How to Find the Net Magnetic Field Between Two Parallel Wires

Two parallel wires carry current in the same direction. You place a point halfway between them and try to predict the magnetic field. Should the fields reinforce because the currents flow in the same direction?

No. At the midpoint, the magnetic field from the left wire points upwards, while the field from the right wire points downwards. The fields oppose each other, even though the currents flow in the same direction.

This is the key to finding the net magnetic field between two parallel wires: calculate the magnetic field produced by each wire separately, determine both directions using the right-hand grip rule, and then add the magnetic fields as vectors.

The calculation is straightforward once you get the directions right. Let’s build a reliable method.

01The equation for the magnetic field around a straight wire

A current-carrying wire produces a magnetic field that forms concentric circles around the wire.

The magnetic field gets weaker as you move further away from the wire, and stronger as the current increases.

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

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

where:

• \(B\) is the magnetic field strength, measured in tesla (T).
• \(\mu_0\) is the permeability of free space, approximately \(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\).
• \(I\) is the conventional current through the wire, measured in amperes (A).
• \(r\) is the perpendicular distance from the wire to the point where the field is being calculated, measured in metres (m).

For calculations involving wires in air, this equation is normally appropriate.

You can simplify it using:

\[
\frac{\mu_0}{2\pi}=2.0\times10^{-7}
\]

Therefore:

\[
\boxed{B=2.0\times10^{-7}\frac{I}{r}}
\]

This is a particularly useful form for HSC calculations.

Notice that the distance is measured from the individual wire to the point of interest. It is not automatically the separation between the two wires.

What if there are two wires?

Suppose wire 1 produces a magnetic field \(B_1\) at point P, and wire 2 produces a magnetic field \(B_2\) at the same point.

The net magnetic field is the vector sum:

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

This means you cannot simply add the numerical magnitudes without checking their directions.

For a point on the straight line joining the two wire positions, there are only two possibilities:

• The fields point in the same direction, so their magnitudes add.
• The fields point in opposite directions, so their magnitudes subtract.

The direction of the resultant field must also be stated.

02Determine the direction using the right-hand grip rule

Before calculating anything, establish the direction of the magnetic field from each wire.

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 curled fingers show the direction of the magnetic field.

The thumb represents conventional current, not electron flow.

In a top-down cross-sectional diagram, you will commonly see two symbols:

A useful way to remember these symbols is to picture an arrow travelling through the page. You see its pointed tip coming towards you, or its feathers disappearing away from you.

The important pattern between two wires

Consider two wires positioned side by side, with point P somewhere between them.

The direction of the magnetic field at P depends on the relative directions of the currents.

Let’s make this more concrete.

Case 1: Currents in the same direction

Imagine both currents flow out of the page.

At P:

• The field from the left wire points upwards.
• The field from the right wire points downwards.

The magnetic fields oppose each other.

Therefore:

\[
\boxed{B_{\text{net}}=|B_1-B_2|}
\]

The resultant points in the direction of the stronger individual magnetic field.

Case 2: Currents in opposite directions

Now reverse the current in the right wire so that it flows into the page.

At P:

• The field from the left wire points upwards.
• The field from the right wire also points upwards.

The magnetic fields reinforce each other.

Therefore:

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

The resultant points in their common direction.

This gives you a useful shortcut for points directly between the wires:

• Same-direction currents produce opposing magnetic fields between the wires.
• Opposite-direction currents produce reinforcing magnetic fields between the wires.

However, don’t memorise this shortcut without understanding the right-hand grip rule. If a question changes the arrangement, you should be able to reconstruct the directions yourself.

03Don’t confuse magnetic field direction with magnetic force direction

This is one of the most common mistakes in questions involving parallel current-carrying wires.

Magnetic field direction and magnetic force direction are different things.

They answer different questions:

For example, two parallel wires carrying currents in the same direction attract each other.

It is tempting to think that because the wires attract, their magnetic fields between them must point in the same direction.

That conclusion is incorrect.

The attraction describes the force between the wires. The magnetic fields at a point between them still point in opposite directions.

This distinction becomes particularly important when a question asks you to calculate the net magnetic field rather than the force between the wires.

If you need to understand the force calculation separately, see Force Between Parallel Current-Carrying Wires: HSC Physics.

