HSC Physics Right-Hand Rules: Magnetic Fields vs Forces

A current-carrying wire can produce a magnetic field, but it can also experience a magnetic force when placed in another magnetic field. These are two different effects, and confusing them is one of the easiest ways to reverse an answer in an HSC Physics question.

Imagine a straight wire carrying conventional current upwards. First, it produces a magnetic field that circles around the wire. Now place that same wire between the poles of a magnet. The external magnetic field can exert a force on the wire, pushing it in a third direction.

So which way should your right hand point?

The answer depends on what you are trying to find. There are three closely related rules, but they do different jobs:

| Situation | Rule | What your hand tells you |
|---|---|---|
| Magnetic field around a straight wire | Right-hand grip rule | Thumb points along conventional current; curled fingers show the magnetic field. |
| Magnetic field inside a solenoid | Right-hand grip rule | Curled fingers follow conventional current around the coils; thumb points towards the north pole and along the internal magnetic field. |
| Magnetic force on a current-carrying wire | Right-hand cross-product rule | First finger points along conventional current, middle finger along the external magnetic field, and thumb gives the force. |

The key distinction is simple: **the grip rule finds a magnetic field produced by a current, while the cross-product rule finds a force caused by an external magnetic field.**

[IMAGE: Three-panel comparison showing (1) a straight vertical wire with conventional current upwards and circular magnetic field lines, (2) a horizontal solenoid with current circulating around its turns, its north and south poles labelled, and an internal magnetic field arrow, and (3) a straight current-carrying wire between the poles of a horseshoe magnet with mutually perpendicular conventional current, external magnetic field and magnetic force directions. Use consistent right-handed orientations and clearly distinguish the field produced by the wire from the external field acting on it.]

## 1. The first experiment: a current creates a magnetic field

Before comparing the rules, consider the experiment that established why a current-carrying wire needs to be treated as a magnetic object.

In 1820, Danish physicist Hans Christian Ørsted investigated whether an electric current could affect a nearby compass. At the time, electricity and magnetism were often treated as separate phenomena.

His apparatus was straightforward:

- A straight conducting wire was positioned above a compass needle.
- The wire was connected to a battery and a switch.
- With the switch open, no current flowed through the wire.
- Closing the switch allowed current to flow, while the compass needle's direction was observed.

What should happen if a current produces a magnetic field?

The compass should respond to the combined magnetic field of Earth and the current-carrying wire. It should no longer point in exactly the same direction.

That is what happened. The needle deflected when current flowed. Reversing the current reversed the direction of the deflection.

[IMAGE: Historically informed educational illustration of Ørsted's experiment, showing a straight horizontal wire positioned directly above a compass, connected through a switch to a battery. Include two side-by-side states: switch open with the compass aligned north-south, and switch closed with current flowing and the needle deflected. Clearly show the wire's position relative to the compass.]

The important conclusion was that an electric current produces a magnetic field around its conductor.

However, the experiment did not show that the magnetic field points along the wire. The compass response was consistent with a field that circles around it.

### The right-hand grip rule for a straight wire

Picture holding a straight wire in your right hand.

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

[IMAGE: Close-up three-dimensional view of a right hand gripping a straight vertical wire, thumb pointing upwards with an arrow labelled conventional current I, and curled fingers following circular magnetic field arrows around the wire. Include a small end-on view showing current out of the page as a dot and anticlockwise field circulation.]

For a wire perpendicular to the page, two symbols are especially useful:

- A dot, \(\odot\), represents current coming out of the page, towards you.
- A cross, \(\otimes\), represents current going into the page, away from you.

The corresponding field directions are:

| Conventional current | Magnetic field around the wire |
|---|---|
| Out of the page, \(\odot\) | Anticlockwise |
| Into the page, \(\otimes\) | Clockwise |

Try predicting the field direction if the current comes out of the page. Your thumb points towards you, and your fingers curl anticlockwise.

Notice that the magnetic field does not point towards or away from the wire. At any particular position, it points tangentially to the circular field line.

[IMAGE: End-on view of a straight wire carrying current out of the page, with four positions around it labelled top, bottom, left and right. Show anticlockwise circular field lines and tangent arrows at all four positions, demonstrating that the local magnetic field direction changes around the wire.]

This rule is useful when finding the magnetic field direction at a particular point near a conductor. It does not tell you the force on the wire.

For the underlying field pattern and its mathematical relationship with current and distance, 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. The same grip rule works for a solenoid

A solenoid is a long coil of wire containing many circular turns. Each turn produces a magnetic field, and the fields reinforce one another through much of the coil's interior.

