Why a Coil Has a Stronger Central Field Than a Straight Wire

A circular coil can produce a much stronger magnetic field at its centre than a single straight wire carrying the same current. The reason is that the magnetic fields produced by its individual turns reinforce one another. Each turn contributes a field in the same direction along the coil's central axis, so the total field builds as more turns are added.

This is called the **principle of superposition**: magnetic fields combine to produce a resultant magnetic field. The important detail is that the direction of each contribution matters. Fields pointing in the same direction add; fields pointing in opposite directions partially or completely cancel.

For an ideal circular coil with \(N\) closely wound turns, radius \(R\), and steady current \(I\), the magnetic flux density at its centre is

\[
\boxed{B=\frac{\mu_0NI}{2R}}
\]

Here, \(B\) is measured in tesla (T), \(N\) is the number of turns, \(I\) is the conventional current in amperes (A), \(R\) is the coil radius in metres (m), and \(\mu_0\) is the permeability of free space, approximately \(4\pi\times10^{-7}\ \text{T m A}^{-1}\).

The equation reveals three important relationships:

- Doubling the number of turns doubles the central field, provided the current and radius stay constant.
- Doubling the current doubles the central field, provided the geometry stays constant.
- Doubling the radius halves the central field, provided the number of turns and current stay constant.

The rest of the explanation is about understanding why those relationships make physical sense, and knowing when you can use them.

## 1. Why the magnetic fields reinforce one another

Imagine bending a wire into one complete circular loop and placing a compass at its centre. Current flows around the loop.

Each small section of the wire produces a magnetic field around itself. However, at the centre of the loop, the magnetic field contributions from all the sections point along the same central axis.

This is the crucial geometric feature of a circular coil.

[IMAGE: A clear three-dimensional illustration of a circular current-carrying wire loop viewed at an angle, with conventional current arrows around the circumference and several representative magnetic field contribution arrows at the centre, all parallel to the central axis and pointing in the same direction. Label the radius R, centre O, and central magnetic field B.]

Think about what happens when you add another identical turn directly beside the first. The new turn produces another magnetic field at the centre, pointing in exactly the same direction.

The first turn contributes a field \(B_1\). The second contributes another \(B_1\). With two turns, the total field is \(2B_1\).

With \(N\) identical turns:

\[
B_{\text{total}}=NB_1
\]

Since the field at the centre of one ideal circular loop is

\[
B_1=\frac{\mu_0 I}{2R}
\]

the field for \(N\) turns becomes

\[
B_{\text{total}}=\frac{\mu_0NI}{2R}
\]

A coil is therefore not stronger simply because it contains more wire. It is stronger because the geometry makes the magnetic field contributions from its turns reinforce one another at the centre.

[IMAGE: Two side-by-side circular coil diagrams, one with a single turn and one with five closely wound turns. Show conventional current flowing in the same rotational direction in every turn. At each centre, show the individual axial field contributions combining into a larger resultant arrow for the five-turn coil. Label the diagrams 1 turn and 5 turns.]

### Why a straight wire behaves differently

A straight current-carrying wire also produces a magnetic field. However, its field lines form concentric circles around the wire rather than running along a common axis through a central point.

Use the right-hand grip rule: point your right thumb along the conventional current, and your curled fingers show the direction of the magnetic field.

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

\[
B_{\text{wire}}=\frac{\mu_0 I}{2\pi r}
\]

The field becomes weaker as you move further from the wire.

[IMAGE: A straight vertical wire carrying conventional current upwards, viewed in three dimensions. Draw several concentric circular magnetic field lines around it, with directional arrows following the right-hand grip rule. Include a marked observation point at perpendicular distance r and show the local magnetic field direction at that point.]

Notice that the contributions from different sections of a straight wire do not simply disappear or cancel. They also combine. The difference is the geometry: the current in a circular loop is arranged so that every section contributes to the field along the same central axis, while the straight wire produces a different field pattern.

