Where Do Magnetic Fields from Two Parallel Wires Cancel?

For two long, straight, parallel wires, their magnetic fields can cancel along a line parallel to the wires. The position of this zero-field line depends on the directions and magnitudes of the conventional currents.

There are two important cases:

- **Currents in the same direction:** The fields oppose each other between the wires. They cancel at a point closer to the wire carrying the smaller current.
- **Currents in opposite directions:** The fields reinforce each other between the wires. If the currents have different magnitudes, cancellation occurs outside the pair, on the side of the wire carrying the smaller current. If the currents have equal magnitudes, there is no finite cancellation point.

The important distinction is that magnetic field direction and magnetic force direction are different things. Two wires can attract each other even though their magnetic fields oppose each other between them.

Let's work out exactly where cancellation occurs, why the location changes, and how an experiment could test the prediction.

## Where do two magnetic fields cancel?

Start with two parallel wires separated by a distance \(d\). Each carries a steady conventional current, and each produces a circular magnetic field around itself.

At any point, the resultant magnetic field is the vector sum:

\[
\vec B_{\text{net}}=\vec B_1+\vec B_2
\]

For the resultant to be zero, the two individual fields must have:

1. Equal magnitudes.
2. Opposite directions.

For a long, straight wire, the magnetic field at perpendicular distance \(r\) is:

\[
B=\frac{\mu_0 I}{2\pi r}
\]

Here, \(B\) is magnetic flux density in tesla (T), \(I\) is conventional current in amperes (A), \(r\) is perpendicular distance in metres (m), and \(\mu_0\) is the permeability of free space.

This gives us the key relationship:

\[
B\propto\frac{I}{r}
\]

A larger current produces a stronger field. Moving further from the wire produces a weaker field.

At a cancellation point, the two fields must therefore satisfy:

\[
\frac{I_1}{r_1}=\frac{I_2}{r_2}
\]

But this equation is only half the job. We must first establish whether the fields actually point in opposite directions at the location we're considering.

[IMAGE: Cross-sectional comparison of two long parallel wires carrying conventional currents in the same direction. Show circular magnetic field lines around both wires, opposite tangent field arrows at points between them, and a marked zero-field position between the wires. Include current symbols, wire separation d, and distances r1 and r2. Do not show a cancellation point outside the pair.]

Assume the wires are long, straight, parallel conductors carrying steady currents. We will ignore end effects and other sources of magnetic fields when calculating the ideal cancellation position.

The current magnitudes \(I_1\) and \(I_2\) are positive quantities. Their directions are specified separately.

### First, get the field directions right

Use the right-hand grip rule:

1. Point your right thumb in the direction of conventional current.
2. Curl your fingers around the wire.
3. Your fingers show the direction of the magnetic field.

Remember that conventional current is opposite to electron flow in a metal wire. Use conventional current for the right-hand grip rule, even if a question describes electrons moving through the conductor.

Two symbols are particularly useful in cross-sectional diagrams:

| Symbol | Meaning |
|---|---|
| \(\odot\) | Current out of the page, towards you |
| \(\otimes\) | Current into the page, away from you |

For a refresher on the circular field pattern and the right-hand grip rule, see [Magnetic Field Around a Current-Carrying Wire: HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-magnetic-field-around-a-current-carrying-wire/).

Now consider the two possible current arrangements.

### Case 1: Currents flow in the same direction

Suppose both currents flow out of the page.

At a point between the wires:

- The left wire produces an upwards magnetic field.
- The right wire produces a downwards magnetic field.

The fields oppose each other. This makes cancellation possible between the wires.

Outside the pair, the fields point in the same direction, so they reinforce each other.

[IMAGE: Cross-sectional comparison showing two parallel wires with currents both out of the page. Include circular field lines, arrows pointing upwards and downwards between the wires, and arrows pointing in the same direction outside the pair. Clearly distinguish the left and right wire contributions.]

### Case 2: Currents flow in opposite directions

Now reverse the current in the right-hand wire.

At a point between the wires, both magnetic fields point upwards. They reinforce each other.

There cannot be a cancellation point between the wires, regardless of how large or small the currents are.

However, outside the pair, the fields point in opposite directions. Cancellation is possible there if the current magnitudes are unequal.

The cancellation position is on the outside of the wire carrying the smaller current.

[IMAGE: Cross-sectional comparison showing the left wire carrying current out of the page and the right wire carrying current into the page. Show reinforcing field arrows between the wires and opposing field arrows outside. Mark the possible cancellation region on the outside of the weaker-current wire, without specifying its exact position.]

