultant, but their turning effects reinforce one another.
There is an important experimental detail here. If only one vertical side is inside the strong magnetic field, the forces will not balance. The coil may experience both a resultant force and a torque. That arrangement does not isolate the effect we want to demonstrate.
The improved arrangement places both opposite sides fully inside the same uniform field. A Hall probe can check that the magnetic flux density is approximately equal at both positions.
The coil's rotation demonstrates the torque. The equal-and-opposite force calculation, supported by the measured field uniformity and the symmetry of the coil, establishes why the resultant magnetic force is zero. A fixed axle alone would not prove zero resultant force because its support could also exert a reaction force.
### A small experimental scale estimate
Consider the following illustrative measurements:
| Quantity | Illustrative value |
|---|---:|
| Magnetic flux density, \(B\) | 0.24 T |
| Current, \(I\) | 1.5 A |
| Active vertical length, \(L\) | 0.10 m |
| Separation between vertical sides, \(w\) | 0.080 m |
| Number of turns, \(N\) | 1 |
The magnetic force on each vertical side is:
\[
F_B=BIL
\]
\[
F_B=0.24\times1.5\times0.10
\]
\[
F_B=0.036\text{ N}
\]
The forces are equal in magnitude and opposite in direction, giving:
\[
F_{\text{net}}=0\text{ N}
\]
But the torque is:
\[
\tau=F_Bw
\]
\[
\tau=0.036\times0.080
\]
\[
\boxed{\tau=2.88\times10^{-3}\text{ N m}}
\]
The loop experiences zero resultant magnetic force but a torque of \(2.88\times10^{-3}\text{ N m}\).
The experiment demonstrates something that is easy to miss when drawing force diagrams: **the locations of forces matter when calculating torque**.
---
## Generalising the torque equation
The previous arrangement gave maximum torque because the normal to the loop was perpendicular to the magnetic field.
A loop can be placed at other angles, though. Its torque then changes.
The general equation for the magnitude of the torque on a current-carrying coil in a uniform magnetic field is:
\[
\boxed{\tau=NIAB\sin\alpha}
\]
Where:
- \(\tau\) is the torque in newton metres (N m).
- \(N\) is the number of turns.
- \(I\) is the current in amperes (A).
- \(A\) is the area enclosed by one turn in square metres (\(\text{m}^2\)).
- \(B\) is the magnetic flux density in tesla (T).
- \(\alpha\) is the angle between the normal to the coil's plane and the magnetic field.
Be careful: \(\alpha\) is not the same angle as \(\theta\) in the straight-conductor equation \(F_B=BIL\sin\theta\).
The angle \(\theta\) describes the relationship between the current direction and the magnetic field for an individual conductor segment. The angle \(\alpha\) describes the orientation of the entire coil.
### What happens when the coil rotates?
Consider three possible orientations:
| Angle \(\alpha\) | Torque | Physical interpretation |
|---|---|---|
| \(0^\circ\) | Zero | The coil's normal is parallel to the field. |
| \(90^\circ\) | Maximum | The coil's normal is perpendicular to the field. |
| \(180^\circ\) | Zero | The coil's normal is antiparallel to the field. |
[IMAGE: Three rectangular coil orientations in the same uniform magnetic field, shown side by side. Illustrate the coil normal and magnetic field direction for alpha = 0 degrees, 90 degrees, and 180 degrees. Clearly show that the normal is parallel, perpendicular, and antiparallel to B respectively. Include a compact accompanying graph of torque magnitude divided by maximum torque against alpha from 0 to 180 degrees, following the sine curve.]
At \(90^\circ\), the torque is greatest because the forces produce their maximum combined turning effect.
At \(0^\circ\) and \(180^\circ\), the torque is zero. The forces may still act on the conductor segments, but their lines of action produce no net turning effect about the coil's axis.
For a freely rotating coil, the parallel orientation is a stable equilibrium: a small displacement produces a torque tending to restore the alignment. The antiparallel orientation is an unstable equilibrium: a small displacement produces a torque that moves the coil further away from that alignment.
Notice that the net force remains zero throughout these orientations, provided the field is uniform and the complete loop is considered.
### Worked example: Calculate the torque on a rectangular coil
A rectangular coil has 40 turns and dimensions \(0.10\text{ m}\times0.080\text{ m}\). It carries a current of \(0.50\text{ A}\) in a uniform magnetic field of \(0.25\text{ T}\). The normal to the coil makes an angle of \(35^\circ\) with the magnetic field.
Calculate the magnitude of the torque on the coil.
[IMAGE: Rectangular 40-turn coil in a uniform magnetic field, viewed in perspective. Show the magnetic field B as parallel horizontal arrows, the coil's normal vector at 35 degrees to B, and the rectangular dimensions 0.10 m and 0.080 m. Do not draw force arrows.]