A useful comparison makes the difference quantitative.

| Conductor | Magnetic flux density at the comparison point |
|---|---|
| One circular loop of radius \(R\) | \(\displaystyle B=\frac{\mu_0 I}{2R}\) |
| Circular coil with \(N\) turns | \(\displaystyle B=\frac{\mu_0 NI}{2R}\) |
| Ideal long straight wire, distance \(R\) away | \(\displaystyle B=\frac{\mu_0 I}{2\pi R}\) |

For the same current and the same characteristic distance \(R\), the ratio between an \(N\)-turn circular coil and an ideal long straight wire is

\[
\frac{B_{\text{coil}}}{B_{\text{wire}}}
=
\frac{\mu_0NI}{2R}
\times
\frac{2\pi R}{\mu_0 I}
\]

Therefore,

\[
\boxed{\frac{B_{\text{coil}}}{B_{\text{wire}}}=\pi N}
\]

For example, a 10-turn circular coil produces a central field approximately 31.4 times the field of an ideal long straight wire carrying the same current at the same perpendicular distance.

This is an idealised comparison, not a universal multiplier for every wire arrangement. The straight-wire equation assumes a wire long enough for end effects to be negligible. A finite straight segment requires its actual length and position to be considered.

For a closer look at the circular field pattern around a straight conductor, see [Magnetic Field Around a Current-Carrying Wire: HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-magnetic-field-around-a-current-carrying-wire/).

## 2. Finding the direction of the coil's magnetic field

Knowing the field is stronger is only half the job. You also need to know which way it points.

Use the **right-hand grip rule for a coil**:

1. Curl the fingers of your right hand in the direction of conventional current around the turns.
2. Your thumb points along the magnetic field through the centre of the coil.
3. The end towards which your thumb points is the coil's north pole.

This rule uses conventional current, which is defined as the direction positive charge would move. In a metal wire, electrons drift in the opposite direction.

[IMAGE: Two face-on circular coil diagrams. In the first, conventional current travels anticlockwise as viewed by the reader, with the central magnetic field pointing out of the page, marked with a dot. In the second, current travels clockwise, with the field pointing into the page, marked with a cross. Clearly label conventional current and the viewing direction.]

For a coil viewed directly from one end:

| Conventional current viewed from that end | Magnetic field at the centre | Pole facing the observer |
|---|---|---|
| Anticlockwise | Towards the observer | North |
| Clockwise | Away from the observer | South |

A dot (\(\odot\)) represents a direction out of the page, towards you. A cross (\(\otimes\)) represents a direction into the page, away from you.

Be careful when the viewpoint changes. Looking at the same coil from the opposite end makes its current appear to rotate in the opposite direction. The physical magnetic field has not changed. You have simply changed where you are looking from.

### Field direction rules are not force rules

This is a common source of mistakes in HSC Physics.

The right-hand grip rule tells you the direction of the magnetic field produced by a current.

The motor effect rule, commonly represented by Fleming's left-hand rule, tells you the direction of the force on a current-carrying conductor placed in an external magnetic field.

These are different questions.

| What you need to find | Appropriate rule |
|---|---|
| Magnetic field around a straight current-carrying wire | Right-hand grip rule |
| Magnetic field through the centre of a coil | Right-hand grip rule |
| North and south poles of a solenoid | Right-hand grip rule |
| Force on a current-carrying wire in an external magnetic field | Motor effect rule (Fleming's left-hand rule) |

The force magnitude for a straight conductor in an external magnetic field is

\[
F=BIL\sin\theta
\]

Here, \(F\) is force in newtons (N), \(B\) is the external magnetic flux density in tesla, \(I\) is current in amperes, \(L\) is the length of conductor in the field in metres, and \(\theta\) is the angle between the current direction and the external magnetic field.

Do not use this force equation to calculate the magnetic field produced by the coil itself. They describe different physical effects.

For the next step, see [Torque on a Current-Carrying Coil for HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-torque-on-a-current-carrying-coil/).

## 3. Worked examples

### Worked example: Finding the central field of a circular coil

A circular coil has 25 turns, a radius of \(0.060\ \text{m}\), and carries a steady conventional current of \(0.80\ \text{A}\). Calculate the magnetic flux density at its centre. The current appears clockwise when viewed directly from the left-hand end of the coil.

