How to Find Net Force on a Rectangular Current Loop

A rectangular loop carries a current while sitting in a magnetic field. Each of its four sides can experience a magnetic force. So should you add the four force magnitudes and conclude that the loop gets pushed?

No. If the entire loop is inside the same uniform magnetic field, the net magnetic force is zero. The forces on opposite sides are equal in magnitude and opposite in direction, so they cancel.

The important skill is knowing when that cancellation applies, calculating the force on individual conductor segments, and recognising what changes when only part of the loop is inside the field.

01The answer: calculate each side, then add the forces as vectors

For a straight current-carrying conductor in a uniform magnetic field, the magnitude of the magnetic force is

\[ F=BIL\sin\theta \]

For a rectangular loop, calculate this force separately for each side. Then add the forces, taking their directions into account.

There are two situations to distinguish:

• The entire loop is inside one uniform magnetic field: opposite sides experience equal and opposite forces, giving a net force of zero.
• Only part of the loop is inside the field, or different parts experience different magnetic fields: the forces might not cancel. You must calculate each contribution and find the resultant.

This distinction is the key to solving questions about the net magnetic force on a rectangular current loop.

02Start with the force on one conductor segment

Before considering the whole loop, focus on just one straight side.

The equation is:

\[ F=BIL\sin\theta \]

Each symbol represents a different physical quantity.

Don’t confuse B with F

This is a surprisingly common mistake.

• \(B\) describes the magnetic field. It is measured in tesla.
• \(F\) describes the force experienced by the conductor. It is measured in newtons.

A magnetic field of \(0.40\text{ T}\) does not mean that the conductor experiences a force of \(0.40\text{ N}\).

You still need the current, conductor length, and angle to calculate the force.

Which length do you use?

The length \(L\) is the actual length of conductor inside the magnetic field.

If a \(20\text{ cm}\) wire has only \(8\text{ cm}\) inside the field, use

\[ L=0.080\text{ m} \]

If the entire side of a rectangular loop is inside the field, use its full length.

A conductor outside the magnetic field does not experience a force from that field.

Which angle do you use?

The angle \(\theta\) is measured between the direction of conventional current and the magnetic field.

It is not necessarily the angle between the wire and the edge of the magnetic-field region.

Remember these three cases:

The equation gives the magnitude. To determine the direction, use the motor-effect direction rule:

\[ \vec F=I\vec L\times\vec B \]

Here, \(\vec L\) points in the conventional current direction.

For example, if current flows to the right and the magnetic field points into the page, the force is upwards.

If the current reverses, the force reverses too.

If you need more practice identifying the correct angle for an individual segment, see How to Find Magnetic Force on an Angled Wire | HSC Physics.

03Why does a complete rectangular loop have zero net force?

Consider a rectangular loop lying flat in the plane of the page.

A uniform magnetic field points into the page, and conventional current flows clockwise.

The loop has:

• Width \(w\), measured horizontally.
• Height \(h\), measured vertically.
• Current \(I\), flowing continuously around the loop.
• Magnetic field \(B\), with the same magnitude and direction everywhere around the loop.

Let’s examine each side.

The top and bottom sides

The top side carries current to the right.

With the magnetic field directed into the page, the force on this side points upwards.

The bottom side carries current to the left. Its magnetic force points downwards.

Both sides:

• Have the same length \(w\).
• Carry the same current \(I\).
• Experience the same magnetic field \(B\).
• Are perpendicular to the field.

Therefore, their force magnitudes are equal:

\[ F_{\text{top}}=F_{\text{bottom}}=BIw \]

But their directions are opposite.

So,

\[ \vec F_{\text{top}}+\vec F_{\text{bottom}}=0 \]

The left and right sides

The left side carries current upwards, so its magnetic force points left.

The right side carries current downwards, so its magnetic force points right.

Both sides have the same length \(h\), current \(I\), and magnetic field \(B\).

Their force magnitudes are therefore

\[ F_{\text{left}}=F_{\text{right}}=BIh \]

Again, their directions are opposite:

\[ \vec F_{\text{left}}+\vec F_{\text{right}}=0 \]

Add all four forces

The net magnetic force is the vector sum:

\[ \begin{aligned} \vec F_{\text{net}}={}& \vec F_{\text{top}}+ \vec F_{\text{bottom}}\\ &+\vec F_{\text{left}}+ \vec F_{\text{right}} \end{aligned} \]

Both pairs cancel, giving

\[ \boxed{\vec F_{\text{net}}=0} \]

This is the result you should expect whenever a complete closed current loop is entirely inside one uniform magnetic field.

