Why Opposite Sides of a Current Loop Experience Opposite Forces

Place a rectangular current loop between the north and south poles of a magnet. The magnetic field across the gap points from left to right. Now imagine conventional current flowing upwards along the left side of the loop and downwards along the right side.

Both sides are sitting in the same magnetic field. So, would you expect them to experience forces in the same direction?

No. The left side experiences a force into the page, while the right side experiences a force out of the page. The forces are equal in magnitude but opposite in direction, assuming a uniform magnetic field and equal conductor lengths.

The reason is that the magnetic field stays the same, but the current reverses direction as it travels around the loop. That reversal changes the direction of the magnetic force.

Let’s work through why this happens, and then look at the more interesting result: the forces can cancel as a net force while still making the loop rotate.

01Start with one side of the loop

Before considering the entire loop, focus on a single straight conductor.

A current-carrying conductor placed in a magnetic field experiences a magnetic force. This is the motor effect.

The equation for the magnitude of that force is:

\[ F=B I\ell\sin\theta \]

There are two quantities here that students sometimes mix up: \(B\) and \(F\).

They are not the same thing.

The magnetic field \(B\) describes the field at the conductor’s position. The force \(F\) is the effect that the field has on the current-carrying conductor.

Think of \(B\) as describing the magnetic environment and \(F\) as describing the resulting push.

The equation calculates the force’s magnitude. To find its direction, you need the motor-effect direction rule.

For vector calculations, the relationship is:

\[ \vec F=I\vec{\ell}\times\vec B \]

Here, the vector \(\vec{\ell}\) points in the direction of conventional current. The cross product tells us that the force is perpendicular to both the current and the magnetic field.

If you need to revise the force equation or identify the active conductor length, see Force on a Current-Carrying Conductor: HSC Physics Guide.

What happens on each side of the loop?

Return to our rectangular loop.

Assume:

• The magnetic field points horizontally from left to right.
• The field is uniform across the loop.
• The left vertical side carries current upwards.
• The right vertical side carries current downwards.
• Both vertical sides have the same length.
• The top and bottom sections are parallel or antiparallel to the magnetic field.

We can analyse each section separately.

Notice something important.

The magnetic field points in the same direction at both vertical sides. It does not reverse when the current reverses.

Instead, the current direction determines which way the magnetic force acts.

For the left side, the current points upwards. For the right side, it points downwards. Applying the motor-effect direction rule gives opposite force directions.

This is why opposite sides of a current loop experience opposite magnetic forces.

The key is not that the magnetic fields are opposite. They are the same. It is the direction of the current that changes.

02Why the forces can cancel without cancelling the turn

Here’s where things get interesting.

If the two vertical sides experience equal forces in opposite directions, what happens to the loop as a whole?

You might predict that the forces simply cancel, leaving the loop stationary.

That prediction is only partly correct.

The two forces do cancel as a resultant translational force. However, they act at different positions on the loop.

This separation creates a turning effect called a torque.

The two forces form a couple

A couple consists of two equal, opposite, parallel forces acting along different lines.

Imagine pushing the top and bottom edges of a steering wheel in opposite directions. The forces cancel as a net push, but the wheel rotates.

The same principle applies to the current loop.

The force on the left side acts into the page. The force on the right side acts out of the page. Because the sides are separated, the forces produce a turning effect about the loop’s central axis.

For the arrangement where the magnetic field is parallel to the plane of the loop, the force on each active side is:

\[ F=BI\ell \]

This is because the current is perpendicular to the magnetic field, so \(\sin90^\circ=1\).

If the perpendicular separation between the two forces is \(w\), the maximum torque on a single-turn rectangular loop is:

\[ \tau=Fw \]

Substituting the magnetic force:

\[ \tau=BI\ell w \]

Since the area of the rectangular loop is \(A=\ell w\):

\[ \tau=BIA \]

For a coil with \(N\) turns, each turn contributes to the total torque:

\[ \tau_{\max}=NBIA \]

Here:

• \(\tau\) is torque, measured in newton metres (\(\text{N m}\)).
• \(N\) is the number of turns.
• \(B\) is magnetic flux density, in tesla.
• \(I\) is current, in amperes.
• \(A\) is the area enclosed by one turn, in square metres.

So the loop can have zero resultant force but a non-zero torque.