**Step 1**: Identify the relevant equation.

For a closely wound circular coil:

\[
B=\frac{\mu_0NI}{2R}
\]

**Step 2**: List the known quantities.

- \(N=25\)
- \(I=0.80\ \text{A}\)
- \(R=0.060\ \text{m}\)
- \(\mu_0=4\pi\times10^{-7}\ \text{T m A}^{-1}\)

All measurements are already in SI units.

**Step 3**: Substitute the values.

\[
\begin{aligned}
B&=\frac{(4\pi\times10^{-7})(25)(0.80)}
{2(0.060)}\\
&=\frac{8\pi\times10^{-6}}{0.12}\\
&=2.09\times10^{-4}\ \text{T}
\end{aligned}
\]

**Step 4**: Determine the direction.

The current is clockwise when viewed from the left-hand end. Using the right-hand grip rule, the magnetic field points away from the observer, along the coil's axis towards the right.

Therefore, the central magnetic flux density is

\[
\boxed{B=2.09\times10^{-4}\ \text{T}}
\]

directed towards the right-hand end of the coil.

The magnitude comes from the coil equation. The direction comes from the current orientation. Keep those two decisions separate.

### Worked example: Two coils producing opposing fields

Two closely wound circular coils are placed concentrically in the same plane. Both have a radius of \(0.080\ \text{m}\).

- Coil A has 40 turns and carries \(0.50\ \text{A}\) anticlockwise when viewed from the front.
- Coil B has 15 turns and carries \(0.80\ \text{A}\) clockwise when viewed from the front.

Calculate the resultant magnetic flux density at the shared centre and state its direction.

[IMAGE: Two concentric circular coils in the same plane, viewed face-on. Label the outer coil A with 40 turns and an anticlockwise conventional current, and the inner coil B with 15 turns and a clockwise conventional current. Mark their shared centre O and common radius of 0.080 m. Do not show the resultant field.]

**Step 1**: Recognise that the two magnetic fields point in opposite directions.

Coil A carries anticlockwise conventional current, so its field points out of the page.

Coil B carries clockwise conventional current, so its field points into the page.

The fields therefore oppose each other. We must subtract their magnitudes rather than add them.

**Step 2**: Calculate the contribution from each coil.

For coil A:

\[
\begin{aligned}
B_A&=\frac{\mu_0N_AI_A}{2R}\\
&=\frac{(4\pi\times10^{-7})(40)(0.50)}
{2(0.080)}\\
&=1.57\times10^{-4}\ \text{T}
\end{aligned}
\]

For coil B:

\[
\begin{aligned}
B_B&=\frac{\mu_0N_BI_B}{2R}\\
&=\frac{(4\pi\times10^{-7})(15)(0.80)}
{2(0.080)}\\
&=9.42\times10^{-5}\ \text{T}
\end{aligned}
\]

**Step 3**: Subtract the opposing contributions.

\[
\begin{aligned}
B_{\text{net}}&=B_A-B_B\\
&=1.57\times10^{-4}-9.42\times10^{-5}\\
&=6.28\times10^{-5}\ \text{T}
\end{aligned}
\]

**Step 4**: State the direction.

Coil A produces the larger field, so the resultant points in its direction: out of the page, towards the observer.

\[
\boxed{B_{\text{net}}=6.28\times10^{-5}\ \text{T}}
\]

The important lesson is that adding turns does not guarantee a stronger resultant field. The direction of the current in each turn determines whether its contribution reinforces or opposes the others.

## 4. What changes the central field?

The equation

\[
B=\frac{\mu_0NI}{2R}
\]

gives a useful way to predict how a coil responds when its properties change.

| Change | Effect on central field | Reason |
|---|---|---|
| Double \(N\) | Doubles \(B\) | Twice as many reinforcing turns |
| Double \(I\) | Doubles \(B\) | Each turn produces a stronger field |
| Double \(R\) | Halves \(B\) | The centre is further from each section of wire |
| Reverse \(I\) | Same magnitude, opposite direction | Every turn reverses its field contribution |
| Reverse the current in only some turns | Depends on the opposing contributions | Some fields now subtract instead of add |

One detail is worth emphasising: the equation assumes the turns have approximately the same radius and are close together. It describes the field at the centre of the circular coil, not the field at every point around it.