There is also a useful mathematical reason why this works for any orientation of the rectangular loop.

The directed lengths of the four sides add to zero because the loop returns to its starting point. When the magnetic field is constant, the force contributions therefore add to zero as well:

\[ \vec F_{\text{net}} =I\left(\sum \vec L_i\right)\times\vec B =0 \]

This confirms that the result does not depend on the loop being perfectly aligned with the page.

Important: zero net force does not always mean zero torque

A loop can have zero net magnetic force and still experience a turning effect.

Equal and opposite forces can act at different positions, forming a couple that rotates the loop.

For the particular arrangement above, where the magnetic field is perpendicular to the loop’s plane, the torque is also zero.

However, if the loop is tilted relative to the magnetic field, a non-zero torque can occur even though the net force remains zero.

This is a separate calculation. You can explore it in Why a Current Loop Can Have Zero Net Force but Nonzero Torque.

04How an experiment can test the cancellation

The mathematical prediction can be tested by comparing a complete rectangular loop inside a uniform field with a loop that is only partly inside it.

Imagine an experimental setup with:

• A rectangular conducting loop mounted on a sensitive force sensor.
• Two large magnetic pole pieces producing a uniform magnetic field across a gap.
• A DC power supply connected to the loop.
• An ammeter in series to monitor the current.
• Connecting leads arranged outside the magnetic-field region.

The loop is positioned centrally between the pole pieces, with the magnetic field perpendicular to its plane.

What should the experiment show?

Start with the complete loop inside the uniform field.

Each side experiences a magnetic force, but opposite sides experience equal and opposite forces.

The predicted resultant force on the loop is zero.

The force sensor should therefore show no change in its measured resultant force when the current is switched on, assuming the loop is held in the stated orientation and other effects are accounted for.

Now change the arrangement.

Move the loop until one vertical side remains inside the field while the opposite vertical side is outside it.

The two horizontal sides still experience forces in opposite directions, provided equal lengths of each are inside the field. These cancel.

However, the remaining vertical side experiences a magnetic force without an opposing force from the other vertical side.

The force sensor should now register a non-zero resultant.

A small measurement estimate

Suppose the remaining vertical side has:

• Magnetic field \(B=0.30\text{ T}\).
• Current \(I=2.0\text{ A}\).
• Length \(L=0.10\text{ m}\).

The current is perpendicular to the magnetic field.

The predicted force is

\[ \begin{aligned} F&=BIL\sin90^\circ\\ &=(0.30)(2.0)(0.10)\\ &=0.060\text{ N} \end{aligned} \]

The partially immersed loop should therefore experience a resultant magnetic force of \(0.060\text{ N}\).

This is a predicted value for the illustrative setup, not a reported experimental measurement.

What could make the original experiment unreliable?

A major problem is that the magnetic field near the edges of ordinary magnets is not perfectly uniform.

If one side experiences a slightly different field from its opposite side, their forces might not cancel exactly.

The connecting leads can also experience magnetic forces if they pass through the field.

A better design uses a sufficiently wide, mapped uniform-field region, keeps the complete loop well inside that region for the first trial, and routes the connecting leads outside it.

The loop can then be moved deliberately towards the field boundary for the contrasting trial.

The important conclusion is that cancellation depends on the magnetic field being the same across the entire loop, not merely on the loop being rectangular.

06When the loop is only partly inside the magnetic field

Now consider a different arrangement.

Suppose the rectangular loop is only partially immersed in a magnetic field.

The field is still uniform wherever it exists, but it does not cover the entire loop.

This changes the calculation.

You can no longer assume that opposite sides experience equal forces.

Instead, use this decision process:

1. Identify which parts of the loop are inside the magnetic field.
2. Determine the actual conductor length inside the field for each side.
3. Calculate the force on each active segment using \(F=BIL\sin\theta\).
4. Determine the direction of each force.
5. Add the forces as vectors.

The most important detail is that the length inside the field may differ between opposite sides.

If the top and bottom sides have equal active lengths, their forces may cancel.

If their active lengths differ, they leave a resultant force.

Likewise, if one vertical side is inside the field while the other is completely outside, there is no opposing vertical-side force.

08Questions and solutions

All numerical values below are illustrative. Assume steady currents, idealised field boundaries, and negligible fringe effects unless stated otherwise.

Question 1

A rectangular loop has a width of \(0.20\text{ m}\) and a height of \(0.12\text{ m}\).

It carries a clockwise current of \(2.0\text{ A}\) and is completely immersed in a uniform magnetic field of \(0.35\text{ T}\), directed into the page.