That distinction is central to understanding how electric motors work.

03What happens when the loop changes orientation?

The torque on a current loop depends on its orientation relative to the magnetic field.

To describe this, we introduce another angle, \(\alpha\).

This is the angle between:

• The magnetic field direction.
• The normal to the plane of the loop (an imaginary line perpendicular to the loop).

The general torque equation for a planar coil in a uniform magnetic field is:

\[ \tau=NBIA\sin\alpha \]

Be careful not to confuse this angle with the angle in the straight-conductor force equation.

• In \(F=BI\ell\sin\theta\), \(\theta\) is between the current and the magnetic field.
• In \(\tau=NBIA\sin\alpha\), \(\alpha\) is between the loop’s normal and the magnetic field.

These angles describe different physical relationships.

Case 1: The plane of the loop is parallel to the magnetic field

The normal to the loop is perpendicular to the magnetic field.

Therefore:

\[ \alpha=90^\circ \]

And:

\[ \tau=NBIA \]

The torque is at its maximum.

This is the arrangement we have been using. The two vertical sides experience equal and opposite forces, producing the strongest turning effect.

Case 2: The plane of the loop is perpendicular to the magnetic field

Now the normal to the loop is parallel to the magnetic field.

Therefore:

\[ \alpha=0^\circ \]

And:

\[ \tau=NBIA\sin0^\circ=0 \]

The torque is zero.

But does this mean the magnetic force on every side disappears?

No. In this orientation, all four sides of the loop are perpendicular to the magnetic field, so each side can still experience a magnetic force.

The important difference is that the forces now act along lines passing through the loop’s centre. They produce no net turning effect.

This is a useful distinction for HSC questions:

Zero torque does not necessarily mean zero magnetic force.

It means the forces produce no net turning effect about the chosen axis.

04How a simple experiment tests the explanation

We can test the behaviour of the loop using a straightforward motor-effect demonstration.

The apparatus

Imagine a rectangular conducting coil mounted on a low-friction vertical axle.

The coil sits between two magnet poles:

• The north pole is on the left.
• The south pole is on the right.
• The magnetic field points horizontally across the gap.
• The two vertical sides of the coil sit inside the field.
• The top and bottom sections run parallel to the field.

Connect the coil to a DC power supply through an ammeter and switch. Flexible electrical leads allow the coil to move through a small angle.

A marker attached to the axle allows the initial direction of rotation to be observed.

What should happen?

Before switching on the current, predict the result.

If the current flows upwards on the left side and downwards on the right, the motor-effect rule predicts:

• A force into the page on the left side.
• A force out of the page on the right side.
• A turning effect about the central axle.

When the current is switched on, the coil begins to rotate.

Now reverse the current while keeping the magnetic field unchanged.

The current direction on both vertical sides reverses. Consequently, both magnetic forces reverse, and the direction of the torque changes.

What if you reverse the magnetic field instead?

The torque also reverses.

If you reverse both the current and the magnetic field, the original force directions are restored.

This gives us a useful prediction table.

What does the experiment actually prove?

The rotating coil demonstrates that the magnetic forces produce a turning effect.

However, observing rotation alone does not directly measure the magnitude of the force on each side. It shows the combined effect of the forces.

A more controlled investigation could hold the coil stationary and use calibrated force probes attached to the two active vertical sides. A magnetic field probe could first check that the field is approximately uniform at both positions.

The readings could then be compared.

If both sides have:

• The same active length,
• The same current magnitude,
• The same angle to the field, and
• The same magnetic flux density,

their force magnitudes should be equal.

This also identifies a limitation of the simple model. Near the edges of real magnet poles, the field can become non-uniform. If the two sides experience different values of \(B\), their forces will not necessarily have equal magnitudes.

05Worked examples

The following numerical values are illustrative. Assume uniform magnetic fields unless stated otherwise, and use conventional current directions.

06Questions and solutions

The following questions build from basic force calculations to more demanding reasoning about current direction, non-uniform fields, and torque.

Question 1

A single-turn rectangular loop is placed in a uniform magnetic field of \(0.25\text{ T}\), directed from left to right.

The two vertical sides are each \(0.10\text{ m}\) long. The loop carries a current of \(3.0\text{ A}\), upwards along the left side and downwards along the right side.