A long solenoid uses a different idealised expression:

\[
B\approx\frac{\mu_0NI}{L}
\]

where \(L\) is the solenoid's length. The number of turns per metre matters because the turns are distributed along the solenoid's axis. Do not substitute the solenoid equation into a circular-loop question.

[IMAGE: Side-by-side comparison of a flat circular multi-turn coil and a long cylindrical solenoid. Show the circular coil's central axis perpendicular to its plane, and the solenoid's central axis running along its length. Draw the respective central field directions and label the circular radius R and solenoid length L.]

There is also an important practical limitation. The equations assume steady current, an air core, and idealised winding geometry. A magnetic core can increase the field, while heating, non-uniform winding, and end effects can make real measurements differ from the ideal prediction.

## Questions and solutions

### Question 1

A circular coil has 30 turns, a radius of \(0.075\ \text{m}\), and carries a conventional current of \(0.60\ \text{A}\). The current flows anticlockwise when viewed directly from the front.

Calculate the magnetic flux density at the centre of the coil and state its direction.

[IMAGE: Face-on circular multi-turn coil with 30 turns, radius 0.075 m, conventional current arrows anticlockwise, and centre O marked. Leave the central magnetic field direction for the student to determine.]

### Solution 1

The central magnetic flux density is \(1.51\times10^{-4}\ \text{T}\), directed out of the page towards the observer.

**Step 1**: Select the circular coil equation.

\[
B=\frac{\mu_0NI}{2R}
\]

**Step 2**: Substitute the known values.

\[
\begin{aligned}
B&=\frac{(4\pi\times10^{-7})(30)(0.60)}
{2(0.075)}\\
&=\frac{7.2\pi\times10^{-6}}{0.15}\\
&=1.51\times10^{-4}\ \text{T}
\end{aligned}
\]

**Step 3**: Determine the direction.

The conventional current is anticlockwise when viewed from the front.

Curl the fingers of your right hand in this direction. Your thumb points towards you, so the magnetic field is out of the page.

\[
\boxed{B=1.51\times10^{-4}\ \text{T}}
\]

The magnitude and direction must both appear in a complete answer.

### Question 2

Two concentric circular coils are placed in the same plane. Both have a radius of \(0.10\ \text{m}\).

- Coil A has 50 turns and carries a current of \(0.40\ \text{A}\) clockwise when viewed from the front.
- Coil B has 30 turns and carries a current of \(0.50\ \text{A}\) anticlockwise when viewed from the front.

Calculate the resultant magnetic flux density at their shared centre. State its direction.

[IMAGE: Face-on view of two concentric circular coils. Label the outer coil A, 50 turns, 0.40 A clockwise, and the inner coil B, 30 turns, 0.50 A anticlockwise. Show the common radius of 0.10 m and centre O. Do not show the resultant field.]

### Solution 2

The resultant magnetic flux density is \(3.14\times10^{-5}\ \text{T}\), directed into the page.

**Step 1**: Determine the direction of each field.

Coil A carries clockwise conventional current, so its magnetic field points into the page.

Coil B carries anticlockwise conventional current, so its magnetic field points out of the page.

The fields oppose each other. We must subtract their magnitudes.

**Step 2**: Calculate the field from coil A.

\[
\begin{aligned}
B_A&=\frac{\mu_0N_AI_A}{2R}\\
&=\frac{(4\pi\times10^{-7})(50)(0.40)}
{2(0.10)}\\
&=1.26\times10^{-4}\ \text{T}
\end{aligned}
\]

**Step 3**: Calculate the field from coil B.

\[
\begin{aligned}
B_B&=\frac{\mu_0N_BI_B}{2R}\\
&=\frac{(4\pi\times10^{-7})(30)(0.50)}
{2(0.10)}\\
&=9.42\times10^{-5}\ \text{T}
\end{aligned}
\]

**Step 4**: Find the resultant field.

\[
\begin{aligned}
B_{\text{net}}&=B_A-B_B\\
&=1.26\times10^{-4}-9.42\times10^{-5}\\
&=3.14\times10^{-5}\ \text{T}
\end{aligned}
\]

Coil A produces the stronger field, so the resultant points in its direction.

\[
\boxed{B_{\text{net}}=3.14\times10^{-5}\ \text{T}}
\]

The tempting mistake is to add the two fields because both coils carry current. However, magnetic fields are vectors. Their directions determine whether they reinforce or oppose each other.

