Why Only the Wire Inside a Magnetic Field Contributes to Force

A current can flow through an entire 30 cm wire, yet a magnetic balance might register a force caused by only 6 cm of it. Why doesn’t the rest of the wire contribute?

The answer is that magnetic force acts where a current-carrying conductor overlaps an external magnetic field. In the standard HSC model, the field is concentrated between the magnet’s pole pieces, so wire outside that region experiences negligible magnetic force.

This distinction matters when calculating force from \(F=BIL\sin\theta\). You must use the length of wire inside the magnetic field, not the total length of wire in the circuit.

Before going further, make a prediction: if we double the length of wire outside the magnet gap but leave the section inside the gap unchanged, will the magnetic force double?

No. The force stays the same. The extra wire carries current, but it isn’t in the region where the external magnetic field acts.

01What the magnetic force equation actually counts

Imagine a straight wire passing through the gap between two magnets.

The current flows through the entire wire. However, the external magnetic field only occupies a particular region of space. The interaction responsible for the magnetic force occurs where the current-carrying wire and the external field overlap.

A useful comparison is friction. A moving object experiences friction at the surfaces in contact, not along every part of the object. Similarly, the magnetic force on a wire depends on the field acting at the location of each conductor segment.

The analogy has a limit: magnetic fields aren’t physical surfaces, and magnetic forces act through space rather than through contact.

The equation

For a straight conductor in a uniform magnetic field, the magnitude of the magnetic force is:

\[ F=BIL\sin\theta \]

Each quantity has a specific meaning:

The important detail is \(L\).

It represents the actual length of the conductor inside the field, measured along the wire. It does not represent the total length of the circuit.

For example, suppose a 40 cm wire passes through a magnetic field, but only 8 cm sits between the pole pieces.

You use:

\[ L=8.0\text{ cm}=0.080\text{ m} \]

You do not use \(L=0.40\text{ m}\).

What is the difference between B and F?

These two quantities are easy to confuse.

• \(B\) describes the magnetic field at a particular location. It is called magnetic flux density and is measured in tesla.
• \(F\) describes the force experienced by the current-carrying conductor. It is measured in newtons.

The magnetic field doesn’t automatically produce a force on every object nearby. The current-carrying conductor must also have an appropriate orientation relative to that field.

For example, a wire carrying current parallel to the magnetic field experiences zero magnetic force, even when it is completely inside the field.

This is why the equation includes \(\sin\theta\).

Why does the angle matter?

Consider three arrangements:

• At \(90^\circ\), the wire is perpendicular to the magnetic field. The magnetic force is maximum.
• At \(30^\circ\), only part of the wire’s direction is perpendicular to the field. The force is smaller.
• At \(0^\circ\), the wire is parallel to the field. The magnetic force is zero.

Mathematically:

\[ \begin{aligned} F_{\max}&=BIL\sin90^\circ\\ &=BIL \end{aligned} \]

But when the wire is parallel to the field:

\[ F=BIL\sin0^\circ=0 \]

So being inside the magnetic field is necessary for a magnetic force in this model, but it isn’t sufficient. The wire must also have a component of its direction perpendicular to the field.

02How a magnetic balance experiment demonstrates this

A magnetic balance experiment provides a useful way to investigate which parts of a wire contribute to the measured force.

The apparatus consists of:

1. Two magnet pole pieces facing each other across a narrow gap.
2. A straight horizontal wire passing through the gap.
3. A DC power supply connected to the wire through an ammeter.
4. A separate support holding the wire in position.
5. A magnet assembly resting on an electronic balance.

The straight wire extends beyond the magnet gap, and the connecting leads are routed away from the strong magnetic field.

Notice that the wire and magnet are supported independently. The wire doesn’t rest on the magnet.

What does the balance actually measure?

With the current switched off, record the balance reading.

Now switch on the current.

The wire experiences a magnetic force. According to Newton’s third law, the magnet experiences an equal and opposite force.