**Step 1**: Identify the relevant quantities.
\[
\begin{aligned}
N&=40\\
I&=0.50\text{ A}\\
B&=0.25\text{ T}\\
A&=0.10\times0.080\\
&=0.0080\text{ m}^2\\
\alpha&=35^\circ
\end{aligned}
\]
**Step 2**: Select the torque equation.
Because the question gives the angle between the coil's normal and the magnetic field, use:
\[
\tau=NIAB\sin\alpha
\]
**Step 3**: Substitute the values.
\[
\begin{aligned}
\tau&=40\times0.50\times0.0080\\
&\quad\times0.25\times\sin35^\circ\\
&=0.040\times0.574
\end{aligned}
\]
**Step 4**: State the result.
\[
\boxed{\tau=0.0230\text{ N m}}
\]
The coil experiences a torque of \(0.0230\text{ N m}\). Its torque is less than the maximum possible torque of \(0.040\text{ N m}\) because its normal is not perpendicular to the magnetic field.
Importantly, the resultant magnetic force on the complete coil is still zero under the uniform-field assumption.
## When does the net force stop being zero?
So far, we have assumed a uniform magnetic field. This assumption is essential to the equal-and-opposite force argument.
Imagine moving one side of the loop towards the edge of the magnet's field. The magnetic flux density may become weaker at that side.
The currents on the two opposite sides still flow in opposite directions, but the magnetic forces no longer have equal magnitudes.
For example:
\[
F_{\text{left}}=B_{\text{left}}IL
\]
\[
F_{\text{right}}=B_{\text{right}}IL
\]
If the two magnetic flux densities differ, the resultant force is:
\[
F_{\text{net}}=
I L(B_{\text{right}}-B_{\text{left}})
\]
Here, the sign indicates the direction relative to the chosen force convention.
The loop can now experience both a net force and a net torque. It may translate as well as rotate.
This is an important distinction:
| Situation | Net force | Net torque |
|---|---|---|
| Complete loop in a uniform field | Zero | May be nonzero |
| Loop in a non-uniform field | May be nonzero | May be nonzero |
| Loop with its normal parallel to a uniform field | Zero | Zero |
| Loop with its normal perpendicular to a uniform field | Zero | Maximum magnitude |
The last two rows refer to a complete loop in a uniform field, not to an isolated wire segment.
For more practice with the force on individual wire sections, see [How to Analyse a Wire Suspended by Magnetic Force](https://tutorgum.com/learn/resources/hsc-physics-how-to-analyse-a-wire-suspended-against-gravity-by-magnetic-force/).
A rectangular wire loop sits in a uniform magnetic field. You switch on the current, and the two vertical sides of the loop experience magnetic forces. One force points into the page, and the other points out of it.
You might expect the forces to cancel, leaving the loop doing nothing. But the loop starts rotating.
How can the net force be zero while the loop still experiences a turning effect?
The answer is that **equal and opposite forces can cancel as a resultant force without cancelling their torque**. The forces act at different positions on the loop, so they form a couple. The loop has no net magnetic force, but it does have a nonzero net torque.
This is the key distinction: net force determines translational acceleration, while net torque determines rotational acceleration.
## Why opposite magnetic forces can still rotate a loop
### Start with one conductor segment
Before considering the whole loop, look at just one straight section of wire inside the magnetic field.
Imagine a vertical conductor placed between the north and south poles of a magnet. Conventional current flows upwards through the wire, while the magnetic field points horizontally from left to right.
The conductor experiences a magnetic force perpendicular to both the current and the magnetic field. Using the right-hand rule, the force points into the page.
[IMAGE: Close-up annotated diagram of a straight vertical current-carrying wire between horizontal north and south magnet pole faces. Show conventional current I vertically upwards, uniform magnetic flux density B horizontally from left to right, and magnetic force F_B directed into the page using a cross symbol. Label the active conductor length L and the angle theta between current and B.]
The magnitude of this magnetic force is:
\[
F_B=BIL\sin\theta
\]
Here is what each symbol represents:
| Symbol | Meaning | SI unit |
|---|---|---|
| \(F_B\) | Magnetic force on the conductor | N |
| \(B\) | Magnetic flux density | T |
| \(I\) | Conventional current | A |
| \(L\) | Length of conductor inside the magnetic field | m |
| \(\theta\) | Angle between the current direction and the magnetic field | Degrees |
There is an important distinction between \(B\) and \(F_B\).
- \(B\) describes the magnetic field at the conductor's position. It is measured in tesla.
- \(F_B\) is the force experienced by the conductor because of its interaction with that field. It is measured in newtons.
The magnetic field is not the force. A stronger magnetic field produces a greater force if the current, conductor length, and angle remain unchanged.
For a conductor perpendicular to the field, \(\theta=90^\circ\), so:
\[
F_B=BIL
\]
For a conductor parallel to the field, \(\theta=0^\circ\), so:
\[
F_B=0
\]
Notice that \(L\) refers only to the section of conductor inside the magnetic field, not necessarily the total length of wire in the circuit.