The pattern is worth keeping in a table:

| Current directions | Between the wires | Outside the wires | Possible cancellation |
|---|---|---|---|
| Same direction | Fields oppose | Fields reinforce | Between |
| Opposite directions | Fields reinforce | Fields oppose | Outside, on the weaker-current side |

There is an important detail in the second row. Although the fields oppose on both outer sides, a zero-field position exists on only one side: the side of the weaker-current wire.

## Don't confuse magnetic field direction with magnetic force

Here is a particularly tempting mistake.

Two parallel wires carrying currents in the same direction attract each other. A student might conclude that their magnetic fields must therefore point towards each other, or reinforce each other between the wires.

Neither conclusion is correct.

Magnetic field direction and magnetic force direction describe different things.

| What are you finding? | Rule | What it tells you |
|---|---|---|
| Magnetic field around a wire | Right-hand grip rule | Direction of the magnetic field |
| Force on a current-carrying wire | Motor-effect rule | Direction of the magnetic force |
| Force between parallel wires | Parallel-wire force relationship | Same-direction currents attract; opposite-direction currents repel |

The force on one wire is caused by the magnetic field produced by the other wire at its position.

It does not depend on whether the two fields cancel at some separate point between the wires.

[IMAGE: Two side-by-side cross-sectional illustrations. In the first, two wires carrying currents out of the page have opposing magnetic field arrows at the midpoint but attractive force arrows on the wires. In the second, opposite currents have reinforcing field arrows at the midpoint and repulsive force arrows on the wires. Clearly distinguish magnetic field arrows from force arrows.]

This distinction is useful in HSC questions because a question can ask about the field at a point, the force on a wire, or both. Identify the quantity before choosing a direction rule.

## Finding the exact cancellation position

Now that we know where cancellation is possible, we can calculate its position.

### Same-direction currents: cancellation between the wires

Let:

- \(d\) be the separation between the wires, in metres.
- \(r_1\) be the distance from wire 1 to the cancellation point, in metres.
- \(r_2\) be the distance from wire 2 to the cancellation point, in metres.
- \(I_1\) and \(I_2\) be the current magnitudes, in amperes.

Because the point lies between the wires:

\[
r_1+r_2=d
\]

For the fields to cancel, their magnitudes must be equal:

\[
B_1=B_2
\]

Using the long-wire equation:

\[
\frac{\mu_0 I_1}{2\pi r_1}
=
\frac{\mu_0 I_2}{2\pi r_2}
\]

Cancel the common constants:

\[
\frac{I_1}{r_1}=\frac{I_2}{r_2}
\]

Combining this with \(r_1+r_2=d\) gives:

\[
\boxed{r_1=\frac{I_1d}{I_1+I_2}}
\]

and

\[
\boxed{r_2=\frac{I_2d}{I_1+I_2}}
\]

These equations give the exact cancellation position for two ideal long parallel wires carrying currents in the same direction.

Notice the relationship:

\[
\frac{r_1}{r_2}=\frac{I_1}{I_2}
\]

The cancellation point is closer to the wire with the smaller current.

If the currents are equal, the cancellation point is halfway between the wires.

### Worked example: Two currents in the same direction

Two long, straight, parallel wires are separated by \(18\ \text{cm}\). Wire A carries \(3.0\ \text{A}\) out of the page, and wire B carries \(6.0\ \text{A}\) out of the page.

Calculate the position where their magnetic fields cancel.

[IMAGE: Question setup showing two parallel wires A and B, 18 cm apart, both carrying conventional current out of the page. Label the currents 3.0 A and 6.0 A. Do not show the cancellation position or magnetic field arrows.]

**Step 1**: Identify where cancellation is possible.

Both currents flow in the same direction, so their magnetic fields oppose each other between the wires.

The cancellation point must therefore lie between A and B.

**Step 2**: Write the equal-field condition.

At the cancellation point:

\[
\frac{I_A}{r_A}=\frac{I_B}{r_B}
\]

We also know:

\[
r_A+r_B=0.18
\]

**Step 3**: Substitute the current values.

\[
\frac{3.0}{r_A}=\frac{6.0}{r_B}
\]

Rearranging:

\[
r_B=2r_A
\]

Substitute into the separation equation:

\[
r_A+2r_A=0.18
\]

\[
3r_A=0.18
\]

\[
r_A=0.060\ \text{m}
\]

Therefore:

\[
r_B=0.120\ \text{m}
\]

**Step 4**: State the position.

The magnetic fields cancel \(6.0\ \text{cm}\) from wire A and \(12.0\ \text{cm}\) from wire B.

The point is closer to wire A because it carries the smaller current.