### Question 3

A 10-turn circular coil has a radius of \(0.040\ \text{m}\) and carries a current of \(0.50\ \text{A}\).

An ideal long straight wire carries the same current. A point is located \(0.040\ \text{m}\) perpendicular to the wire.

Calculate:

a) The magnetic flux density at the centre of the circular coil.

b) The magnetic flux density at the specified point near the straight wire.

c) The ratio of the coil's central field to the straight wire's field.

d) Why the comparison does not establish one universal ratio for every possible coil and straight wire arrangement.

[IMAGE: Two clearly separated comparison panels. Left: a face-on 10-turn circular coil with radius 0.040 m, current 0.50 A, and central observation point O. Right: a long straight vertical wire carrying 0.50 A, with an observation point at perpendicular distance 0.040 m. Show the geometries and measurement distances without showing calculated field values.]

### Solution 3

The circular coil produces a central field of \(7.85\times10^{-5}\ \text{T}\), which is 31.4 times the field of the ideal long straight wire at the specified distance.

**Step 1**: Calculate the field at the centre of the coil.

\[
\begin{aligned}
B_{\text{coil}}&=\frac{\mu_0NI}{2R}\\
&=\frac{(4\pi\times10^{-7})(10)(0.50)}
{2(0.040)}\\
&=7.85\times10^{-5}\ \text{T}
\end{aligned}
\]

**Step 2**: Calculate the field near the straight wire.

For an ideal long straight wire:

\[
B_{\text{wire}}=\frac{\mu_0I}{2\pi r}
\]

Substitute:

\[
\begin{aligned}
B_{\text{wire}}&=\frac{(4\pi\times10^{-7})(0.50)}
{2\pi(0.040)}\\
&=2.50\times10^{-6}\ \text{T}
\end{aligned}
\]

**Step 3**: Calculate the ratio.

\[
\begin{aligned}
\frac{B_{\text{coil}}}{B_{\text{wire}}}
&=\frac{7.85\times10^{-5}}{2.50\times10^{-6}}\\
&=31.4
\end{aligned}
\]

Therefore,

\[
\boxed{\frac{B_{\text{coil}}}{B_{\text{wire}}}=31.4}
\]

**Step 4**: Explain the physical difference.

The coil has 10 turns, with every turn contributing to the central field in the same direction.

The straight wire produces a circular magnetic field pattern around itself. Its field at the observation point is determined by its geometry and perpendicular distance.

The ratio is consistent with

\[
\frac{B_{\text{coil}}}{B_{\text{wire}}}
=\pi N=10\pi=31.4
\]

This result assumes an ideal long straight wire and compares the field at distance \(R\) with the field at the centre of a coil of radius \(R\).

A shorter straight segment would require its actual geometry to be considered. The result is not a universal multiplier for arbitrary conductors.

### Question 4

A 40-turn circular coil has a radius of \(0.080\ \text{m}\) and carries a conventional current of \(0.50\ \text{A}\). Its plane lies in the page, and the current is anticlockwise when viewed from the front.

The coil is placed in a uniform external magnetic field of \(0.20\ \text{T}\), directed horizontally towards the right.

At the rightmost point of the coil, a short conductor segment has a length of \(0.030\ \text{m}\). The conventional current in this segment flows upwards.

Calculate:

a) The magnetic flux density produced by the coil at its centre, including its direction.

b) The magnitude and direction of the magnetic force on the specified segment due to the external magnetic field.

Assume the segment is perpendicular to the external field. Ignore the force due to the coil's own magnetic field.

[IMAGE: Face-on circular coil in the plane of the page, with anticlockwise current arrows. Show a horizontal external magnetic field arrow pointing right across the page. Highlight the rightmost conductor segment, label its length 0.030 m, and show conventional current upwards. Mark the coil centre O. Do not show the magnetic force direction.]

### Solution 4

The coil produces a central magnetic flux density of \(1.57\times10^{-4}\ \text{T}\), directed out of the page. The specified conductor experiences a force of \(3.0\times10^{-3}\ \text{N}\), directed into the page.