If you want to practise identifying the correct angle and active conductor length, see [How to Find Magnetic Force on an Angled Wire](https://tutorgum.com/learn/resources/hsc-physics-how-to-find-magnetic-force-on-an-angled-wire-segment/).
### Now consider the complete rectangular loop
Place a rectangular loop in the same uniform magnetic field. The plane of the loop is aligned with the field, so the field runs horizontally through the rectangle.
Conventional current travels upwards along the left vertical side and downwards along the right vertical side. The current directions on these two sides are opposite.
Apply the right-hand rule to each side:
- The left vertical side experiences a force into the page.
- The right vertical side experiences a force out of the page.
The forces have equal magnitudes because both sides have the same length, carry the same current, and experience the same uniform magnetic field.
What happens to the other two sides? They are parallel to the magnetic field, so they experience zero magnetic force.
[IMAGE: Large, clearly annotated rectangular single-turn current loop viewed face-on. Show a uniform magnetic field B directed horizontally to the right across the entire loop. Label the left vertical side with upward current I and a magnetic force into the page (cross symbols); label the right vertical side with downward current I and a magnetic force out of the page (dot symbols). Show the top and bottom conductors parallel to B, the vertical active length L, the horizontal separation w between the two force lines, and a central vertical rotation axis. Keep the two force arrows on opposite sides of the rectangle.]
The two magnetic forces cancel when added as vectors:
\[
\vec F_{\text{net}}=
\vec F_{\text{left}}+\vec F_{\text{right}}=0
\]
There is no resultant magnetic force on the loop.
However, notice where the forces act. They are separated by the horizontal width of the rectangle. The left force pushes one side into the page, while the right force pushes the other side out of the page.
The two sides therefore try to rotate the loop in the same rotational direction.
This arrangement is called a **couple**: two equal, opposite, parallel forces acting along different lines of action.
Think about turning a steering wheel. One hand pushes upwards on one side while the other pushes downwards on the opposite side. The forces can cancel as a resultant, but both contribute to turning the wheel.
The analogy is about the positions and directions of the forces. In the magnetic loop, the forces come from the interaction between current and magnetic field rather than from your hands.
The important point is that cancelling forces do not necessarily cancel their turning effects.
### Why the turning effects add
Torque measures the turning effect of a force about an axis.
For a force acting perpendicular to a lever arm, its torque magnitude is:
\[
\tau=Fd
\]
Here, \(F\) is the force in newtons, and \(d\) is the perpendicular distance from the axis of rotation to the force's line of action, measured in metres.
For the rectangular loop, each force acts at a distance of \(w/2\) from the central rotation axis, where \(w\) is the loop's width.
Each side produces a torque in the same rotational direction:
\[
\tau_{\text{net}}
=
F_B\frac{w}{2}
+
F_B\frac{w}{2}
\]
Therefore:
\[
\tau_{\text{net}}=F_Bw
\]
Substitute the magnetic force equation, using a conductor length \(L\) perpendicular to the field:
\[
\tau_{\text{net}}=BILw
\]
Since the rectangular loop has area \(A=Lw\):
\[
\boxed{\tau_{\text{net}}=BIA}
\]
This is the maximum torque for a single-turn rectangular loop in this arrangement.
For a coil containing \(N\) turns, each turn experiences the same torque, so:
\[
\boxed{\tau=NIAB}
\]
The loop can therefore have:
- Zero resultant magnetic force.
- A nonzero resultant torque.
- Rotational acceleration without translational acceleration.
That is the exact answer to the original puzzle.
## How an experiment demonstrates the difference
The idea can be tested using a rectangular coil placed between the poles of a strong magnet.
Consider the apparatus in the following order:
1. A magnet has two broad pole pieces facing each other, producing a magnetic field across the gap.
2. A rectangular conducting coil sits between the pole pieces, with both vertical sides inside the field.
3. The coil is mounted on a low-friction axle through its centre.
4. Slip-ring contacts connect the rotating coil to a power supply, allowing current to continue flowing as the coil turns.
5. A pointer attached to the axle indicates the coil's angular displacement.
A Hall probe can also be used to check the magnetic flux density at the positions of the two vertical sides.
[IMAGE: Experimental apparatus showing a rectangular conducting coil mounted on a central low-friction vertical axle between the north and south pole pieces of a magnet. Show both vertical sides of the coil inside the uniform field, horizontal magnetic field arrows across the gap, slip-ring electrical contacts connected to a DC power supply, and a pointer or angular scale at the axle. Include two Hall-probe measurement positions beside the opposite vertical sides.]
### What should the experiment show?
Suppose the two vertical sides experience equal and opposite magnetic forces.
A student might predict that the coil cannot move because the forces cancel. That prediction correctly identifies the zero resultant force, but it ignores the positions at which the forces act.
The actual prediction is that the coil experiences a torque and begins rotating.
When the current is switched on, the coil turns. The forces on the opposite sides still cancel as a resultant, but their turning effects reinforce one another.