**Step 5**: Check that the fields have equal magnitudes.

Using \(B=2.0\times10^{-7}I/r\):

\[
\begin{aligned}
B_A&=2.0\times10^{-7}\frac{3.0}{0.060}\\
&=1.0\times10^{-5}\ \text{T}
\end{aligned}
\]

For wire B:

\[
\begin{aligned}
B_B&=2.0\times10^{-7}\frac{6.0}{0.120}\\
&=1.0\times10^{-5}\ \text{T}
\end{aligned}
\]

The magnitudes are equal, and their directions are opposite. Therefore, the resultant magnetic field is zero.

[IMAGE: Solution diagram for the 3.0 A and 6.0 A wires, showing the cancellation position 6.0 cm from A and 12.0 cm from B. Include equal and opposite magnetic field arrows of 10 microtesla at the cancellation position.]

### Opposite-direction currents: cancellation outside the wires

The calculation changes because the cancellation point is no longer between the wires.

Suppose the weaker-current wire is on the left. The zero-field point will be somewhere to its left.

Let:

- \(I_w\) be the smaller current magnitude.
- \(I_s\) be the larger current magnitude.
- \(d\) be the wire separation.
- \(x\) be the distance from the weaker-current wire to the cancellation point, measured outwards.

The distance from the stronger-current wire is \(d+x\).

For cancellation:

\[
\frac{I_w}{x}=\frac{I_s}{d+x}
\]

Cross-multiplying:

\[
I_w(d+x)=I_sx
\]

Expanding and rearranging:

\[
I_wd=I_sx-I_wx
\]

Therefore:

\[
\boxed{x=\frac{I_wd}{I_s-I_w}}
\]

This formula applies when the currents flow in opposite directions and their magnitudes are unequal.

If the currents have equal magnitudes, the denominator becomes zero. There is no finite cancellation position.

Why? On either outside region, the nearer wire produces the stronger field. Between the wires, the fields reinforce each other.

### Worked example: Finding an external cancellation point

Two parallel wires are separated by \(20\ \text{cm}\).

Wire A carries \(4.0\ \text{A}\) out of the page, while wire B carries \(6.0\ \text{A}\) into the page.

Find the position where their magnetic fields cancel, considering the entire straight line containing the wires.

[IMAGE: Question setup showing wire A on the left carrying 4.0 A out of the page and wire B on the right carrying 6.0 A into the page. Label the separation as 20 cm. Leave the external cancellation position unmarked.]

**Step 1**: Identify the possible cancellation region.

The currents flow in opposite directions.

Their magnetic fields reinforce each other between the wires, so cancellation cannot occur in the gap.

Outside the pair, the fields oppose each other.

Since wire A carries the smaller current, the cancellation position must be to the left of wire A.

**Step 2**: Identify the current magnitudes.

\[
I_w=4.0\ \text{A}
\]

\[
I_s=6.0\ \text{A}
\]

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

**Step 3**: Calculate the external distance.

Use:

\[
x=\frac{I_wd}{I_s-I_w}
\]

Substitute:

\[
\begin{aligned}
x&=\frac{4.0\times0.20}{6.0-4.0}\\
&=\frac{0.80}{2.0}\\
&=0.40\ \text{m}
\end{aligned}
\]

**Step 4**: State the position.

The magnetic fields cancel \(0.40\ \text{m}\), or \(40\ \text{cm}\), to the left of wire A.

The distance from this point to wire B is:

\[
r_B=0.40+0.20=0.60\ \text{m}
\]

**Step 5**: Verify the result.

For wire A:

\[
\begin{aligned}
B_A&=2.0\times10^{-7}\frac{4.0}{0.40}\\
&=2.0\times10^{-6}\ \text{T}
\end{aligned}
\]

For wire B:

\[
\begin{aligned}
B_B&=2.0\times10^{-7}\frac{6.0}{0.60}\\
&=2.0\times10^{-6}\ \text{T}
\end{aligned}
\]

The fields have equal magnitudes and opposite directions, confirming the cancellation position.

[IMAGE: Solution diagram showing the two opposite-current wires, with the zero-field position 40 cm to the left of wire A. Label the distances from the zero-field point as 40 cm to A and 60 cm to B. Show equal and opposite field arrows at the zero-field point.]

## Why can't the fields cancel anywhere else?

There is one more geometrical detail worth understanding.

So far, we have considered points along the straight line joining the two wires. Could a zero-field point exist somewhere above or below that line?

For two ideal parallel wires carrying nonzero currents, the answer is no.

Each wire produces circular magnetic field lines around itself. At any point, the magnetic field is tangent to the circular field line.

For two fields to cancel, their arrows must point in exactly opposite directions.