If the magnetic force on the wire is upwards, the force on the magnet is downwards. The balance therefore registers an increased apparent weight.

The balance doesn’t directly measure the force on the wire. It measures the corresponding change in the force acting on the magnet.

This is an important distinction when interpreting experimental results.

What should happen when the in-field length changes?

Suppose we keep the magnetic field strength and current constant, and gradually increase the length of wire inside the magnetic field.

The equation predicts:

\[ F=BIL \]

assuming the wire remains perpendicular to the field.

Therefore:

\[ F\propto L \]

Doubling the in-field length should double the magnetic force.

However, increasing the length outside the field should make no appreciable difference.

Consider the following illustrative data. These are hypothetical values calculated to demonstrate the expected relationship, not measurements from a recorded experiment.

Assume:

• Magnetic flux density: \(B=0.30\text{ T}\)
• Current: \(I=2.0\text{ A}\)
• Wire perpendicular to the field
• Gravitational acceleration: \(g=9.8\text{ m s}^{-2}\)

The balance changes approximately in proportion to the length inside the field.

Why are the readings given in grams rather than newtons?

An electronic balance normally displays an equivalent mass. To convert the change in reading into force, use:

\[ F=\Delta m g \]

For example, at an in-field length of 6.0 cm:

\[ \begin{aligned} \Delta m&=3.7\text{ g}\\ &=0.0037\text{ kg}\\[4pt] F&=\Delta m g\\ &=0.0037\times9.8\\ &=0.0363\text{ N} \end{aligned} \]

We can use this force to estimate the magnetic flux density:

\[ B=\frac{F}{IL} \]

Substituting:

\[ \begin{aligned} B&=\frac{0.0363}{2.0\times0.060}\\ &=0.3025\text{ T} \end{aligned} \]

Therefore, the estimated magnetic flux density is approximately:

\[ \boxed{B=0.30\text{ T}} \]

The important experimental observation is that the measured force changes with the length of wire inside the field.

Why would the original experimental design need improving?

Imagine starting with a complete rectangular loop of wire placed inside the magnetic field.

It seems reasonable to measure the force on the whole loop. However, there is a problem: several sections of the loop are now inside the field, and each can experience a magnetic force.

Some of these forces may act in opposite directions.

The balance measures the resultant force, not necessarily the force on the particular section we want to investigate.

A better design uses one straight test section inside the uniform field and routes the remaining wire outside the strong-field region.

Reversing the current provides another useful control. The magnetic force reverses direction, while the ordinary gravitational loading of the apparatus remains unchanged.

This makes it easier to distinguish the magnetic effect from a fixed balance offset.

An important limitation: real magnetic fields have edges

The HSC model usually treats the magnetic field as uniform inside the gap and negligible outside it.

A real magnetic field is not so tidy.

Near the edges of the magnet, the field spreads out. This is called a fringe field.

Consequently, a small magnetic force may act on wire just outside the visible pole gap.

The more accurate statement is that each wire segment experiences a magnetic force according to the external field at its location. The force becomes negligible where the external field becomes negligible.

For ordinary HSC calculations, follow the assumptions stated in the question. Unless fringe effects are explicitly included, use the uniform-field approximation.

03Worked examples

04Questions and solutions

Question 1

A straight wire carries a current of \(4.0\text{ A}\). Its total length is \(0.30\text{ m}\), but only \(5.0\text{ cm}\) lies inside a uniform magnetic field of \(0.20\text{ T}\).

The section inside the field is perpendicular to the magnetic field.

Calculate the magnitude of the magnetic force on the wire.

Solution 1

The magnetic force is:

\[ \boxed{F=0.040\text{ N}} \]

Only \(5.0\text{ cm}\) of the wire is inside the magnetic field, so:

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

Since the wire is perpendicular to the field:

\[ \theta=90^\circ \]

Using:

\[ F=BIL\sin\theta \]

Substitute:

\[ \begin{aligned} F&=0.20\times4.0\times0.050\times\sin90^\circ\\ &=0.040\text{ N} \end{aligned} \]

The total wire length of \(0.30\text{ m}\) is not used.