There is an important experimental detail here. If only one vertical side is inside the strong magnetic field, the forces will not balance. The coil may experience both a resultant force and a torque. That arrangement does not isolate the effect we want to demonstrate.
The improved arrangement places both opposite sides fully inside the same uniform field. A Hall probe can check that the magnetic flux density is approximately equal at both positions.
The coil's rotation demonstrates the torque. The equal-and-opposite force calculation, supported by the measured field uniformity and the symmetry of the coil, establishes why the resultant magnetic force is zero.
A fixed axle alone would not prove zero resultant force because its support could also exert a reaction force. If the experiment needs to measure the resultant magnetic force directly, the support force or translational acceleration must also be measured.
### A small experimental scale estimate
Consider the following illustrative measurements:
| Quantity | Illustrative value |
|---|---:|
| Magnetic flux density, \(B\) | 0.24 T |
| Current, \(I\) | 1.5 A |
| Active vertical length, \(L\) | 0.10 m |
| Separation between vertical sides, \(w\) | 0.080 m |
| Number of turns, \(N\) | 1 |
The magnetic force on each vertical side is:
\[
F_B=BIL
\]
\[
F_B=0.24\times1.5\times0.10
\]
\[
F_B=0.036\text{ N}
\]
The forces are equal in magnitude and opposite in direction, giving:
\[
F_{\text{net}}=0\text{ N}
\]
But the torque is:
\[
\tau=F_Bw
\]
\[
\tau=0.036\times0.080
\]
\[
\boxed{\tau=2.88\times10^{-3}\text{ N m}}
\]
The loop experiences zero resultant magnetic force but a torque of \(2.88\times10^{-3}\text{ N m}\).
The experiment demonstrates something that is easy to miss when drawing force diagrams: **the locations of forces matter when calculating torque**.
---
## Generalising the torque equation
The previous arrangement gave maximum torque because the normal to the loop was perpendicular to the magnetic field.
A loop can be placed at other angles, though. Its torque then changes.
The general equation for the magnitude of the torque on a current-carrying coil in a uniform magnetic field is:
\[
\boxed{\tau=NIAB\sin\alpha}
\]
Where:
- \(\tau\) is the torque in newton metres (N m).
- \(N\) is the number of turns.
- \(I\) is the current in amperes (A).
- \(A\) is the area enclosed by one turn in square metres (\(\text{m}^2\)).
- \(B\) is the magnetic flux density in tesla (T).
- \(\alpha\) is the angle between the normal to the coil's plane and the magnetic field.
Be careful: \(\alpha\) is not the same angle as \(\theta\) in the straight-conductor equation \(F_B=BIL\sin\theta\).
The angle \(\theta\) describes the relationship between the current direction and the magnetic field for an individual conductor segment. The angle \(\alpha\) describes the orientation of the entire coil.
### What happens when the coil rotates?
Consider three possible orientations:
| Angle \(\alpha\) | Torque | Physical interpretation |
|---|---|---|
| \(0^\circ\) | Zero | The coil's normal is parallel to the field. |
| \(90^\circ\) | Maximum | The coil's normal is perpendicular to the field. |
| \(180^\circ\) | Zero | The coil's normal is antiparallel to the field. |
[IMAGE: Three rectangular coil orientations in the same uniform magnetic field, shown side by side. Illustrate the coil normal and magnetic field direction for alpha = 0 degrees, 90 degrees, and 180 degrees. Clearly show that the normal is parallel, perpendicular, and antiparallel to B respectively. Include a compact accompanying graph of torque magnitude divided by maximum torque against alpha from 0 to 180 degrees, following the sine curve.]
At \(90^\circ\), the torque is greatest because the forces produce their maximum combined turning effect.
At \(0^\circ\) and \(180^\circ\), the torque is zero. The forces may still act on the conductor segments, but their lines of action produce no net turning effect about the coil's axis.
For a freely rotating coil, the parallel orientation is a stable equilibrium: a small displacement produces a torque tending to restore the alignment. The antiparallel orientation is an unstable equilibrium: a small displacement produces a torque that moves the coil further away from that alignment.
Notice that the net force remains zero throughout these orientations, provided the field is uniform and the complete loop is considered.
### Worked example: Calculate the torque on a rectangular coil
A rectangular coil has 40 turns and dimensions \(0.10\text{ m}\times0.080\text{ m}\). It carries a current of \(0.50\text{ A}\) in a uniform magnetic field of \(0.25\text{ T}\). The normal to the coil makes an angle of \(35^\circ\) with the magnetic field.
Calculate the magnitude of the torque on the coil.
[IMAGE: Rectangular 40-turn coil in a uniform magnetic field, viewed in perspective. Show the magnetic field B as parallel horizontal arrows, the coil's normal vector at 35 degrees to B, and the rectangular dimensions 0.10 m and 0.080 m. Do not draw force arrows.]