That means the two tangent directions must be parallel. Since each tangent is perpendicular to the radius from its wire to the observation point, the two radii must also lie along the same straight line.

This is only possible when the observation point lies on the straight line joining the two wire positions.

[IMAGE: Cross-sectional geometry showing two parallel wires A and B and an observation point P above the line AB. Draw circular field lines around both wires and tangent magnetic field arrows at P. Show that the two field vectors are not parallel or antiparallel. In a separate panel, show a point on line AB where the two tangent field arrows can be opposite. Do not mark a numerical cancellation position.]

This gives us a more precise result:

- In a cross-sectional view, there is at most one cancellation position along the line joining the wires.
- In three dimensions, for ideal infinitely long parallel wires, that position extends into a straight line parallel to both wires.
- Away from that line, the two field vectors cannot cancel exactly.

Equal field magnitudes are not enough. Their directions must also be opposite.

This is the reason that simply solving \(I_1/r_1=I_2/r_2\) at an arbitrary off-axis point is not a valid cancellation calculation.

## Can an experiment locate the cancellation position?

The mathematical prediction is useful, but how could we test it?

Consider an investigation using two long parallel conductors and a Hall probe.

### The apparatus

Imagine two straight conductors mounted vertically, parallel to each other, and separated by \(16\ \text{cm}\).

Connect each wire to a separate low-voltage, current-limited power supply. Place an ammeter in series with each wire so the current can be checked.

Set the currents to:

- Wire A: \(3.0\ \text{A}\) upwards.
- Wire B: \(7.0\ \text{A}\) upwards.

The conventional currents therefore flow in the same direction.

Place a Hall probe on a track that allows it to move horizontally between the wires, at approximately the middle of their length. Its sensitive axis should be aligned with the expected magnetic field direction.

Keep the return leads well away from the measurement region so their magnetic fields do not significantly affect the readings.

[IMAGE: Physical experimental arrangement with two long parallel vertical conductors separated by 16 cm. Show separate current-limited supplies and series ammeters, both conventional currents flowing upwards, a movable Hall probe on a horizontal track between the wires at mid-height, and return leads routed away from the measurement region.]

### What should the experiment show?

Before switching on the currents, measure the background magnetic field.

Then switch on both currents and move the Hall probe along the line joining the wires.

We can predict the cancellation position before collecting any data.

For the same-direction currents:

\[
r_A=\frac{I_Ad}{I_A+I_B}
\]

Substitute:

\[
\begin{aligned}
r_A&=\frac{3.0\times0.16}{3.0+7.0}\\
&=0.048\ \text{m}
\end{aligned}
\]

Therefore, the predicted cancellation position is:

- \(4.8\ \text{cm}\) from wire A.
- \(11.2\ \text{cm}\) from wire B.

At this position, each wire produces:

\[
\begin{aligned}
B_A&=2.0\times10^{-7}\frac{3.0}{0.048}\\
&=1.25\times10^{-5}\ \text{T}
\end{aligned}
\]

The field from wire B has the same magnitude:

\[
B_B=1.25\times10^{-5}\ \text{T}
\]

The fields point in opposite directions, so their resultant is zero.

### What should the measurements look like?

The following table contains illustrative model predictions, not actual experimental measurements.

| Position from A | Field from A | Field from B | Resultant field |
|---|---:|---:|---:|
| 4.0 cm | 15.0 µT up | 11.7 µT down | 3.3 µT up |
| 4.8 cm | 12.5 µT up | 12.5 µT down | 0 µT |
| 6.0 cm | 10.0 µT up | 14.0 µT down | 4.0 µT down |

Notice the change in direction.

Before the cancellation position, the field from wire A is stronger. After the cancellation position, the field from wire B is stronger.

The signed field reading should therefore pass through zero as the probe moves across the predicted position.

### Why isn't a compass enough?

A compass can indicate the direction of the resultant magnetic field, but it is not a precise instrument for measuring magnetic field strength.

There is another complication: Earth has its own magnetic field.

Even if the two wire fields cancel perfectly, a Hall probe can still register Earth's magnetic field. A raw reading of zero is therefore not necessarily what we should expect.

A better investigation uses a calibrated three-axis Hall probe and accounts for the background field.

One useful method is to measure the field with both currents flowing in their original directions, then reverse both currents while keeping their magnitudes unchanged.

The wire-generated field reverses, but Earth's background field does not.

If the two readings are \(\vec B_+\) and \(\vec B_-\), respectively, the field produced by the test wires can be isolated using:

\[
\boxed{
\vec B_{\text{wires}}
=
\frac{\vec B_+-\vec B_-}{2}
}
\]

At the predicted cancellation position, this difference should be approximately zero.