Also, remember that the right-hand grip rule uses conventional current. In a metal wire, electron flow is in the opposite direction. For a refresher, see Electron Flow vs Conventional Current in HSC Magnetic Questions.

04Worked examples: calculating the net magnetic field

Now let’s put the method into practice.

05Can you experimentally test magnetic field cancellation?

There is a useful prediction hidden in the first worked example.

If two wires carry equal currents in the same direction, what happens at the exact midpoint between them?

The two magnetic fields have equal magnitudes but opposite directions. The net magnetic field should therefore be zero.

This is a prediction of the long-wire model. We can design an experiment to test it.

Setting up the experiment

Imagine two long, straight conductors mounted parallel to each other, separated by \(8.0\ \mathrm{cm}\).

Connect each conductor to a low-voltage, current-limited supply. Use ammeters in series with the conductors to check that both carry the same conventional current, \(5.0\ \mathrm{A}\), in the same direction.

Place a Hall probe halfway between the wires. Its sensitive axis should be aligned with the magnetic field direction at that point.

The return leads should be positioned well away from the measurement region so their magnetic fields do not significantly affect the reading.

What should the experiment show?

First, calculate the magnetic field produced by either wire at the midpoint.

The distance from each wire is:

\[
r=0.040\ \mathrm{m}
\]

Each wire produces:

\[
\begin{aligned}
B&=2.0\times10^{-7}\frac{5.0}{0.040}\\
&=2.5\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

The two fields have equal magnitudes and opposite directions.

Therefore:

\[
B_{\text{net}}=2.5\times10^{-5}-2.5\times10^{-5}
\]

\[
\boxed{B_{\text{net}}=0\ \mathrm{T}}
\]

Now reverse the current in one wire while keeping both current magnitudes unchanged.

The two magnetic fields will point in the same direction. The predicted resultant becomes:

\[
\begin{aligned}
B_{\text{net}}&=2.5\times10^{-5}+2.5\times10^{-5}\\
&=5.0\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

The expected observation is a near-zero background-corrected reading in the first arrangement, followed by a reading of approximately \(5.0\times10^{-5}\ \mathrm{T}\) after reversing one current.

These are predicted values, not claimed experimental measurements.

Why isn’t a compass enough?

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

If the compass needle does not deflect at the midpoint, you cannot conclude that neither wire produces a magnetic field. Their individual fields may simply be cancelling.

A Hall probe provides a more useful quantitative test. Even then, you need to account for Earth’s magnetic field, probe orientation, unequal currents, and the magnetic fields from the return leads.

The key experimental distinction is the same as the calculation: a zero resultant field does not mean zero individual fields.

06What if the point isn’t halfway between the wires?

A common assumption is that the magnetic field contributions will always be equal when the wires carry equal currents.

That is only true when the point is equally distant from both wires.

For example, if two wires carry equal currents in the same direction, but P is closer to the left wire, the left wire produces the stronger magnetic field.

The field magnitudes depend on:

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

This means both current and distance matter.

There is also an important geometrical limitation to the simple addition and subtraction rules.

The rules in this article apply directly when P lies on the straight line joining the two wire positions. If P is somewhere else between the wires, the magnetic field directions at P will generally not be parallel or opposite.

In that situation, you must add the magnetic field vectors using vector addition rather than simply adding or subtracting their magnitudes.

This is why drawing the magnetic field directions before calculating their magnitudes is such a reliable habit.

07Questions and solutions

The following questions move from standard HSC calculations towards more demanding problems involving field cancellation, experimental evidence, and changing current conditions.

Assume all wires are long, straight, parallel conductors in air, with steady currents. Unless otherwise stated, ignore external magnetic fields and end effects.

Question 1

Two long, parallel wires are separated by \(10\ \mathrm{cm}\).

The left wire carries a current of \(4.0\ \mathrm{A}\) out of the page, and the right wire carries a current of \(2.0\ \mathrm{A}\) out of the page.

Point P is located halfway between the two wires.

Calculate the net magnetic field strength and direction at P.

Solution 1

The net magnetic field is \(8.0\times10^{-6}\ \mathrm{T}\), directed upwards on the page.

Both currents flow out of the page, so the magnetic fields at P point in opposite directions.

Step 1: Find the distance from each wire to P.

Since P is halfway between the wires:

\[
r_1=r_2=\frac{0.10}{2}=0.050\ \mathrm{m}
\]

Step 2: Calculate the magnetic field from each wire.