**Step 1**: Identify the relevant quantities.
\[
\begin{aligned}
N&=40\\
I&=0.50\text{ A}\\
B&=0.25\text{ T}\\
A&=0.10\times0.080\\
&=0.0080\text{ m}^2\\
\alpha&=35^\circ
\end{aligned}
\]
**Step 2**: Select the torque equation.
Because the question gives the angle between the coil's normal and the magnetic field, use:
\[
\tau=NIAB\sin\alpha
\]
**Step 3**: Substitute the values.
\[
\begin{aligned}
\tau&=40\times0.50\times0.0080\\
&\quad\times0.25\times\sin35^\circ\\
&=0.040\times0.574
\end{aligned}
\]
**Step 4**: State the result.
\[
\boxed{\tau=0.0230\text{ N m}}
\]
The coil experiences a torque of \(0.0230\text{ N m}\). Its torque is less than the maximum possible torque of \(0.040\text{ N m}\) because its normal is not perpendicular to the magnetic field.
Importantly, the resultant magnetic force on the complete coil is still zero under the uniform-field assumption.
## When does the net force stop being zero?
So far, we have assumed a uniform magnetic field. This assumption is essential to the equal-and-opposite force argument.
Imagine moving one side of the loop towards the edge of the magnet's field. The magnetic flux density may become weaker at that side.
The currents on the two opposite sides still flow in opposite directions, but the magnetic forces no longer have equal magnitudes.
For example:
\[
F_{\text{left}}=B_{\text{left}}IL
\]
\[
F_{\text{right}}=B_{\text{right}}IL
\]
If the two magnetic flux densities differ, the resultant force is:
\[
F_{\text{net}}=
IL(B_{\text{right}}-B_{\text{left}})
\]
Here, the sign indicates the direction relative to the chosen force convention.
The loop can now experience both a net force and a net torque. It may translate as well as rotate.
This is an important distinction:
| Situation | Net force | Net torque |
|---|---|---|
| Complete loop in a uniform field | Zero | May be nonzero |
| Loop in a non-uniform field | May be nonzero | May be nonzero |
| Loop with its normal parallel to a uniform field | Zero | Zero |
| Loop with its normal perpendicular to a uniform field | Zero | Maximum magnitude |
The last two rows refer to a complete loop in a uniform field, not to an isolated wire segment.
## Questions and solutions
### Question 1
A single-turn rectangular loop has a vertical height of \(0.10\text{ m}\) and a horizontal width of \(0.060\text{ m}\). It carries a current of \(2.0\text{ A}\) in a uniform magnetic field of \(0.30\text{ T}\).
The magnetic field lies in the plane of the loop and is perpendicular to the two vertical sides. The top and bottom sections are parallel to the field.
Calculate:
a) The magnetic force on each vertical side.
b) The resultant magnetic force on the complete loop.
c) The magnitude of the torque about the central vertical axis.
[IMAGE: Face-on rectangular single-turn loop with height 0.10 m and width 0.060 m. Show B pointing horizontally right, current upwards on the left side and downwards on the right side, and opposite magnetic force directions using into-page crosses and out-of-page dots. Label the central vertical rotation axis.]
### Solution 1
The loop experiences a magnetic force of \(0.060\text{ N}\) on each vertical side, zero resultant force, and a torque of \(3.6\times10^{-3}\text{ N m}\).
**Part a: Magnetic force on each vertical side**
**Step 1**: Identify the active conductor length.
Only the vertical sections experience a magnetic force because the top and bottom sections are parallel to the magnetic field.
Therefore:
\[
L=0.10\text{ m}
\]
The current is perpendicular to the field, so \(\sin90^\circ=1\).
**Step 2**: Apply the magnetic force equation.
\[
F_B=BIL
\]
Substitute:
\[
F_B=0.30\times2.0\times0.10
\]
\[
\boxed{F_B=0.060\text{ N}}
\]
Each vertical side experiences a force of \(0.060\text{ N}\), but the directions are opposite.
**Part b: Resultant magnetic force**
**Step 3**: Add the forces as vectors.
Taking out of the page as positive:
\[
F_{\text{net}}=+0.060-0.060
\]
\[
\boxed{F_{\text{net}}=0\text{ N}}
\]
The forces cancel because they have equal magnitudes and opposite directions.
**Part c: Torque**
**Step 4**: Calculate the perpendicular separation between the force lines.
The two forces are separated by the width of the rectangle:
\[
d=0.060\text{ m}
\]
**Step 5**: Calculate the torque produced by the couple.
\[
\tau=F_Bd
\]
\[
\tau=0.060\times0.060
\]
\[
\boxed{\tau=3.6\times10^{-3}\text{ N m}}
\]
Both forces produce torque in the same rotational direction, so their turning effects add.
The key result is that zero resultant force does not mean zero torque.
### Question 2
Two rectangular coils, A and B, are placed in identical uniform magnetic fields. Both coils have 20 turns, carry a current of \(0.50\text{ A}\), and have their normals perpendicular to the magnetic field.
The magnetic flux density is \(0.30\text{ T}\).