The revised design also needs to control:

- The current magnitudes, checked using the ammeters.
- The Hall probe's position and orientation.
- Magnetic fields from the return leads.
- End effects, by measuring near the middle of sufficiently long wires.

This is a proposed investigation and a prediction from the long-wire model, not a report of measured experimental results.

The important experimental idea is that cancellation concerns the resultant field produced by the two wires, not necessarily the total field detected by an instrument.

## Questions and solutions

For all five questions, assume long, straight, parallel wires carrying steady currents in air. Take \(\mu_0=4\pi\times10^{-7}\ \text{T\,m\,A}^{-1}\), and ignore end effects unless stated otherwise.

### Question 1

Two long, parallel wires are separated by \(12\ \text{cm}\).

Each wire carries a conventional current of \(4.0\ \text{A}\) out of the page.

Determine where their magnetic fields cancel.

If the separation is doubled while both currents remain unchanged, where will the new cancellation position be?

[IMAGE: Question setup showing two parallel wires 12 cm apart, both carrying 4.0 A out of the page. Include the wire separation and current symbols, but do not mark the cancellation position.]

### Solution 1

The magnetic fields cancel halfway between the wires, \(6.0\ \text{cm}\) from each wire. If the separation doubles, the cancellation position moves to the new midpoint, \(12.0\ \text{cm}\) from each wire.

**Step 1**: Identify the field directions.

Both currents flow in the same direction, so their magnetic fields oppose each other between the wires.

**Step 2**: Compare the field magnitudes.

The currents are equal, so the two fields have equal magnitudes when the distances from the wires are equal.

Therefore:

\[
r_1=r_2=\frac{12}{2}=6.0\ \text{cm}
\]

**Step 3**: Consider doubling the separation.

The new separation is:

\[
d=24\ \text{cm}
\]

The currents are still equal, so the cancellation position remains halfway between the wires:

\[
r_1=r_2=\frac{24}{2}=12.0\ \text{cm}
\]

The cancellation position moves further apart, but the magnetic fields still cancel at the midpoint.

[IMAGE: Solution diagram showing equal currents in the two wires and opposing field arrows at the midpoint, first for a 12 cm separation and then for a 24 cm separation. Mark the midpoint distances of 6 cm and 12 cm respectively.]

### Question 2

Two parallel wires, A and B, are separated by \(24\ \text{cm}\).

Both carry conventional currents out of the page.

Wire A carries \(4.0\ \text{A}\). A Hall probe identifies a zero-field position \(8.0\ \text{cm}\) from wire A.

Determine:

(a) The current in wire B.

(b) The magnitude of the magnetic field produced individually by each wire at the zero-field position.

[IMAGE: Question setup showing parallel wires A and B separated by 24 cm, both currents out of the page, with A carrying 4.0 A. Mark the measured zero-field position P at 8.0 cm from A. Do not label the unknown current or individual field magnitudes.]

### Solution 2

Wire B carries \(8.0\ \text{A}\), and each wire produces a magnetic field of \(1.0\times10^{-5}\ \text{T}\) at P. The fields have equal magnitudes and opposite directions.

**Step 1**: Find the distance from P to wire B.

The wires are \(24\ \text{cm}\) apart, and P is \(8.0\ \text{cm}\) from A.

Therefore:

\[
r_B=24-8.0=16.0\ \text{cm}
\]

Convert both distances to metres:

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

\[
r_B=0.160\ \text{m}
\]

**Step 2**: Apply the cancellation condition.

At P:

\[
\frac{I_A}{r_A}=\frac{I_B}{r_B}
\]

Rearrange to find the unknown current:

\[
I_B=\frac{I_Ar_B}{r_A}
\]

Substitute:

\[
\begin{aligned}
I_B&=\frac{4.0\times0.160}{0.080}\\
&=8.0\ \text{A}
\end{aligned}
\]

**Step 3**: Calculate the individual field magnitudes.

Use the field equation for wire A:

\[
B_A=2.0\times10^{-7}\frac{I_A}{r_A}
\]

Substitute:

\[
\begin{aligned}
B_A&=2.0\times10^{-7}\frac{4.0}{0.080}\\
&=1.0\times10^{-5}\ \text{T}
\end{aligned}
\]

Because P is a cancellation position:

\[
B_B=B_A
\]

Therefore:

\[
\boxed{B_A=B_B=1.0\times10^{-5}\ \text{T}}
\]

The useful idea here is that a measured cancellation position can reveal an unknown current. You do not always need to know both currents before solving the problem.