**Step 1**: Calculate the coil's central magnetic field.

\[
\begin{aligned}
B_{\text{coil}}&=\frac{\mu_0NI}{2R}\\
&=\frac{(4\pi\times10^{-7})(40)(0.50)}
{2(0.080)}\\
&=1.57\times10^{-4}\ \text{T}
\end{aligned}
\]

The current is anticlockwise when viewed from the front. The right-hand grip rule gives a field directed out of the page.

**Step 2**: Identify the magnetic field responsible for the force.

The force is caused by the stated external magnetic field of \(0.20\ \text{T}\).

Do not substitute the coil's own central field into the force equation. The question specifically asks for the force due to the external field.

**Step 3**: Calculate the force magnitude.

The force equation is

\[
F=B_{\text{external}}IL\sin\theta
\]

The current is perpendicular to the external field, so \(\theta=90^\circ\).

\[
\begin{aligned}
F&=(0.20)(0.50)(0.030)\sin90^\circ\\
&=3.0\times10^{-3}\ \text{N}
\end{aligned}
\]

**Step 4**: Determine the force direction.

The conventional current flows upwards, and the external magnetic field points right.

Using the motor effect rule, the force is directed into the page.

\[
\boxed{F=3.0\times10^{-3}\ \text{N}}
\]

This question uses two different magnetic effects:

- The coil produces its own magnetic field, found using the right-hand grip rule.
- The external magnetic field exerts a force on the current-carrying conductor, found using the motor effect rule.

Confusing these two directions can lead to an incorrect answer even when the calculations are correct.

### Question 5

A 50-turn circular coil of radius \(0.10\ \text{m}\) is centred at point O. Its plane lies in the \(yz\)-plane, so its central axis lies along the \(x\)-axis.

The coil carries a conventional current of \(0.40\ \text{A}\). Viewed from the positive \(x\)-axis, the current flows anticlockwise.

An insulated, ideal long straight wire is positioned parallel to the \(x\)-axis. It is displaced \(0.15\ \text{m}\) from O in the positive \(y\)-direction and carries a conventional current of \(2.0\ \text{A}\) in the positive \(x\)-direction.

Assume both conductors are in air and carry steady currents. Ignore any external magnetic fields.

a) Calculate the magnetic flux density at O due to the circular coil.

b) Calculate the magnetic flux density at O due to the straight wire.

c) Determine the magnitude and direction of the resultant magnetic field at O. Your answer should include the angle between the resultant field and the positive \(x\)-axis.

d) A student argues that because the coil's field is much stronger, the resultant field must point exactly along the positive \(x\)-axis. Explain the error.

e) The current in the coil is now reversed, but the current in the straight wire stays unchanged. Without repeating every calculation, determine whether the magnitude of the resultant field changes and describe its new direction.

[IMAGE: Three-dimensional spatial arrangement of a circular coil lying in the yz-plane, centred at O, with its axis along x. Show the coil radius of 0.10 m and anticlockwise conventional current as viewed from the positive x-axis. Draw a separate insulated straight wire parallel to the x-axis at y = +0.15 m, z = 0, carrying current in the positive x-direction. Mark O, the perpendicular distance from O to the wire, and the positive x, y, and z directions. Do not draw the resultant magnetic field or its angle.]

### Solution 5

The resultant field has a magnitude of approximately \(1.257\times10^{-4}\ \text{T}\) and points \(1.22^\circ\) away from the positive \(x\)-axis towards the negative \(z\)-direction.

The key is that the two magnetic fields are perpendicular, not parallel. Even though the coil produces a much stronger field, the straight wire still changes the direction of the resultant.

**Step 1**: Calculate the magnetic field produced by the coil.

The coil has:

- \(N=50\)
- \(I=0.40\ \text{A}\)
- \(R=0.10\ \text{m}\)

Using the circular coil equation:

\[
\begin{aligned}
B_{\text{coil}}&=\frac{\mu_0NI}{2R}\\
&=\frac{(4\pi\times10^{-7})(50)(0.40)}
{2(0.10)}\\
&=1.257\times10^{-4}\ \text{T}
\end{aligned}
\]

The current is anticlockwise when viewed from the positive \(x\)-axis.

The right-hand grip rule therefore gives a magnetic field in the positive \(x\)-direction.