Calculate:

(a) The magnitude and direction of the force on each horizontal side.

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

(c) The net magnetic force on the loop.

Solution 1

The net magnetic force is zero newtons. Each pair of opposite sides experiences equal forces in opposite directions.

First, calculate the force on the horizontal sides.

Both are perpendicular to the field, so

\[ \begin{aligned} F_{\text{horizontal}} &=BIL\\ &=(0.35)(2.0)(0.20)\\ &=0.14\text{ N} \end{aligned} \]

The top side experiences a force upwards, while the bottom side experiences a force downwards.

Next, calculate the force on the vertical sides:

\[ \begin{aligned} F_{\text{vertical}} &=BIL\\ &=(0.35)(2.0)(0.12)\\ &=0.084\text{ N} \end{aligned} \]

The left side experiences a force to the left, while the right side experiences a force to the right.

Therefore,

\[ \begin{aligned} F_{\text{net}}&=0\\ &=\boxed{0\text{ N}} \end{aligned} \]

The tempting mistake is to add all four magnitudes. However, magnetic force is a vector, so opposite directions must be accounted for.

Question 2

A rectangular loop has a width of \(0.24\text{ m}\) and a height of \(0.12\text{ m}\). It carries a clockwise current of \(3.0\text{ A}\).

A uniform magnetic field of \(0.40\text{ T}\) points into the page and occupies the region to the left of a straight diagonal boundary.

The left vertical side of the loop is completely inside the field. The diagonal boundary intersects:

• The top horizontal side \(0.16\text{ m}\) from its left corner.
• The bottom horizontal side \(0.040\text{ m}\) from its left corner.

The right vertical side is completely outside the field.

Calculate the magnitude and direction of the net magnetic force on the loop.

Solution 2

The net magnetic force is approximately \(0.204\text{ N}\), directed \(45^\circ\) north of west.

The important complication is that the top and bottom sides have different lengths inside the magnetic field. Their forces therefore do not cancel completely.

Step 1: Calculate the force on the left vertical side.

The full vertical side is inside the field.

\[ \begin{aligned} F_{\text{left}} &=BIL\\ &=(0.40)(3.0)(0.12)\\ &=0.144\text{ N} \end{aligned} \]

The current flows upwards, so the force points left.

Step 2: Calculate the force on the top horizontal side.

Only \(0.16\text{ m}\) of the top side is inside the field.

\[ \begin{aligned} F_{\text{top}} &=BIL\\ &=(0.40)(3.0)(0.16)\\ &=0.192\text{ N} \end{aligned} \]

The force points upwards.

Step 3: Calculate the force on the bottom horizontal side.

Only \(0.040\text{ m}\) is inside the field.

\[ \begin{aligned} F_{\text{bottom}} &=BIL\\ &=(0.40)(3.0)(0.040)\\ &=0.048\text{ N} \end{aligned} \]

The force points downwards.

Step 4: Calculate the net vertical force.

The upward force is greater than the downward force.

\[ \begin{aligned} F_y&=0.192-0.048\\ &=0.144\text{ N upwards} \end{aligned} \]

Step 5: Calculate the net horizontal force.

The only horizontal magnetic force comes from the left vertical side.

\[ F_x=0.144\text{ N left} \]

Step 6: Calculate the resultant magnitude.

The horizontal and vertical components are perpendicular.

\[ \begin{aligned} F_{\text{net}} &=\sqrt{F_x^2+F_y^2}\\ &=\sqrt{(0.144)^2+(0.144)^2}\\ &=0.2036\text{ N} \end{aligned} \]

Therefore,

\[ F_{\text{net}}\approx0.204\text{ N} \]

Step 7: Determine the direction.

\[ \begin{aligned} \tan\theta&=\frac{0.144}{0.144}\\ \theta&=45^\circ \end{aligned} \]

The resultant points upwards and to the left, so its direction is \(45^\circ\) north of west.

The tempting approach is to assume that the top and bottom forces cancel because the sides have equal total lengths.

But only the lengths inside the field contribute. The diagonal boundary creates unequal active lengths, leaving an upward resultant.

Question 3

A rectangular loop has a width of \(0.20\text{ m}\) and a height of \(0.12\text{ m}\).

It carries a clockwise current of \(2.5\text{ A}\) and is completely immersed in a uniform magnetic field of \(0.50\text{ T}\), directed horizontally to the right in the plane of the page.

Calculate:

(a) The force on each of the four sides.

(b) The net magnetic force on the loop.