For the left wire:

\[
\begin{aligned}
B_1&=2.0\times10^{-7}\frac{4.0}{0.050}\\
&=1.6\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

For the right wire:

\[
\begin{aligned}
B_2&=2.0\times10^{-7}\frac{2.0}{0.050}\\
&=8.0\times10^{-6}\ \mathrm{T}
\end{aligned}
\]

Step 3: Subtract the magnetic field magnitudes.

\[
\begin{aligned}
B_{\text{net}}&=B_1-B_2\\
&=1.6\times10^{-5}-8.0\times10^{-6}\\
&=8.0\times10^{-6}\ \mathrm{T}
\end{aligned}
\]

Step 4: State the direction.

At P, the field from the left wire points upwards, while the field from the right wire points downwards.

The left wire produces the stronger field, so the resultant points upwards.

\[
\boxed{B_{\text{net}}=8.0\times10^{-6}\ \mathrm{T}\text{ upwards}}
\]

The important point is that equal distances do not guarantee cancellation. The currents must also produce equal magnetic field magnitudes.

Question 2

Two parallel wires are separated by \(14\ \mathrm{cm}\).

The left wire carries a current of \(7.0\ \mathrm{A}\) out of the page, while the right wire carries a current of \(3.0\ \mathrm{A}\) into the page.

A point P is located \(4.0\ \mathrm{cm}\) from the left wire.

Calculate the net magnetic field at P, including its direction. Explain why the two magnetic field magnitudes must be added rather than subtracted.

Solution 2

The net magnetic field is \(4.1\times10^{-5}\ \mathrm{T}\), directed upwards on the page.

The currents flow in opposite directions, which means their magnetic fields reinforce each other at a point between the wires.

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

The distance from the left wire is:

\[
r_1=0.040\ \mathrm{m}
\]

The distance from the right wire is:

\[
r_2=0.14-0.040=0.10\ \mathrm{m}
\]

Step 2: Determine the magnetic field directions.

Using the right-hand grip rule:

• The left wire produces an upwards field at P.
• The right wire also produces an upwards field at P.

Therefore, the fields add.

Step 3: Calculate the individual field strengths.

For the left wire:

\[
\begin{aligned}
B_1&=2.0\times10^{-7}\frac{7.0}{0.040}\\
&=3.5\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

For the right wire:

\[
\begin{aligned}
B_2&=2.0\times10^{-7}\frac{3.0}{0.10}\\
&=6.0\times10^{-6}\ \mathrm{T}
\end{aligned}
\]

Step 4: Add the magnitudes.

\[
\begin{aligned}
B_{\text{net}}&=B_1+B_2\\
&=3.5\times10^{-5}+6.0\times10^{-6}\\
&=4.1\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

Both fields point upwards, so the resultant has the same direction.

\[
\boxed{B_{\text{net}}=4.1\times10^{-5}\ \mathrm{T}\text{ upwards}}
\]

The tempting mistake is to subtract because there are two separate wires. However, subtraction is only appropriate when the magnetic field vectors point in opposite directions. The direction of the currents alone does not tell you to subtract; you must first apply the right-hand grip rule.

Question 3

Two long, parallel wires are separated by \(13\ \mathrm{cm}\).

The left wire carries a current of \(4.0\ \mathrm{A}\) out of the page, and the right wire carries a current of \(9.0\ \mathrm{A}\) out of the page.

(a) Determine the position between the wires where the net magnetic field is zero.

(b) A Hall probe is moved to a point \(1.0\ \mathrm{cm}\) to the right of this zero-field position. Calculate the net magnetic field strength and state its direction.

Solution 3

(a) The zero-field position is \(4.0\ \mathrm{cm}\) from the left wire, or \(9.0\ \mathrm{cm}\) from the right wire.

Step 1: Identify the relationship between the fields.

Both currents flow out of the page, so the magnetic fields at points between the wires oppose each other.

For the net field to be zero, the magnitudes must be equal:

\[
B_1=B_2
\]

Step 2: Let \(x\) be the distance from the left wire to the zero-field position.

The distance from the right wire is \(0.13-x\).