Their dimensions are:
| Coil | Vertical height | Horizontal width |
|---|---:|---:|
| A | 0.12 m | 0.050 m |
| B | 0.15 m | 0.040 m |
a) Calculate the force on one vertical conductor segment of a single turn for each coil.
b) Calculate the total torque on each coil.
c) A student argues that coil B must experience a greater torque because its vertical sides experience a greater magnetic force. Explain why this conclusion is incorrect.
[IMAGE: Two rectangular coils side by side in identical uniform horizontal magnetic fields. Coil A has dimensions 0.12 m by 0.050 m; coil B has dimensions 0.15 m by 0.040 m. Show equal enclosed areas, opposite current directions on their vertical sides, and the corresponding force pairs. Keep the relative dimensions clear.]
### Solution 2
Both coils experience the same maximum torque of \(0.018\text{ N m}\), even though coil B experiences a greater force on each vertical segment.
**Part a: Force on one vertical segment**
**Step 1**: Apply the magnetic force equation to coil A.
The vertical length of coil A is \(0.12\text{ m}\).
\[
F_A=BIL
\]
\[
F_A=0.30\times0.50\times0.12
\]
\[
\boxed{F_A=0.018\text{ N}}
\]
**Step 2**: Repeat for coil B.
Its vertical length is \(0.15\text{ m}\).
\[
F_B=0.30\times0.50\times0.15
\]
\[
\boxed{F_B=0.0225\text{ N}}
\]
Coil B experiences a greater force because its active conductor length is greater.
**Part b: Total torque**
**Step 3**: Calculate the torque on coil A.
Each turn experiences a couple with force separation equal to the horizontal width.
For \(N\) turns:
\[
\tau_A=NF_Aw_A
\]
\[
\tau_A=20\times0.018\times0.050
\]
\[
\boxed{\tau_A=0.018\text{ N m}}
\]
**Step 4**: Calculate the torque on coil B.
\[
\tau_B=NF_Bw_B
\]
\[
\tau_B=20\times0.0225\times0.040
\]
\[
\boxed{\tau_B=0.018\text{ N m}}
\]
The torques are identical.
**Part c: Explain the result**
The student's mistake is comparing the magnetic forces without considering their lever arms.
Coil B experiences a greater force because its vertical sides are longer. However, its horizontal width is smaller, so the perpendicular distance between the force lines is also smaller.
The two effects compensate.
Both coils have the same enclosed area:
\[
A_A=0.12\times0.050
=0.0060\text{ m}^2
\]
\[
A_B=0.15\times0.040
=0.0060\text{ m}^2
\]
The general torque equation confirms the result:
\[
\tau=NIAB
\]
Since both coils have the same \(N\), \(I\), \(A\), and \(B\), their maximum torques must be equal.
This is an important exam lesson: a greater magnetic force does not automatically mean a greater torque. The perpendicular distance from the axis matters too.
### Question 3
A single-turn rectangular loop has a vertical height of \(0.12\text{ m}\) and a width of \(0.080\text{ m}\). It carries a current of \(1.5\text{ A}\).
The loop is placed in a non-uniform magnetic field. The field points horizontally to the right at both vertical sides, but its magnitude differs:
- Left vertical side: \(B_L=0.20\text{ T}\).
- Right vertical side: \(B_R=0.32\text{ T}\).
The current travels upwards along the left side and downwards along the right side. The top and bottom sections are outside the magnetic field.
Calculate:
a) The magnetic force on each vertical side.
b) The resultant magnetic force, including its direction.
c) The net torque about the centre of the loop.
d) Explain why the loop can now experience both translation and rotation.
[IMAGE: Rectangular loop in a non-uniform magnetic field, with the left vertical side in a weaker field and the right vertical side in a stronger field. Show horizontal field arrows pointing right, with visibly greater arrow density near the right side. Label B_L = 0.20 T and B_R = 0.32 T, current directions on both vertical sides, force directions into and out of the page, vertical length 0.12 m, and horizontal separation 0.080 m.]
### Solution 3
The loop experiences a resultant magnetic force of \(0.0216\text{ N}\) out of the page and a torque of \(3.744\times10^{-3}\text{ N m}\).
Take out of the page as the positive force direction. Take the rotational direction produced by the right-side force as positive torque.
**Part a: Force on each side**
**Step 1**: Calculate the force on the left side.
\[
F_L=B_LIL
\]
\[
F_L=0.20\times1.5\times0.12
\]
\[
\boxed{F_L=0.036\text{ N}}
\]
This force acts into the page.
**Step 2**: Calculate the force on the right side.
\[
F_R=B_RIL
\]
\[
F_R=0.32\times1.5\times0.12
\]
\[
\boxed{F_R=0.0576\text{ N}}
\]
This force acts out of the page.
**Part b: Resultant magnetic force**
**Step 3**: Add the forces using the chosen sign convention.
\[
F_{\text{net}}=F_R-F_L
\]
\[
F_{\text{net}}=0.0576-0.036
\]
\[
\boxed{F_{\text{net}}=0.0216\text{ N}}
\]
The positive result means the resultant force acts out of the page.