(c) The magnitude of the torque produced by the magnetic forces about the vertical axis through the centre of the loop.

Explain why the loop can experience a turning effect despite having zero net magnetic force.

Solution 3

The net magnetic force is zero, but the loop experiences a torque of \(0.030\text{ N m}\).

This is a useful example because not all four sides experience a force, yet the loop can still rotate.

Step 1: Examine the horizontal sides.

The top side carries current to the right, parallel to the magnetic field.

Therefore,

\[ F_{\text{top}}=BIL\sin0^\circ=0 \]

The bottom side carries current to the left, opposite to the magnetic field.

Therefore,

\[ F_{\text{bottom}}=BIL\sin180^\circ=0 \]

Neither horizontal side experiences a magnetic force.

Step 2: Calculate the force on the vertical sides.

Both vertical sides are perpendicular to the magnetic field.

\[ \begin{aligned} F_{\text{vertical}} &=BIL\sin90^\circ\\ &=(0.50)(2.5)(0.12)\\ &=0.15\text{ N} \end{aligned} \]

The left side experiences a force into the page.

The right side experiences a force out of the page.

Step 3: Find the net force.

The two forces have equal magnitudes and opposite directions.

Therefore,

\[ \boxed{F_{\text{net}}=0\text{ N}} \]

Step 4: Calculate the torque.

The two forces form a couple because they act at different positions on opposite sides of the loop.

Their perpendicular separation is the loop’s width:

\[ d=0.20\text{ m} \]

The torque magnitude is

\[ \begin{aligned} \tau&=Fd\\ &=(0.15)(0.20)\\ &=0.030\text{ N m} \end{aligned} \]

Therefore,

\[ \boxed{\tau=0.030\text{ N m}} \]

The tempting conclusion is that equal and opposite forces mean the loop cannot move.

That is only true for its translational motion. The forces act at different positions, so they can produce rotation.

Question 4

A rectangular loop has a width of \(0.18\text{ m}\) and a height of \(0.10\text{ m}\). It carries a clockwise current of \(2.0\text{ A}\).

The loop is pulled from left to right through a finite rectangular region containing a uniform magnetic field of \(0.40\text{ T}\), directed into the page.

The field region is large enough to contain the entire loop at one stage.

An ideal current supply maintains the clockwise current at \(2.0\text{ A}\) throughout the movement. Ignore any induced-current changes.

Consider five positions:

• A: The entire loop is outside the field, before entering.
• B: The right vertical side is inside the field, while the left vertical side remains outside. Equal lengths of the top and bottom sides are inside.
• C: The entire loop is inside the field.
• D: The left vertical side remains inside the field, while the right vertical side is outside. Equal lengths of the top and bottom sides are inside.
• E: The entire loop is outside the field, after leaving.

Determine the magnitude and direction of the net magnetic force at each position. Explain why the force changes as the loop moves.

Solution 4

The net magnetic forces are:

The important point is that the force depends on which vertical sides are inside the field.

Step 1: Calculate the force on one vertical side.

The vertical side has length \(0.10\text{ m}\).

\[ \begin{aligned} F&=BIL\\ &=(0.40)(2.0)(0.10)\\ &=0.080\text{ N} \end{aligned} \]

Step 2: Analyse position A.

The entire loop is outside the magnetic field.

Therefore, no part of the loop experiences a force from this field:

\[ F_{\text{net}}=0 \]

Step 3: Analyse position B.

The right vertical side is inside the field, while the left vertical side is outside.

The current on the right side flows downwards.

Using the motor-effect direction rule, the force points to the right.

The top and bottom sides have equal active lengths and experience equal opposite forces, so they cancel.

Therefore,

\[ F_{\text{net}}=0.080\text{ N right} \]

Step 4: Analyse position C.

The complete loop is inside the uniform magnetic field.

Both vertical sides experience equal forces in opposite directions. The horizontal forces also cancel.

Therefore,

\[ F_{\text{net}}=0 \]

Step 5: Analyse position D.

The right vertical side is now outside the field, while the left vertical side remains inside.

The current on the left side flows upwards, so its force points to the left.

The horizontal forces again cancel.

Therefore,

\[ F_{\text{net}}=0.080\text{ N left} \]

Step 6: Analyse position E.

The entire loop has left the magnetic-field region.

Therefore,

\[ F_{\text{net}}=0 \]

The key pattern is that a non-zero net force occurs when one vertical side experiences a magnetic force without an equal opposing force on the other vertical side.

The assumption that the current remains clockwise is essential here. If the loop were a passive circuit with an induced current, electromagnetic induction would also need to be considered.