[IMAGE: Solution diagram showing the two fields at P as equal and opposite, with distances 8 cm and 16 cm and individual magnitudes of 10 microtesla.]

### Question 3

Two long parallel wires are separated by \(12\ \text{cm}\).

Wire A, on the left, carries \(3.0\ \text{A}\) out of the page. Wire B, on the right, carries \(6.0\ \text{A}\) into the page.

Point P is located \(24\ \text{cm}\) to the left of wire A.

(a) Calculate the net magnetic field at P, including its direction.

(b) Determine the position where the two fields cancel.

(c) State whether the wires attract or repel. Explain why this does not contradict your answer to part (a).

[IMAGE: Question setup showing wire A on the left carrying 3.0 A out of the page, wire B on the right carrying 6.0 A into the page, and P located 24 cm to the left of A. Label the 12 cm wire separation. Do not show field arrows or the cancellation position.]

### Solution 3

The resultant field at P is \(8.3\times10^{-7}\ \text{T}\) upwards. The cancellation position is \(12\ \text{cm}\) to the left of wire A, and the two wires repel each other.

**Step 1**: Find the distances from P to both wires.

The distance to wire A is:

\[
r_A=0.24\ \text{m}
\]

The distance to wire B is:

\[
r_B=0.24+0.12=0.36\ \text{m}
\]

**Step 2**: Determine the field directions at P.

The current in wire A is out of the page. At a point to its left, its magnetic field points downwards.

The current in wire B is into the page. At a point to its left, its magnetic field points upwards.

The two fields therefore oppose each other.

**Step 3**: Calculate the individual field magnitudes.

For wire A:

\[
\begin{aligned}
B_A&=2.0\times10^{-7}\frac{3.0}{0.24}\\
&=2.50\times10^{-6}\ \text{T}
\end{aligned}
\]

For wire B:

\[
\begin{aligned}
B_B&=2.0\times10^{-7}\frac{6.0}{0.36}\\
&=3.33\times10^{-6}\ \text{T}
\end{aligned}
\]

**Step 4**: Calculate the resultant.

The fields oppose, so subtract:

\[
\begin{aligned}
B_{\text{net}}&=B_B-B_A\\
&=3.33\times10^{-6}-2.50\times10^{-6}\\
&=8.3\times10^{-7}\ \text{T}
\end{aligned}
\]

Wire B produces the stronger field, so the resultant points upwards.

**Step 5**: Find the actual cancellation position.

The currents flow in opposite directions, so cancellation must occur outside the pair, on the side of the weaker current.

Wire A carries the smaller current.

Use:

\[
x=\frac{I_wd}{I_s-I_w}
\]

Substitute:

\[
\begin{aligned}
x&=\frac{3.0\times0.12}{6.0-3.0}\\
&=0.12\ \text{m}
\end{aligned}
\]

The cancellation position is therefore \(12\ \text{cm}\) to the left of wire A.

Notice that P is \(24\ \text{cm}\) to the left of A, so it is not at the cancellation position.

**Step 6**: Determine the force between the wires.

The currents flow in opposite directions.

Parallel wires carrying opposite-direction currents repel each other.

This does not contradict the field calculation. The field at P is determined by vector addition at P, whereas the force between the wires is determined by the magnetic field each wire experiences due to the other.

[IMAGE: Solution diagram showing the opposing fields at P, with the stronger upward contribution from wire B, and a separate external cancellation point 12 cm to the left of A. Include repulsive force arrows on the two wires in a separate visual panel.]

### Question 4

Two long parallel wires are separated by \(21\ \text{cm}\).

Wire A carries \(2.0\ \text{A}\), and wire B carries \(5.0\ \text{A}\). Both currents flow upwards.

A Hall probe is placed at the predicted cancellation position.

With the currents switched off, the probe reads a background magnetic field of \(34\ \mu\text{T}\) towards the east.

With both currents switched on, the probe still reads \(34\ \mu\text{T}\) towards the east.

A student concludes:

"The experiment has failed because the Hall probe did not read zero."

(a) Calculate the predicted cancellation position.

(b) Calculate the individual magnetic field magnitudes produced by the wires at this position.

(c) Explain whether the student's conclusion is justified.

(d) Describe a suitable experimental method for separating the field produced by the wires from the background field. Identify two other important experimental controls.

[IMAGE: Experimental question setup showing two parallel vertical wires separated by 21 cm, A carrying 2.0 A upwards and B carrying 5.0 A upwards. Include a movable Hall probe between them, ammeters, power supplies, and return leads routed away. Do not mark the cancellation position.]

### Solution 4

The student's conclusion is incorrect. The readings are consistent with perfect cancellation of the two wire-generated fields because the Hall probe also detects the background magnetic field.