Instead of a field that simply circles a straight wire, the combined field inside a long solenoid is approximately uniform and points along its central axis.

[IMAGE: Three-dimensional cutaway of a horizontal solenoid with closely spaced wire turns, conventional current arrows following the winding, a labelled north pole at the right end and south pole at the left end. Show dense, straight, parallel magnetic field arrows inside pointing from left to right, with weaker curved return field lines outside.]

The right-hand grip rule is almost identical to the straight-wire version:

1. Curl the fingers of your right hand in the direction of conventional current around the solenoid's turns.
2. Your thumb points along the magnetic field inside the solenoid.
3. Your thumb also identifies the solenoid's north pole.

This gives you two useful pieces of information at once: the internal magnetic field direction and the location of the north pole.

### A useful end-on shortcut

Sometimes a question shows a solenoid viewed directly from one end, rather than showing the full winding.

You can use this shortcut:

| Current viewed from the end | End of the solenoid |
|---|---|
| Anticlockwise | North |
| Clockwise | South |

Why does this work?

If the current appears anticlockwise, curl your right-hand fingers anticlockwise. Your thumb points towards you, so the magnetic field inside the solenoid points out through the end you are viewing. That end is north.

If the current appears clockwise, your thumb points away from you. That end is south.

[IMAGE: Two end-on cross-sectional views of a solenoid, side by side. In the left view, show anticlockwise conventional current around the circular turns, with the thumb and internal field pointing towards the viewer and the end labelled NORTH. In the right view, show clockwise current, with the thumb and internal field pointing away from the viewer and the end labelled SOUTH.]

### Worked example: Identify the poles of a solenoid

A horizontal solenoid carries conventional current anticlockwise when viewed from its left-hand end. Identify its north and south poles and state the direction of the magnetic field inside it.

**Step 1**: Identify the viewing direction.

The question specifies that you are looking directly at the left-hand end of the solenoid. The current appears anticlockwise from this viewpoint.

**Step 2**: Apply the solenoid end-on rule.

Anticlockwise conventional current means the end being viewed is north.

Therefore, the left-hand end is north.

**Step 3**: Identify the other pole and internal field direction.

The opposite end must be south. Inside a solenoid, the magnetic field points from south to north.

Therefore:

- Left-hand end: north.
- Right-hand end: south.
- Internal magnetic field: from right to left.

The important detail is the viewing direction. Clockwise and anticlockwise only have meaning when you know which end you are looking at.

For a more detailed treatment of field strength, turn density, and the ideal-solenoid approximation, see [Magnetic Field Inside a Solenoid: HSC Physics Explained](https://tutorgum.com/learn/resources/hsc-physics-magnetic-field-inside-a-solenoid/).

## 3. Magnetic force: a different question needs a different rule

Now consider a different experiment.

A straight wire is suspended between the poles of a horseshoe magnet. The magnetic field in the gap points from the north pole towards the south pole. The wire is connected to a battery so that conventional current flows through it.

What happens when the switch is closed?

The wire experiences a magnetic force. If the current-carrying section is free to move, it will be pushed sideways.

This is called the motor effect.

[IMAGE: Educational apparatus illustration of a straight horizontal conducting wire suspended between the north and south pole faces of a horseshoe magnet. Show the conventional current direction along the wire, the external magnetic field across the gap from N to S, electrical leads connected to a battery and switch, and space around the wire for its perpendicular force direction. Do not pre-draw a force arrow in the apparatus view.]

This experiment addresses a different question from Ørsted's compass experiment.

- Ørsted's experiment asks: what magnetic field does a current produce?
- The motor-effect experiment asks: what force does an external magnetic field exert on a current-carrying conductor?

That distinction determines which right-hand rule you use.

### The right-hand cross-product rule for magnetic force

For a straight wire carrying conventional current in a uniform external magnetic field, use the right-hand cross-product rule.

Arrange your right hand so that:

1. Your first finger points in the direction of conventional current.
2. Your middle finger points in the direction of the external magnetic field.
3. Your thumb points in the direction of the magnetic force.

All three directions are mutually perpendicular when the current and field are perpendicular.

[IMAGE: Clear three-dimensional right-hand cross-product demonstration with the index finger labelled conventional current I, middle finger labelled external magnetic field B, and thumb labelled magnetic force F. Show all three arrows at right angles, with the labels attached to the correct fingers. Include a small adjacent wire-between-magnet-poles illustration linking the hand orientation to the motor effect.]