The forces no longer cancel because the magnetic field is stronger at the right side.
**Part c: Net torque**
**Step 4**: Calculate the perpendicular distance from the centre to either force line.
\[
d=\frac{0.080}{2}=0.040\text{ m}
\]
**Step 5**: Calculate the torque from each force.
Both forces produce torque in the same rotational direction.
\[
\tau_{\text{net}}=F_Rd+F_Ld
\]
\[
\tau_{\text{net}}=(0.0576+0.036)\times0.040
\]
\[
\boxed{\tau_{\text{net}}=3.744\times10^{-3}\text{ N m}}
\]
Notice that the forces are subtracted when calculating the resultant force, but their torque magnitudes are added.
This is because the forces point in opposite directions but act on opposite sides of the centre. Their turning effects reinforce one another.
**Part d: Explain the motion**
The loop now experiences both a nonzero resultant force and a nonzero resultant torque.
The resultant force produces translational acceleration of the centre of mass. The torque produces rotational acceleration about the centre.
The tempting assumption is that the current returning around a closed loop must always cancel the magnetic force. That is only guaranteed for a complete loop in a uniform magnetic field.
Here, the two sides experience different field strengths, so their magnetic forces are unequal.
### Question 4
A rectangular loop has a vertical height of \(0.12\text{ m}\) and a horizontal width of \(0.10\text{ m}\). It carries a current of \(1.0\text{ A}\).
The loop is positioned at the edge of a magnetic field. Only its right vertical side is inside the field. The other three sides, including the left vertical side, are outside the field and experience negligible magnetic force.
The magnetic flux density inside the field is \(0.40\text{ T}\), perpendicular to the active conductor.
The magnetic force on the active side points out of the page.
a) Calculate the resultant magnetic force on the loop.
b) Calculate the torque about the geometric centre of the loop.
c) Explain why the fact that the wire forms a closed circuit does not guarantee zero resultant magnetic force.
d) If the horizontal width of the loop is doubled while the active vertical length and all other conditions remain unchanged, what happens to the resultant force and torque?
[IMAGE: Rectangular closed conducting loop positioned partly inside the uniform magnetic field between magnet poles. Only the right vertical side lies inside the field; the left vertical side and top and bottom sections are clearly outside it. Show the active side experiencing an out-of-page magnetic force, current circulating around the complete loop, vertical length 0.12 m, horizontal width 0.10 m, and the loop's geometric centre.]
### Solution 4
The loop experiences a resultant force of \(0.048\text{ N}\) out of the page and a torque of \(0.0024\text{ N m}\) about its centre.
**Part a: Resultant magnetic force**
**Step 1**: Identify which section experiences a magnetic force.
Only the right vertical side is inside the magnetic field. The other sections experience negligible magnetic force.
Therefore, the resultant force is simply the force on the active conductor:
\[
F_{\text{net}}=BIL
\]
**Step 2**: Substitute the values.
\[
F_{\text{net}}=0.40\times1.0\times0.12
\]
\[
\boxed{F_{\text{net}}=0.048\text{ N}}
\]
The resultant force acts out of the page.
**Part b: Torque about the centre**
**Step 3**: Calculate the perpendicular distance from the centre to the active force line.
The active side is at the right edge, so:
\[
d=\frac{0.10}{2}=0.050\text{ m}
\]
**Step 4**: Calculate the torque.
\[
\tau=F_{\text{net}}d
\]
\[
\tau=0.048\times0.050
\]
\[
\boxed{\tau=0.0024\text{ N m}}
\]
The magnetic force produces both a resultant force and a turning effect.
**Part c: Explain the closed-circuit misconception**
The current does return around the loop, but the return section is outside the magnetic field.
A current-carrying conductor only experiences the relevant magnetic force where it interacts with the external magnetic field. The return current cannot produce an equal and opposite magnetic force if its conductor section is outside the field.
A closed circuit alone does not guarantee zero resultant magnetic force.
The uniform-field condition matters.
**Part d: Double the width**
The resultant force remains:
\[
F_{\text{net}}=0.048\text{ N}
\]
This is because the magnetic force depends on \(B\), \(I\), and the active vertical length, none of which has changed.
However, the distance from the centre to the active side doubles:
\[
d_{\text{new}}=0.10\text{ m}
\]
Therefore:
\[
\tau_{\text{new}}=0.048\times0.10
\]
\[
\boxed{\tau_{\text{new}}=0.0048\text{ N m}}
\]
The torque doubles, even though the magnetic force stays the same.
This reinforces the distinction between force and torque: changing the position at which a force acts can change its turning effect without changing its magnitude.
### Question 5
A rectangular coil is mounted on a frictionless axle and placed in a uniform magnetic field.
An ideal current supply connects to the coil through slip rings, maintaining a constant current in the same direction relative to the coil as it rotates.