The tempting mistake is to assume that because current flows through the entire wire, the entire wire contributes to the force. However, the external magnetic field is negligible outside the gap.

Question 2

A magnet assembly rests on an electronic balance. A straight conductor is supported independently and passes through the magnetic field between the magnet’s pole pieces.

With the current switched off, the balance reads \(180.0\text{ g}\).

When a current of \(1.8\text{ A}\) is switched on, the balance reads \(182.4\text{ g}\).

The length of conductor inside the field is \(0.065\text{ m}\), and the conductor is perpendicular to the magnetic field.

Assuming the magnetic force is vertical, calculate the magnetic flux density.

Take \(g=9.8\text{ m s}^{-2}\).

Solution 2

The magnetic flux density is approximately:

\[ \boxed{B=0.20\text{ T}} \]

First, determine the change in the balance reading:

\[ \Delta m=182.4-180.0 \]

\[ \Delta m=2.4\text{ g} \]

Convert to kilograms:

\[ \Delta m=0.0024\text{ kg} \]

The balance measures an equivalent mass change, so convert this into force:

\[ F=\Delta m g \]

\[ \begin{aligned} F&=0.0024\times9.8\\ &=0.02352\text{ N} \end{aligned} \]

Now rearrange the magnetic force equation:

\[ F=BIL \]

\[ B=\frac{F}{IL} \]

Substitute:

\[ \begin{aligned} B&=\frac{0.02352}{1.8\times0.065}\\ &=0.2010\text{ T} \end{aligned} \]

Therefore:

\[ \boxed{B=0.20\text{ T}} \]

The increased balance reading means that the magnet experiences an additional downward force. The conductor experiences the equal and opposite upward force.

The key trap is treating the balance reading in grams as though it were already a force in newtons. Always convert the mass change using \(F=\Delta mg\).

Question 3

A student investigates the magnetic force on a straight conductor using a magnetic balance.

The balance readings are:

The conductor has an in-field length of \(0.045\text{ m}\), carries a current of \(2.0\text{ A}\), and is perpendicular to the magnetic field.

Calculate the magnetic flux density.

Explain why the difference between the two current-on readings is not the mass-equivalent of a single magnetic force.

Solution 3

The magnetic flux density is approximately:

\[ \boxed{B=0.29\text{ T}} \]

The crucial detail is that reversing the current reverses the direction of the magnetic force.

The two current-on readings therefore represent opposite magnetic effects.

Calculate the difference between the readings:

\[ 152.7-147.3=5.4\text{ g} \]

However, this difference represents twice the mass-equivalent of the magnetic force.

The actual mass-equivalent of one force is:

\[ \Delta m=\frac{5.4}{2} \]

\[ \Delta m=2.7\text{ g} \]

Convert to kilograms:

\[ \Delta m=0.0027\text{ kg} \]

Calculate the magnetic force:

\[ \begin{aligned} F&=\Delta m g\\ &=0.0027\times9.8\\ &=0.02646\text{ N} \end{aligned} \]

Now calculate the magnetic flux density:

\[ B=\frac{F}{IL} \]

\[ \begin{aligned} B&=\frac{0.02646}{2.0\times0.045}\\ &=0.294\text{ T} \end{aligned} \]

Therefore:

\[ \boxed{B=0.29\text{ T}} \]

Why divide the difference by two?

One current direction produces a positive change of \(2.7\text{ g}\), while reversing the current produces a negative change of \(2.7\text{ g}\).

The separation between these readings is therefore \(5.4\text{ g}\), representing two opposite force effects.

This is also useful experimentally because taking half the difference between reversed-current readings removes a constant balance offset.