The top and bottom sections are parallel and antiparallel to the magnetic field.

Calculate:

(a) The magnitude of the magnetic force on each vertical side.

(b) The direction of the force on each side.

(c) The resultant magnetic force on the loop.

Solution 1

The magnetic force on each vertical side is \(0.075\text{ N}\). The forces act in opposite directions, so the resultant force on the loop is zero.

Part (a): Calculate the force magnitude.

The vertical sides are perpendicular to the magnetic field:

\[ \theta=90^\circ \]

Using:

\[ F=BI\ell\sin\theta \]

Substitute:

\[ \begin{aligned} F&=(0.25)(3.0)(0.10)\sin90^\circ\\ &=0.075\text{ N} \end{aligned} \]

Each side experiences a force of \(0.075\text{ N}\).

Part (b): Determine the directions.

Using the motor-effect direction rule:

• Left side: current upwards, force into the page.
• Right side: current downwards, force out of the page.

The magnetic field is the same at both sides. The opposite force directions result from the opposite current directions.

Part (c): Calculate the resultant force.

The two forces are equal and opposite:

\[ F_{\text{resultant}}=0.075-0.075 \]

\[ F_{\text{resultant}}=0\text{ N} \]

The loop has no resultant translational force from these two sides. However, because the forces act at different positions, they can still produce torque.

Question 2

A rectangular coil has 20 turns. Its two active vertical sides are each \(0.16\text{ m}\) long and are separated by \(0.050\text{ m}\).

The coil carries a current of \(1.8\text{ A}\) in a uniform magnetic field of \(0.35\text{ T}\).

The plane of the coil is parallel to the magnetic field, and the current flows in opposite directions along the two vertical sides.

Calculate:

(a) The total magnetic force acting on each vertical side of the 20-turn coil.

(b) The resultant force on the coil.

(c) The torque acting on the coil.

(d) Explain what happens to the torque if the current is reversed without changing the magnetic field.

Solution 2

The total force on each vertical side is \(2.016\text{ N}\), and the torque is \(0.101\text{ N m}\). Reversing the current reverses the torque direction without changing its magnitude.

Part (a): Calculate the total force on each side.

Each turn experiences a force:

\[ F=BI\ell \]

For 20 turns, the forces add:

\[ F_{\text{side}}=NB I\ell \]

Substitute:

\[ \begin{aligned} F_{\text{side}} &=(20)(0.35)(1.8)(0.16)\\ &=2.016\text{ N} \end{aligned} \]

Therefore, each vertical side experiences a total force of approximately:

\[ F_{\text{side}}=2.02\text{ N} \]

The two forces have opposite directions.

Part (b): Calculate the resultant force.

Since the magnetic field is uniform and both sides have equal lengths, the two forces are equal in magnitude and opposite in direction.

Therefore:

\[ F_{\text{resultant}}=0\text{ N} \]

Part (c): Calculate the torque.

The perpendicular separation between the forces is:

\[ w=0.050\text{ m} \]

Using:

\[ \tau=F_{\text{side}}w \]

Substitute:

\[ \begin{aligned} \tau&=(2.016)(0.050)\\ &=0.1008\text{ N m} \end{aligned} \]

Therefore:

\[ \tau\approx0.101\text{ N m} \]

Part (d): Explain the effect of reversing the current.

Reversing the current reverses the direction of the magnetic force on both vertical sides.

The two forces still have equal magnitudes and opposite directions, but their positions relative to their force directions have changed.

Consequently, the torque reverses direction.

Its magnitude remains \(0.101\text{ N m}\), because neither \(B\), \(I\), \(N\), nor the coil dimensions have changed.

Question 3

Two identical rectangular current loops, A and B, are positioned in identical orientations.

Both loops have the same dimensions and carry the same current magnitude.

For loop A:

• The magnetic field points horizontally to the right.
• Current flows upwards along the left vertical side.

For loop B:

• The magnetic field points horizontally to the left.
• Current flows downwards along the left vertical side.

A student argues:

“Loop B must rotate in the opposite direction because its current has been reversed.”

Is the student correct?

Determine whether the magnetic force directions and torque directions on the two loops are the same or opposite. Explain your reasoning using the relationship between current, magnetic field, and force.

Solution 3

The student is incorrect. Both loops experience the same magnetic force pattern and therefore the same torque direction.