Initially:
- The coil's normal is perpendicular to the magnetic field, so \(\alpha=90^\circ\).
- The coil is stationary.
- The current is switched on.
The coil begins rotating towards the orientation where its normal is parallel to the magnetic field.
At the first instant the coil reaches \(\alpha=0^\circ\), a sensor records zero magnetic torque.
A student concludes:
"The torque is zero, so the coil must stop rotating at this position."
Assume negligible friction and air resistance, and that the external magnetic field remains uniform throughout the motion.
a) Is the student's conclusion correct? Explain what happens to the coil's angular velocity as it reaches \(\alpha=0^\circ\).
b) What happens immediately after the coil passes through this orientation if the current direction remains unchanged?
c) Does the resultant magnetic force become nonzero as the coil passes through the aligned orientation?
d) Compare this situation with a second coil that is placed at \(\alpha=0^\circ\) while stationary and then released.
e) Explain how these situations demonstrate the difference between zero torque and zero angular velocity.
[IMAGE: Three sequential side-view illustrations of a rectangular current loop rotating about a fixed central axle in a uniform magnetic field. Show its initial position with normal perpendicular to B at alpha = 90 degrees, its aligned position with normal parallel to B at alpha = 0 degrees, and its position just after passing the aligned orientation. Include curved arrows showing continuous angular motion, the magnetic torque direction before and after alignment, and a clear indication that the resultant magnetic force remains zero. Do not imply that the coil stops at the aligned position.]
### Solution 5
The student's conclusion is incorrect. The coil passes through the zero-torque orientation because it already has angular velocity. With constant current and negligible damping, the magnetic torque subsequently reverses direction, slowing the coil and eventually causing it to rotate back.
**Part a: What happens at \(\alpha=0^\circ\)?**
The magnetic torque is:
\[
\tau=NIAB\sin\alpha
\]
At the aligned orientation:
\[
\tau=NIAB\sin0^\circ=0
\]
The magnetic torque is zero.
However, the coil has been accelerating throughout its initial rotation. By the time it reaches the aligned position, it has acquired angular velocity.
In fact, under the ideal lossless assumptions, its angular speed is greatest as it passes through this stable equilibrium position.
The mistake is assuming that zero torque means zero angular velocity.
Torque determines angular acceleration, not angular velocity itself.
A zero resultant torque means that angular velocity is not changing at that instant. It does not mean that angular velocity must be zero.
**Part b: What happens after the coil passes alignment?**
Once the coil moves beyond \(\alpha=0^\circ\), the magnetic torque acts in the opposite rotational direction.
It now opposes the coil's existing angular motion.
The coil therefore slows down. Eventually, it reaches a turning point where its angular velocity becomes zero momentarily.
The torque then accelerates it back towards the aligned orientation.
With no friction or other energy losses, the coil oscillates about the stable aligned position.
If friction were present, the oscillations would gradually decrease until the coil settled into alignment.
**Part c: Does the resultant magnetic force become nonzero?**
No. The resultant magnetic force remains zero throughout the motion.
The coil is still completely inside a uniform magnetic field. Its opposite conductor segments experience equal and opposite magnetic forces.
Changing the coil's orientation changes its torque, but it does not create a resultant magnetic force.
This is another important distinction: the force balance and the torque balance are separate calculations.
**Part d: What if the coil starts aligned and stationary?**
The second coil remains stationary.
At \(\alpha=0^\circ\), its magnetic torque is zero. Because it is initially at rest, there is no angular velocity to maintain.
With no external disturbance, it stays in that orientation.
Compare the two situations:
| Situation | Torque at alignment | Angular velocity | Result |
|---|---|---|---|
| Coil arrives at alignment while rotating | Zero | Nonzero | Passes through |
| Coil starts aligned and stationary | Zero | Zero | Remains stationary |
Both coils have the same orientation and experience the same magnetic torque.
Their motion differs because their angular velocities are different.
**Part e: Explain the underlying principle**
The central principle is:
\[
\tau_{\text{net}}=0
\quad\Rightarrow\quad
\text{angular acceleration}=0
\]
It does not follow that:
\[
\omega=0
\]
Here, \(\omega\) represents angular velocity.
This is the rotational equivalent of Newton's first law. An object with zero resultant force can continue moving at constant velocity. Similarly, a rotating object with zero resultant torque can continue rotating at constant angular velocity.
The coil's motion depends on both its current rotational state and the torque acting on it.
This is particularly important when studying electric motors. A motor uses a split-ring commutator to reverse the current at appropriate points during rotation, allowing the magnetic torque to continue acting in the same rotational direction rather than simply pulling the coil back towards alignment.
## Where to go next
The current loop is the starting point for understanding the DC electric motor.
You now know why opposite magnetic forces can produce a turning effect, why the torque depends on coil orientation, and why zero torque does not necessarily mean zero rotational motion.
The next step is to understand how a split-ring commutator reverses the current to maintain continuous rotation.
That brings together the motor effect, magnetic torque, current direction, and rotational motion into one working device.