**Step 2**: Calculate the magnetic field produced by the straight wire.

The wire has:

- \(I=2.0\ \text{A}\)
- \(r=0.15\ \text{m}\)

Using the long straight wire equation:

\[
\begin{aligned}
B_{\text{wire}}&=\frac{\mu_0I}{2\pi r}\\
&=\frac{(4\pi\times10^{-7})(2.0)}
{2\pi(0.15)}\\
&=2.67\times10^{-6}\ \text{T}
\end{aligned}
\]

The wire's current flows in the positive \(x\)-direction.

At O, the point is displaced from the wire in the negative \(y\)-direction. Applying the right-hand grip rule, the magnetic field points in the negative \(z\)-direction.

The two fields are therefore perpendicular.

**Step 3**: Combine the magnetic fields as vectors.

The coil produces a field along positive \(x\), while the straight wire produces a field along negative \(z\).

Because these directions are perpendicular, we use Pythagoras:

\[
B_{\text{resultant}}
=\sqrt{B_{\text{coil}}^2+B_{\text{wire}}^2}
\]

Substitute:

\[
\begin{aligned}
B_{\text{resultant}}
&=\sqrt{(1.257\times10^{-4})^2
+(2.67\times10^{-6})^2}\\
&=1.257\times10^{-4}\ \text{T}
\end{aligned}
\]

The straight wire makes only a small contribution to the magnitude, but it still affects the direction.

**Step 4**: Calculate the angle of the resultant.

Let \(\alpha\) be the angle between the resultant and the positive \(x\)-axis.

\[
\tan\alpha
=\frac{B_{\text{wire}}}{B_{\text{coil}}}
\]

Therefore,

\[
\begin{aligned}
\alpha&=\tan^{-1}
\left(\frac{2.67\times10^{-6}}
{1.257\times10^{-4}}\right)\\
&=1.22^\circ
\end{aligned}
\]

The resultant field is directed \(1.22^\circ\) from positive \(x\), towards negative \(z\).

**Step 5**: Evaluate the student's claim.

The student is incorrect.

A stronger magnetic field does not automatically determine the exact direction of the resultant. Both fields contribute, and their directions must be included in the vector addition.

The straight wire's contribution is small, which is why the resultant lies very close to the coil's axis. But it is not exactly along that axis.

**Step 6**: Consider reversing the coil current.

Reversing the coil current reverses its magnetic field.

The coil now produces a field along negative \(x\), while the straight wire continues to produce a field along negative \(z\).

The two components remain perpendicular and retain their original magnitudes.

Consequently, the resultant magnitude is unchanged:

\[
\boxed{B_{\text{resultant}}
=1.257\times10^{-4}\ \text{T}}
\]

However, its direction changes. It now points \(1.22^\circ\) away from the negative \(x\)-axis towards negative \(z\).

This is the important distinction: reversing one current changes the direction of its magnetic field, but it does not necessarily reverse the entire resultant field.

The tempting assumption was that a field 47 times stronger would completely determine the resultant. The perpendicular component exposes why that reasoning fails. A smaller perpendicular vector always changes the resultant's direction, even when its effect on the magnitude is barely noticeable.

## 5. The decision process to use in an HSC question

When a question involves a coil and another current-carrying conductor, work through these decisions in order.

1. **Identify the geometry.** Is it a circular loop, a multi-turn circular coil, a long solenoid, or a straight wire?
2. **Choose the correct field equation.** Use the circular coil equation for the field at the centre of a circular coil, and the long-wire equation for a point near an ideal long straight wire.
3. **Establish conventional current.** If the diagram shows electron flow, reverse its direction before applying a right-hand rule.
4. **Find each field direction separately.** Do not assume all contributions reinforce one another.
5. **Combine the fields.** Add parallel contributions, subtract opposing contributions, and use vector components when the directions differ.
6. **Check the result.** Make sure the direction, units, and magnitude agree with the physical arrangement.

The central idea is straightforward: a coil produces a strong magnetic field because its turns are arranged so that their contributions reinforce one another. Once you understand that, the next step is to apply the same reasoning to solenoids, magnetic field interactions, and the forces and torques experienced by current-carrying coils.