Using the magnetic field equation:

\[
2.0\times10^{-7}\frac{4.0}{x}
=
2.0\times10^{-7}\frac{9.0}{0.13-x}
\]

Cancel the common factor:

\[
\frac{4.0}{x}=\frac{9.0}{0.13-x}
\]

Cross-multiply:

\[
4.0(0.13-x)=9.0x
\]

\[
0.52-4x=9x
\]

\[
13x=0.52
\]

\[
x=0.040\ \mathrm{m}
\]

Therefore, the zero-field position is \(4.0\ \mathrm{cm}\) from the left wire.

The remaining distance is:

\[
0.13-0.040=0.090\ \mathrm{m}
\]

(b) The net magnetic field \(1.0\ \mathrm{cm}\) to the right of the zero-field position is \(6.5\times10^{-6}\ \mathrm{T}\), directed downwards on the page.

Step 1: Find the new distances.

The measurement point is:

\[
x=0.050\ \mathrm{m}
\]

Therefore:

\[
r_1=0.050\ \mathrm{m}
\]

\[
r_2=0.13-0.050=0.080\ \mathrm{m}
\]

Step 2: Calculate both magnetic field strengths.

For the left wire:

\[
\begin{aligned}
B_1&=2.0\times10^{-7}\frac{4.0}{0.050}\\
&=1.60\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

For the right wire:

\[
\begin{aligned}
B_2&=2.0\times10^{-7}\frac{9.0}{0.080}\\
&=2.25\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

Step 3: Determine the resultant.

The magnetic field from the left wire points upwards, while the field from the right wire points downwards.

The right wire now produces the stronger field.

\[
\begin{aligned}
B_{\text{net}}&=B_2-B_1\\
&=2.25\times10^{-5}-1.60\times10^{-5}\\
&=6.5\times10^{-6}\ \mathrm{T}
\end{aligned}
\]

The resultant therefore points downwards.

\[
\boxed{B_{\text{net}}=6.5\times10^{-6}\ \mathrm{T}\text{ downwards}}
\]

This demonstrates why the zero-field position is not simply the midpoint. It depends on the ratio of the currents as well as the separation.

Question 4

Two parallel wires are separated by \(8.0\ \mathrm{cm}\). Each carries a current of \(5.0\ \mathrm{A}\) out of the page.

A Hall probe is placed exactly halfway between the wires. After accounting for the background magnetic field, the probe reads approximately zero.

A student concludes:

“Neither wire produces a magnetic field at the midpoint because the Hall probe reads zero.”

(a) Explain why the student’s conclusion is incorrect.

(b) Calculate the individual magnetic field strengths at the midpoint.

(c) The current in the right-hand wire is reversed without changing its magnitude. Predict the new net magnetic field strength and direction at the midpoint.

(d) Identify two experimental controls that would help ensure the Hall probe reading represents the magnetic field produced by the two test wires.

Solution 4

(a) The student’s conclusion is incorrect because the two wires each produce a magnetic field, but those fields cancel at the midpoint.

A zero net field does not mean that the individual fields are zero.

Step 1: Identify the directions.

Both currents flow out of the page.

At the midpoint:

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

The fields have equal magnitudes because the currents and distances are equal.

(b) Each wire produces a magnetic field of \(2.5\times10^{-5}\ \mathrm{T}\).

Step 2: Calculate the distance from each wire.

\[
r=\frac{0.080}{2}=0.040\ \mathrm{m}
\]

Step 3: Calculate the magnetic field from either wire.

\[
\begin{aligned}
B&=2.0\times10^{-7}\frac{5.0}{0.040}\\
&=2.5\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

Therefore:

\[
\boxed{B_1=B_2=2.5\times10^{-5}\ \mathrm{T}}
\]

The fields are equal and opposite, so:

\[
B_{\text{net}}=0\ \mathrm{T}
\]

(c) The new net magnetic field is \(5.0\times10^{-5}\ \mathrm{T}\), directed upwards on the page.

Step 4: Determine the new field directions.

The left current still flows out of the page, while the right current now flows into the page.

Using the right-hand grip rule, both fields at the midpoint now point upwards.

Step 5: Add the field magnitudes.

\[
\begin{aligned}
B_{\text{net}}&=B_1+B_2\\
&=2.5\times10^{-5}+2.5\times10^{-5}\\
&=5.0\times10^{-5}\ \mathrm{T}
\end{aligned}
\]

Therefore:

\[
\boxed{B_{\text{net}}=5.0\times10^{-5}\ \mathrm{T}\text{ upwards}}
\]

(d) Two suitable experimental controls are:

1. Keep the Hall probe at exactly the same position and orientation when reversing the current. Otherwise, a change in reading could be caused by probe movement rather than the current reversal.
2. Measure the background magnetic field with the test currents switched off, then account for it in the readings.