The important detail is that both the current and the magnetic field have been reversed.

For loop A, the force direction is determined by:

\[ \vec F=I\vec{\ell}\times\vec B \]

For loop B, both vectors reverse:

\[ \vec F'=I(-\vec{\ell})\times(-\vec B) \]

Rearranging:

\[ \vec F'=I\vec{\ell}\times\vec B \]

Therefore:

\[ \vec F'=\vec F \]

The magnetic forces on corresponding segments are unchanged.

The tempting mistake is to remember that reversing current reverses magnetic force. That statement is correct only when the magnetic field remains unchanged.

Here, the field also reverses, so the two reversals cancel each other.

Because the force pattern is unchanged and both loops have the same geometry, their torque directions are also the same.

This is a useful reminder that magnetic force depends on the relationship between two directions, not on current direction alone.

Question 4

A rectangular current loop has two vertical conductor sections positioned in separate magnetic field regions.

The left vertical side is \(0.12\text{ m}\) long and carries a current of \(2.0\text{ A}\) upwards.

The right vertical side is also \(0.12\text{ m}\) long and carries the same current downwards.

The magnetic field at each side points horizontally to the right, but its magnitude is different:

• Left side: \(B_L=0.30\text{ T}\).
• Right side: \(B_R=0.50\text{ T}\).

The horizontal connecting sections lie outside the magnetic field regions.

Calculate:

(a) The magnetic force on each vertical side.

(b) The magnitude and direction of the resultant force on the two active sections.

(c) Explain why the forces are not equal and opposite, even though the current magnitudes and conductor lengths are identical.

Solution 4

The left side experiences a force of \(0.072\text{ N}\) into the page, while the right side experiences a force of \(0.120\text{ N}\) out of the page. The resultant force is \(0.048\text{ N}\) out of the page.

Part (a): Calculate the force on each side.

Both vertical sides are perpendicular to their local magnetic fields.

For the left side:

\[ F_L=B_LI\ell \]

\[ \begin{aligned} F_L&=(0.30)(2.0)(0.12)\\ &=0.072\text{ N} \end{aligned} \]

For the right side:

\[ F_R=B_RI\ell \]

\[ \begin{aligned} F_R&=(0.50)(2.0)(0.12)\\ &=0.120\text{ N} \end{aligned} \]

Part (b): Calculate the resultant force.

The left-side force acts into the page, while the right-side force acts out of the page.

Therefore:

\[ \begin{aligned} F_{\text{resultant}} &=F_R-F_L\\ &=0.120-0.072\\ &=0.048\text{ N} \end{aligned} \]

The resultant force is \(0.048\text{ N}\) out of the page.

Part (c): Explain why the forces differ.

The tempting prediction is that the forces must be equal because both sides have the same current magnitude and conductor length.

However, the magnetic field strengths are different.

From:

\[ F=BI\ell\sin\theta \]

the force magnitude is directly proportional to \(B\), provided the other quantities remain constant.

The right side experiences a stronger magnetic field, so it experiences a larger force.

This means the forces are opposite in direction but unequal in magnitude.

The two forces therefore do not cancel as a resultant force.

This is why the equal-and-opposite force result requires an approximately uniform magnetic field across the active sides of the loop.

Question 5

A single-turn rectangular current loop has dimensions:

\[ 0.12\text{ m}\times0.080\text{ m} \]

It carries a steady current of \(2.0\text{ A}\) in a uniform magnetic field of \(0.50\text{ T}\).

The loop is mounted on a frictionless axle through its centre and is free to rotate.

Initially, the plane of the loop is parallel to the magnetic field. The current direction is chosen so that the loop’s magnetic moment tends to align with the field.

The loop is released from rest. Assume weak damping, so that after oscillations have died away, it settles into the stable orientation where its magnetic moment is parallel to the magnetic field.

A student makes the following claim:

“The loop eventually stops because the magnetic force on every conductor segment becomes zero. If any segment still experienced a force, the loop would have to keep rotating.”

Evaluate this claim.

In your answer:

(a) Calculate the magnetic force on each pair of opposite sides in the final orientation.

(b) Determine the resultant force on the loop.

(c) Explain why the final torque is zero even though the individual conductor segments still experience magnetic forces.