**Step 1**: Calculate the cancellation position.

Both currents flow in the same direction, so the cancellation point lies between the wires.

Using:

\[
r_A=\frac{I_Ad}{I_A+I_B}
\]

Substitute:

\[
\begin{aligned}
r_A&=\frac{2.0\times0.21}{2.0+5.0}\\
&=\frac{0.42}{7.0}\\
&=0.060\ \text{m}
\end{aligned}
\]

Therefore, the cancellation position is:

\[
\boxed{6.0\ \text{cm from A}}
\]

The distance from B is:

\[
r_B=21-6=15\ \text{cm}
\]

**Step 2**: Calculate the individual fields.

For wire A:

\[
\begin{aligned}
B_A&=2.0\times10^{-7}\frac{2.0}{0.060}\\
&=6.67\times10^{-6}\ \text{T}
\end{aligned}
\]

For wire B:

\[
\begin{aligned}
B_B&=2.0\times10^{-7}\frac{5.0}{0.150}\\
&=6.67\times10^{-6}\ \text{T}
\end{aligned}
\]

The two fields have equal magnitudes and opposite directions.

Therefore:

\[
B_{\text{wires}}=0\ \text{T}
\]

**Step 3**: Interpret the Hall probe reading.

The probe measures the total magnetic field at its position.

This includes:

- The field produced by wire A.
- The field produced by wire B.
- The background magnetic field, including Earth's field.

When the wire fields cancel, the background field remains.

Consequently, a reading of \(34\ \mu\text{T}\) towards the east is entirely consistent with the predicted cancellation.

**Step 4**: Describe a suitable revised measurement.

One approach is to measure the background field with both currents switched off, then subtract this vector reading from the reading with both currents switched on.

Another approach is to reverse both currents simultaneously.

The magnetic field produced by the wires reverses, but the background field remains unchanged.

Taking half the vector difference between the two readings isolates the field produced by the wires.

**Step 5**: Identify experimental controls.

Two important controls are:

1. Keep the Hall probe at exactly the same position and orientation throughout the measurements.
2. Use the ammeters to confirm that both currents have the intended magnitudes.

Other suitable controls include using sufficiently long wires, keeping the return leads away from the probe, and measuring near the middle of the wires to reduce end effects.

[IMAGE: Solution diagram showing the Hall probe at the predicted cancellation position, with equal opposing fields from wires A and B and a separate background field pointing east. Show that the total probe reading remains directed east even though the wire-generated resultant is zero.]

### Question 5

Two long, straight, parallel wires A and B are separated by \(12.0\ \text{cm}\).

Both carry conventional currents of \(4.0\ \text{A}\) out of the page.

Point P is located \(5.0\ \text{cm}\) above the midpoint between the wires.

Use the horizontal direction along AB as the x-axis and the vertical direction as the y-axis.

(a) Calculate the distance from P to each wire and the magnetic field magnitude produced individually by each wire.

(b) Resolve both magnetic field vectors into horizontal and vertical components. Calculate the resultant magnetic field at P, including its direction.

(c) A student argues that the two fields must cancel because they have equal magnitudes. Explain precisely why this is incorrect.

(d) The current in wire B is reversed so that it flows into the page, while its magnitude remains \(4.0\ \text{A}\). Calculate the new resultant field at P.

(e) Could you adjust the magnitude of the current in B, keeping the current in A at \(4.0\ \text{A}\), to make the total magnetic field zero at P? Consider both possible current directions in B and justify your answer using vector components.

[IMAGE: Question setup showing two parallel wires A and B, 12.0 cm apart, both carrying 4.0 A out of the page. Place point P 5.0 cm directly above their midpoint. Include horizontal separation, vertical height, and current symbols. Do not draw magnetic field arrows or components.]

### Solution 5

The magnetic fields do not cancel at P. With both currents out of the page, the resultant is \(1.31\times10^{-5}\ \text{T}\) horizontally to the left. Reversing the current in B produces a resultant of \(1.57\times10^{-5}\ \text{T}\) upwards. Neither arrangement can produce a true zero field at P by adjusting the magnitude of B alone.

**Step 1**: Calculate the distance from P to each wire.

The horizontal distance from P to either wire is:

\[
\frac{12.0}{2}=6.0\ \text{cm}
\]

The vertical distance is \(5.0\ \text{cm}\).

Using Pythagoras:

\[
\begin{aligned}
r&=\sqrt{6.0^2+5.0^2}\\
&=\sqrt{61}\\
&=7.81\ \text{cm}
\end{aligned}
\]

Convert to metres:

\[
r=0.0781\ \text{m}
\]

**Step 2**: Calculate the individual magnetic field magnitudes.

Both wires carry \(4.0\ \text{A}\), and P is equally distant from them.