Other valid controls include checking that both currents have the intended magnitudes with ammeters and routing return leads away from the measurement region.

The important experimental idea is that a Hall probe measures the resultant field at its location. It cannot independently tell you the individual contributions unless you use the current arrangement and field model to work them out.

Question 5

This final problem explores what happens to a zero-field position when the current magnitudes change.

Two long, parallel wires are separated by \(20\ \mathrm{cm}\).

The left wire carries a current \(I_1=4.0\ \mathrm{A}\) out of the page. The right wire carries a current \(I_2=9.0\ \mathrm{A}\) into the page.

Let \(x=0\) represent the position of the left wire, with positive \(x\) directed towards the right wire. The right wire is therefore located at \(x=0.20\ \mathrm{m}\).

(a) Determine whether a zero-field position exists between the wires. If not, explain why.

(b) Find the position along the line joining the wires where the net magnetic field is zero.

(c) The current in the left wire is gradually increased from \(4.0\ \mathrm{A}\) to \(12.0\ \mathrm{A}\), while the right current remains fixed at \(9.0\ \mathrm{A}\), and neither current changes direction.

Describe how the zero-field position changes throughout this process. In particular, explain what happens when the left current reaches \(9.0\ \mathrm{A}\), and calculate the final zero-field position when the left current reaches \(12.0\ \mathrm{A}\).

Assume the wires remain infinitely long for the purpose of the model, and consider the entire straight line containing both wires, including positions outside the gap.

Solution 5

(a) There is no zero-field position between the wires because the magnetic fields point in the same direction throughout the gap.

Step 1: Determine the field directions between the wires.

The left current flows out of the page, while the right current flows into the page.

At any point between the wires:

• The field from the left wire points upwards.
• The field from the right wire also points upwards.

Therefore, the fields reinforce each other.

There is no cancellation point between the wires, regardless of the current magnitudes.

This is the first important constraint. A zero-field position cannot exist in a region where the two field vectors always point in the same direction.

(b) The zero-field position is \(0.16\ \mathrm{m}\) to the left of the left wire.

Step 2: Determine where cancellation is possible.

Between the wires, the fields reinforce each other.

Outside the wires, however, their field directions oppose each other.

Consider a point to the left of both wires. The magnetic field from the left wire points downwards, while the magnetic field from the right wire points upwards.

Cancellation is therefore possible in this region.

Because the left current is smaller, the zero-field position will be on the left side of the weaker-current wire.

Step 3: Define the distances from the zero-field position.

Let \(x\) be the coordinate of the zero-field point relative to the left wire.

Since the point is to the left of the left wire:

\[
x<0
\]

Its distance from the left wire is \(-x\).

Its distance from the right wire is \(0.20-x\).

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

\[
2.0\times10^{-7}\frac{4.0}{-x}
=
2.0\times10^{-7}\frac{9.0}{0.20-x}
\]

Cancel the common factor:

\[
\frac{4.0}{-x}=\frac{9.0}{0.20-x}
\]

Cross-multiply:

\[
4.0(0.20-x)=9.0(-x)
\]

\[
0.80-4x=-9x
\]

\[
5x=-0.80
\]

\[
x=-0.16\ \mathrm{m}
\]

The negative sign confirms that the point is to the left of the left wire.

Therefore:

\[
\boxed{x=-0.16\ \mathrm{m}}
\]

The zero-field position is \(16\ \mathrm{cm}\) to the left of the left wire.

(c) The zero-field position moves towards negative infinity as the left current approaches \(9.0\ \mathrm{A}\). At equal currents, no finite zero-field position exists. Once the left current exceeds \(9.0\ \mathrm{A}\), a zero-field position reappears to the right of the right wire.

This is the most interesting part of the problem.

Step 4: Establish a general relationship for the zero-field position.

For currents flowing in opposite directions, a zero-field position can only exist outside the gap.