Question 5

A rectangular loop has a width of \(0.20\text{ m}\) and a height of \(0.12\text{ m}\). It carries a clockwise current of \(3.0\text{ A}\).

The loop is positioned across a vertical boundary separating two adjacent magnetic-field regions.

The left half of the loop is inside region A, which contains a uniform magnetic field of \(0.30\text{ T}\) directed into the page.

The right half is inside region B, which contains a uniform magnetic field of \(0.20\text{ T}\) directed out of the page.

The boundary passes through the centre of the loop, so each horizontal side has half its length in each region.

Calculate:

(a) The magnitude and direction of the net magnetic force on the loop.

(b) The net magnetic force if the field in region B were instead directed into the page, with its magnitude unchanged.

(c) Explain why the usual zero-net-force result for a complete loop does not apply automatically to this arrangement.

Assume the current is maintained at a constant value and neglect fringe effects at the boundary.

Solution 5

With the magnetic fields pointing in opposite directions, the net force is \(0.180\text{ N}\) to the left.

If both fields point into the page, the net force becomes \(0.036\text{ N}\) to the left.

The important detail is that the magnetic field is not uniform across the complete loop. Although each region has a uniform field, the field vector changes at the boundary.

Step 1: Calculate the force on the left vertical side.

The left side is entirely inside region A.

Its current flows upwards, and the magnetic field points into the page.

The force is therefore directed leftwards.

\[ \begin{aligned} F_A &=B_AIL\\ &=(0.30)(3.0)(0.12)\\ &=0.108\text{ N} \end{aligned} \]

Step 2: Calculate the force on the right vertical side.

The right side is entirely inside region B.

Its current flows downwards, and the magnetic field points out of the page.

Using the motor-effect direction rule, the force also points leftwards.

\[ \begin{aligned} F_B &=B_BIL\\ &=(0.20)(3.0)(0.12)\\ &=0.072\text{ N} \end{aligned} \]

Notice the important difference: the two forces point in the same direction.

Step 3: Consider the horizontal sides.

The top side experiences an upwards force in region A and a downwards force in region B.

The bottom side experiences a downwards force in region A and an upwards force in region B.

Within each region, the top and bottom contributions are equal in magnitude and opposite in direction because their active lengths are equal.

Therefore, all horizontal contributions cancel.

Step 4: Calculate the net magnetic force.

Both vertical-side forces point left, so they must be added:

\[ \begin{aligned} F_{\text{net}} &=F_A+F_B\\ &=0.108+0.072\\ &=0.180\text{ N} \end{aligned} \]

Therefore,

\[ \boxed{F_{\text{net}}=0.180\text{ N left}} \]

Step 5: Consider what happens if both fields point into the page.

The left vertical side still experiences a force of \(0.108\text{ N}\) to the left.

However, the current on the right vertical side flows downwards, so its force now points to the right.

The magnitude remains \(0.072\text{ N}\).

The resultant is therefore

\[ \begin{aligned} F_{\text{net}} &=0.108-0.072\\ &=0.036\text{ N} \end{aligned} \]

Hence,

\[ \boxed{F_{\text{net}}=0.036\text{ N left}} \]

The horizontal forces still cancel because the top and bottom have equal active lengths in each region.

Why is this result different from the usual complete-loop calculation?

The standard zero-net-force result assumes that the entire loop experiences the same uniform magnetic field.

In this question, the field reverses direction across the loop.

Consequently, the magnetic forces on the two vertical sides point in the same direction rather than opposing directions.

The tempting approach is to see a complete rectangular loop and immediately write \(F_{\text{net}}=0\).

But the shape of the conductor is not enough to guarantee cancellation. The magnetic field must also be uniform across the complete loop.

09Where this calculation leads next

When solving any rectangular-loop question, make your first decision about the magnetic-field region.

If the entire loop is inside one uniform field, the net magnetic force is zero. If the loop is only partly immersed, or the field changes across it, calculate the forces on the individual active segments and add them as vectors.

Once you can do this reliably, the next useful step is analysing the torque produced by those forces. That is how you move from calculating the resultant force on a loop to understanding how a current-carrying coil rotates in an electric motor.

Syllabus note

This explanation follows the Motor Effect conductor-force content of the 2017 Physics Stage 6 syllabus, which remains the Year 12 syllabus for the 2026 HSC.

The 2025 Physics 11-12 syllabus begins implementation for Year 11 in 2027, with the first HSC examination under the new syllabus scheduled for 2028. See NESA’s Physics Stage 6 syllabus information for the implementation timeline.