Therefore, both fields have the same magnitude:

\[
\begin{aligned}
B&=2.0\times10^{-7}\frac{4.0}{0.0781}\\
&=1.024\times10^{-5}\ \text{T}
\end{aligned}
\]

Thus:

\[
B_A=B_B=10.24\ \mu\text{T}
\]

**Step 3**: Resolve the fields into components.

Take left as the negative x-direction and upwards as the positive y-direction.

For a magnetic field tangent to a circle, the field components depend on the geometry of the radius.

The horizontal component of either field has magnitude:

\[
B_x=B\frac{5.0}{7.81}
\]

\[
B_x=10.24\times\frac{5.0}{7.81}
\]

\[
B_x=6.56\ \mu\text{T}
\]

The vertical component has magnitude:

\[
B_y=B\frac{6.0}{7.81}
\]

\[
B_y=10.24\times\frac{6.0}{7.81}
\]

\[
B_y=7.87\ \mu\text{T}
\]

Using the right-hand grip rule:

- Wire A produces a field pointing left and upwards.
- Wire B produces a field pointing left and downwards.

**Step 4**: Add the horizontal and vertical components separately.

The horizontal components both point left:

\[
B_{\text{net},x}=-6.56-6.56
\]

\[
B_{\text{net},x}=-13.12\ \mu\text{T}
\]

The vertical components are equal and opposite:

\[
B_{\text{net},y}=7.87-7.87=0
\]

Therefore, the resultant field is:

\[
\boxed{B_{\text{net}}=13.12\ \mu\text{T}}
\]

directed horizontally to the left.

**Step 5**: Explain why equal magnitudes do not guarantee cancellation.

The two fields have equal magnitudes, but they are not pointing in opposite directions.

Their vertical components cancel, while their horizontal components reinforce each other.

This is the critical distinction between equal magnitudes and equal, opposite vectors.

[IMAGE: Solution diagram showing point P above the midpoint of two wires carrying equal currents out of the page. Draw the two equal tangent field vectors, one upwards-left and the other downwards-left. Resolve them into equal opposing vertical components and equal leftward horizontal components. Show the resultant pointing left.]

**Step 6**: Reverse the current in wire B.

Reversing the current reverses the direction of the entire magnetic field produced by wire B.

Its horizontal component now points right, and its vertical component points upwards.

Wire A still produces a field pointing left and upwards.

The horizontal components now cancel:

\[
B_{\text{net},x}=-6.56+6.56=0
\]

The vertical components reinforce:

\[
B_{\text{net},y}=7.87+7.87
\]

\[
B_{\text{net},y}=15.74\ \mu\text{T}
\]

Therefore:

\[
\boxed{B_{\text{net}}=15.74\ \mu\text{T}\text{ upwards}}
\]

Reversing one current changes which components reinforce and which oppose. It does not automatically create a zero-field point.

**Step 7**: Decide whether adjusting the current magnitude in B can produce a true zero.

No. Consider the two possible current directions.

If B carries current out of the page, both horizontal field components point left. Increasing or decreasing the current in B changes the magnitude of its leftward component, but it cannot reverse that component.

Wire A always contributes a nonzero leftward component.

If B carries current into the page, its horizontal component points right, but its vertical component points upwards. Wire A also produces an upwards vertical component.

The vertical components therefore reinforce each other, regardless of the magnitude of the reversed current.

In either arrangement, at least one component remains nonzero.

This is the decisive constraint: adjusting the current magnitude can balance one component, but it cannot make two non-antiparallel field vectors cancel completely.

The question is not simply whether the two fields have equal magnitudes. It is whether their directions allow every component to cancel.

[IMAGE: Final comparison diagram with two panels. Panel one shows both currents out of the page, with horizontal components reinforcing to the left and vertical components cancelling. Panel two shows wire B reversed, with horizontal components cancelling and vertical components reinforcing upwards. Label the resultant directions and components.]

## Where this leads next

The location of a zero-field line is determined by two things: where the magnetic field vectors can oppose each other, and where their magnitudes become equal.

Once you can identify that position, the next useful step is to calculate the magnetic force between the wires. That requires evaluating the magnetic field produced by one wire at the position of the other, then applying the force rule rather than the field-direction rule.

For a broader treatment of adding magnetic fields at arbitrary points, see [How to Find the Net Magnetic Field Between Two Parallel Wires](https://tutorgum.com/learn/resources/hsc-physics-how-to-find-the-net-magnetic-field-between-two-parallel-wires/).