Using the coordinate system already defined, equality of the field magnitudes gives:

\[
\frac{I_1}{|x|}
=
\frac{I_2}{|x-0.20|}
\]

Solving for the zero-field coordinate, taking the appropriate outside region, gives:

\[
\boxed{x=\frac{I_1d}{I_1-I_2}}
\]

where:

• \(x\) is the zero-field coordinate measured from the left wire, in metres.
• \(I_1\) is the current in the left wire, in amperes.
• \(I_2\) is the current in the right wire, in amperes.
• \(d\) is the separation between the wires, in metres.

This expression applies when the currents are opposite in direction and \(I_1\ne I_2\). The sign of \(x\) determines which side of the wires contains the zero-field position.

Step 5: Consider what happens as the left current increases.

Initially:

\[
I_1=4.0\ \mathrm{A},\qquad I_2=9.0\ \mathrm{A}
\]

The denominator is negative, so \(x\) is negative. The zero-field point is left of both wires.

As \(I_1\) increases towards \(9.0\ \mathrm{A}\), the denominator becomes smaller in magnitude.

Consequently, the zero-field position moves further and further to the left.

When:

\[
I_1=I_2=9.0\ \mathrm{A}
\]

the denominator becomes zero.

The expression has no finite solution.

This does not mean the magnetic field suddenly becomes zero everywhere. In fact, there is no finite zero-field position anywhere along the line.

Why?

• Between the wires, the fields reinforce each other.
• To the left of both wires, the left wire is closer and produces the stronger field.
• To the right of both wires, the right wire is closer and produces the stronger field.

With equal currents, neither outside region allows the two fields to cancel.

The zero-field position therefore approaches negative infinity as \(I_1\) approaches \(9.0\ \mathrm{A}\) from below. Once \(I_1\) becomes greater than \(9.0\ \mathrm{A}\), the zero-field position is on the opposite side of the arrangement, approaching positive infinity as the currents approach equality from above.

Step 6: Calculate the final position when the left current reaches \(12.0\ \mathrm{A}\).

Substitute:

\[
I_1=12.0\ \mathrm{A}
\]

\[
I_2=9.0\ \mathrm{A}
\]

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

Into the zero-field equation:

\[
\begin{aligned}
x&=\frac{I_1d}{I_1-I_2}\\
&=\frac{12.0\times0.20}{12.0-9.0}\\
&=\frac{2.4}{3.0}\\
&=0.80\ \mathrm{m}
\end{aligned}
\]

Therefore, the zero-field point is located \(0.80\ \mathrm{m}\) to the right of the left wire.

Since the right wire is at \(x=0.20\ \mathrm{m}\), the zero-field point is:

\[
0.80-0.20=0.60\ \mathrm{m}
\]

to the right of the right wire.

We can check the result by calculating the individual fields at this position.

Distance from the left wire:

\[
r_1=0.80\ \mathrm{m}
\]

Distance from the right wire:

\[
r_2=0.60\ \mathrm{m}
\]

For the left wire:

\[
B_1=2.0\times10^{-7}\frac{12.0}{0.80}
=3.0\times10^{-6}\ \mathrm{T}
\]

For the right wire:

\[
B_2=2.0\times10^{-7}\frac{9.0}{0.60}
=3.0\times10^{-6}\ \mathrm{T}
\]

The fields have equal magnitudes and opposite directions, confirming that the net field is zero.

The surprising result is that the zero-field position never travels through the gap between the wires. It moves towards infinity on one side, disappears as a finite solution when the currents are equal, and reappears on the other side when the stronger current changes.

This happens because the directions of the magnetic fields between the wires never change: opposite-direction currents always produce reinforcing fields in the gap.

08What to remember for HSC Physics

When you are asked to find the net magnetic field between two parallel wires, use this sequence:

1. Identify the conventional current direction in each wire.
2. Mark the observation point and calculate its distance from each wire.
3. Apply the right-hand grip rule separately to both wires.
4. Calculate each individual magnetic field using \(B=\frac{\mu_0 I}{2\pi r}\).
5. Add the magnitudes if the fields point in the same direction, or subtract them if they point in opposite directions.
6. State the resultant direction and include the unit tesla (T).

Keep this distinction clear:

Finally, remember that finding the net magnetic field is only one part of analysing two current-carrying wires. If the next question asks whether the wires attract or repel, or asks you to calculate the force between them, you need to move from the right-hand grip rule to the motor-effect force rule.

That is the natural next step: understanding how the magnetic field produced by one wire creates a force on the other.