Question 4

A bent conductor has two vertical sections inside the same uniform magnetic field.

The magnetic field has a flux density of \(0.40\text{ T}\) and points into the page.

The conductor carries a current of \(2.0\text{ A}\).

• The left vertical section has \(0.10\text{ m}\) inside the field, with current flowing upwards.
• The right vertical section has \(0.060\text{ m}\) inside the field, with current flowing downwards.
• The connecting sections are outside the magnetic field and experience negligible magnetic forces.

Calculate the resultant magnetic force on the conductor, including its direction.

Solution 4

The resultant magnetic force is:

\[ \boxed{0.032\text{ N to the left}} \]

The important point is that the two in-field sections experience forces in opposite directions.

We must calculate each force separately and then find the resultant.

Step 1: Determine the force on the left section.

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

Using the right-hand rule, the magnetic force acts towards the left.

The magnitude is:

\[ \begin{aligned} F_L&=BIL_L\\ &=0.40\times2.0\times0.10\\ &=0.080\text{ N} \end{aligned} \]

Step 2: Determine the force on the right section.

The current flows downwards.

Because the current direction is reversed, the magnetic force also reverses direction.

It acts towards the right.

Its magnitude is:

\[ \begin{aligned} F_R&=BIL_R\\ &=0.40\times2.0\times0.060\\ &=0.048\text{ N} \end{aligned} \]

Step 3: Calculate the resultant force.

The forces act in opposite directions, so subtract their magnitudes:

\[ \begin{aligned} F_{\text{net}}&=F_L-F_R\\ &=0.080-0.048\\ &=0.032\text{ N} \end{aligned} \]

The larger force acts towards the left.

Therefore:

\[ \boxed{F_{\text{net}}=0.032\text{ N left}} \]

This question exposes another important misconception.

It isn’t enough to identify every section of wire inside the magnetic field and add all the force magnitudes.

Magnetic force is a vector. You must account for direction.

Only the in-field sections contribute appreciably here, but their contributions don’t necessarily reinforce each other.

Question 5

A uniform magnetic field occupies a slab of space between two parallel faces. The faces are separated by a perpendicular distance of \(5.0\text{ cm}\).

The magnetic field is parallel to the faces.

A straight conductor carries a constant current of \(3.0\text{ A}\) and passes completely through the slab. The magnetic flux density is \(0.40\text{ T}\).

The conductor is rotated in the plane containing the magnetic field direction and the normal to the slab faces.

Initially, the conductor is perpendicular to the magnetic field, so it lies along the slab normal.

It is then tilted until it makes an angle of \(60^\circ\) with the slab normal.

Assume the conductor is long enough to cross the entire slab in both arrangements, and the field is uniform inside the slab and negligible outside it.

(a) Calculate the length of conductor inside the magnetic field in both arrangements.

(b) Calculate the magnetic force in both arrangements.

(c) A student predicts that tilting the wire to \(60^\circ\) will double the force because twice as much wire is now inside the field. Explain precisely why this prediction is incorrect.

(d) Would the same result apply if the conductor were exactly parallel to the slab faces? Explain why the limiting case requires care.

Solution 5

The magnetic force is the same in both arrangements:

\[ \boxed{F_1=F_2=0.060\text{ N}} \]

Tilting the wire doubles the actual length inside the field, but it also reduces the angle between the wire and the magnetic field.

These two changes exactly compensate for each other.

This is the key conceptual interaction in the question.

(a) Calculate the in-field length in both arrangements.

Initially, the wire is perpendicular to the magnetic field and lies along the normal to the slab faces.

Therefore, its in-field length is simply the slab width:

\[ L_1=5.0\text{ cm}=0.050\text{ m} \]

Now tilt the wire to \(60^\circ\) relative to the slab normal.

The perpendicular distance between the faces remains \(0.050\text{ m}\), but the wire travels a longer distance to cross the slab.