(d) Calculate the maximum torque on the loop in its initial orientation.

Solution 5

The student’s claim is incorrect. In the final orientation, every conductor segment experiences a non-zero magnetic force, but the forces produce zero resultant force and zero torque.

This question tests the difference between the force on an individual conductor and the overall turning effect on a loop.

Part (a): Calculate the forces in the final orientation.

In the final orientation, the plane of the loop is perpendicular to the magnetic field.

Therefore, all four conductor segments are perpendicular to the magnetic field.

For each segment:

\[ \theta=90^\circ \]

The force magnitude is:

\[ F=BI\ell \]

For the longer sides, each \(0.12\text{ m}\) long:

\[ \begin{aligned} F_{\text{long}} &=(0.50)(2.0)(0.12)\\ &=0.12\text{ N} \end{aligned} \]

For the shorter sides, each \(0.080\text{ m}\) long:

\[ \begin{aligned} F_{\text{short}} &=(0.50)(2.0)(0.080)\\ &=0.080\text{ N} \end{aligned} \]

Therefore:

• Each longer side experiences a force of \(0.12\text{ N}\).
• Each shorter side experiences a force of \(0.080\text{ N}\).

None of these forces is zero.

Part (b): Determine the resultant force.

Opposite sides of the rectangular loop carry current in opposite directions.

Because the magnetic field is uniform, the forces on each pair of equal-length opposite sides have equal magnitudes and opposite directions.

For the longer sides:

\[ F_{\text{resultant,long}}=0.12-0.12=0 \]

For the shorter sides:

\[ F_{\text{resultant,short}}=0.080-0.080=0 \]

Therefore, the overall resultant force is:

\[ F_{\text{resultant}}=0\text{ N} \]

Part (c): Explain why the torque is zero.

This is the most important part of the question.

Although the individual forces are non-zero, their lines of action now pass through the centre of the loop.

The forces on the longer sides act in opposite directions along the same central line of action when extended. The same is true for the shorter sides.

Consequently, neither pair produces a turning effect about the axle.

We can also verify this using the torque equation:

\[ \tau=NBIA\sin\alpha \]

In the final orientation, the magnetic moment is parallel to the magnetic field.

Therefore:

\[ \alpha=0^\circ \]

Substituting:

\[ \begin{aligned} \tau&=(1)(0.50)(2.0)A\sin0^\circ\\ &=0\text{ N m} \end{aligned} \]

The torque is zero, despite the non-zero magnetic forces on every segment.

Part (d): Calculate the maximum initial torque.

Initially, the plane of the loop is parallel to the magnetic field.

The normal to the loop is therefore perpendicular to the magnetic field:

\[ \alpha=90^\circ \]

This gives maximum torque.

First, calculate the area:

\[ \begin{aligned} A&=(0.12)(0.080)\\ &=0.0096\text{ m}^2 \end{aligned} \]

For a single-turn loop:

\[ \tau_{\max}=BIA \]

Substitute:

\[ \begin{aligned} \tau_{\max} &=(0.50)(2.0)(0.0096)\\ &=0.0096\text{ N m} \end{aligned} \]

Therefore:

\[ \tau_{\max}=9.6\times10^{-3}\text{ N m} \]

The initial torque is non-zero and maximum. As the loop rotates towards alignment, the torque decreases until it reaches zero.

The loop does not stop because the magnetic forces disappear. It settles because there is no longer a net turning effect, while damping removes its rotational energy. Without damping, it would generally continue past the equilibrium position and oscillate.

This is the distinction the student needs to understand: a force can exist without producing a torque about a particular axis.

07Where this leads next

You now have three separate ideas to keep track of:

• The magnetic field \(B\) determines the magnetic environment around the conductor.
• The current direction determines the direction of the magnetic force on each segment.
• The positions of those forces determine the overall torque on the loop.

These ideas lead directly into the operation of a DC electric motor. A split-ring commutator reverses the current every half-turn, allowing the torque to continue acting in the same rotational direction.

For another application of the same motor-effect principle, see Force Between Parallel Current-Carrying Wires: HSC Physics.

When you encounter a current loop in an HSC question, don’t jump straight to the torque equation. First identify the field, trace the current around the loop, and determine the force on each active segment. Once those directions are clear, the overall motion becomes much easier to predict.