Using basic trigonometry:

\[ \cos60^\circ=\frac{0.050}{L_2} \]

Rearranging:

\[ L_2=\frac{0.050}{\cos60^\circ} \]

\[ \boxed{L_2=0.10\text{ m}} \]

The length inside the field has doubled.

(b) Calculate the magnetic force in both arrangements.

For the first arrangement, the wire is perpendicular to the field:

\[ \theta_1=90^\circ \]

Therefore:

\[ \begin{aligned} F_1&=BIL_1\sin90^\circ\\ &=0.40\times3.0\times0.050\\ &=0.060\text{ N} \end{aligned} \]

For the second arrangement, the wire makes an angle of \(60^\circ\) with the slab normal.

The magnetic field is parallel to the slab faces, so the angle between the wire and the magnetic field is:

\[ \theta_2=90^\circ-60^\circ \]

\[ \theta_2=30^\circ \]

Therefore:

\[ \begin{aligned} F_2&=BIL_2\sin30^\circ\\ &=0.40\times3.0\times0.10\times0.5\\ &=0.060\text{ N} \end{aligned} \]

So:

\[ \boxed{F_1=F_2=0.060\text{ N}} \]

(c) Explain why doubling the in-field length doesn’t double the force.

The tempting prediction is that twice the in-field length must produce twice the force.

That would be correct if the angle between the wire and magnetic field stayed constant.

But it doesn’t.

As the wire tilts:

• The actual length inside the field increases.
• The angle between the wire and magnetic field decreases.
• The sine factor in the force equation becomes smaller.

We can demonstrate this algebraically.

Let:

• \(w\) be the perpendicular distance between the slab faces.
• \(\alpha\) be the angle between the wire and the slab normal.

The in-field length is:

\[ L=\frac{w}{\cos\alpha} \]

Because the magnetic field is parallel to the slab faces, the angle between the wire and the field is:

\[ \theta=90^\circ-\alpha \]

Therefore:

\[ \sin\theta=\cos\alpha \]

Substitute these relationships into the force equation:

\[ F=BIL\sin\theta \]

\[ F=BI\frac{w}{\cos\alpha}\cos\alpha \]

The cosine terms cancel:

\[ \boxed{F=BIw} \]

The force depends on the perpendicular separation of the slab faces, not the angle at which the wire crosses them.

The wire’s additional length contributes less force per metre because it is increasingly aligned with the magnetic field.

This is why the initial prediction fails.

(d) What happens if the wire is exactly parallel to the slab faces?

The constant-force result does not extend to an exactly parallel, finite wire.

As the angle \(\alpha\) approaches \(90^\circ\):

\[ L=\frac{w}{\cos\alpha} \]

The required in-field length approaches infinity.

A finite conductor therefore cannot continue crossing the entire slab at angles arbitrarily close to \(90^\circ\).

At exactly \(90^\circ\), a wire lying parallel to the slab faces is also parallel to the magnetic field.

Consequently:

\[ F=BIL\sin0^\circ=0 \]

There is no contradiction.

The constant-force result describes a wire that continues crossing the entire slab, with its in-field length increasing as its angle changes. At exactly \(90^\circ\), that crossing arrangement is no longer possible for a finite wire.

The important lesson is that equations must be interpreted alongside the physical conditions under which they apply.

05Where this idea leads next

When solving magnetic force problems, identifying the correct in-field length is only the first decision.

You also need to consider the angle between the current and magnetic field, the direction of the force, and whether multiple conductor sections produce forces that reinforce or oppose each other.

For further practice with these connected ideas, see:

• How to Find Magnetic Force on an Angled Wire | HSC Physics – developing the angle and direction components of the force equation.
• How to Find Current from a Magnetic Balance Experiment – using balance readings and magnetic force measurements to determine an unknown current.

The key decision to carry into those problems is simple: identify where the external magnetic field acts, determine the actual conductor length in that region, and only then apply the force equation.