The mathematical relationship is:

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

Here:

- \(\vec F\) is the magnetic force on the wire, measured in newtons (N).
- \(I\) is conventional current, measured in amperes (A).
- \(\vec L\) is the length vector of the wire section inside the magnetic field, pointing in the direction of conventional current, measured in metres (m).
- \(\vec B\) is the external magnetic flux density, measured in tesla (T).
- The symbol \(\times\) represents a vector cross product.

The magnitude of this force is:

\[
F = BIL\sin\theta
\]

In this equation, \(\theta\) is the angle between the current direction and the external magnetic field.

This gives three important cases:

| Angle between current and field | Magnetic force |
|---|---|
| \(0^\circ\) | Zero |
| \(90^\circ\) | Maximum, \(F=BIL\) |
| \(180^\circ\) | Zero |

The force is perpendicular to both the current and the magnetic field.

This is the crucial difference from the grip rule. The grip rule tells you how the magnetic field curls around a current. The cross-product rule combines two existing directions to determine a third direction: the force.

### What if the current or magnetic field is reversed?

Consider a wire carrying current to the right, with an external magnetic field pointing upwards.

The right-hand cross-product rule gives the force direction.

Now reverse just the current. The force reverses.

Instead, keep the current unchanged and reverse the magnetic field. The force also reverses.

Reverse both, however, and the force returns to its original direction.

| Current | External field | Force compared with original |
|---|---|---|
| Original | Original | Original direction |
| Reversed | Original | Opposite |
| Original | Reversed | Opposite |
| Reversed | Reversed | Original direction |

[IMAGE: Four-panel comparison of a straight wire in a uniform magnetic field. Keep the original current directed right and field directed upwards in panel one. Panel two reverses current only. Panel three reverses field only. Panel four reverses both. Show the corresponding force directions as mutually perpendicular arrows, with all four cases geometrically consistent.]

This is a useful experimental check. If the force does not reverse when only one of the two directions is reversed, something else may be affecting the setup, or the direction prediction is wrong.

A common mistake is to assume that reversing the current must reverse every magnetic effect in the situation. It reverses the field produced by the wire, but the force depends on the relationship between the current and the external field. These are separate interactions.

### What a controlled motor-effect experiment needs to show

The motor-effect experiment needs a little more care than simply connecting a wire to a battery and putting a magnet nearby.

The investigation has a specific question:

Does an external magnetic field exert a force on a current-carrying wire, and how does the direction of that force depend on the current and field?

A controlled arrangement uses:

- A horseshoe magnet with north and south pole faces opposite each other.
- A straight conducting wire suspended between the pole faces.
- A battery, switch, and ammeter connected in series with the wire.
- Flexible connecting leads routed away from the magnetic field region.

The magnetic field in the gap points from north to south. The short, straight section of wire between the poles is the part we want to investigate.

**Prediction:** If the current is perpendicular to the magnetic field, the wire should experience a force perpendicular to both directions.

Now test the prediction systematically.

| Investigation | What should happen to the force? |
|---|---|
| Switch on the current | The wire experiences a magnetic force. |
| Reverse the current only | The force reverses. |
| Reverse the magnet poles only | The force reverses. |
| Reverse both current and magnet poles | The force returns to its original direction. |

The observation to look for is sideways movement of the suspended wire. Reversing just one of the two directions should reverse that movement.

[IMAGE: Four-stage experimental apparatus showing a freely suspended straight conductor between horseshoe magnet poles, connected to a battery, switch and ammeter. Show the original arrangement, reversed current, reversed magnet poles, and both reversed. Draw the physical wire displacement in each state only in a separate explanatory version; the initial apparatus view should show the experimental connections and supplied current and pole orientations.]

There is an important experimental limitation here.

A poorly arranged circuit might have long connecting leads passing through the magnetic field. Those leads also experience magnetic forces. The magnet's field may also be less uniform near the edges of its pole faces.

These effects can make the observed movement harder to interpret.

A better design keeps the active wire section short, places it in the central region of the magnetic gap, and routes the return leads outside that region. Reversing one variable at a time then provides a much clearer test.

This is a modern controlled version of the motor-effect demonstration, rather than a claim about the exact apparatus used in a historical experiment.

**Conclusion:** A current-carrying conductor experiences a magnetic force when it has a component perpendicular to an external magnetic field. The direction depends on both the conventional current and the external field.

The historical distinction is worth keeping clear. Ørsted's 1820 experiment demonstrated that current affects a nearby compass. The motor-effect investigation examines the force on a conductor itself. The American Physical Society's historical account describes Ørsted's original observations and the reversal of the compass deflection when current was reversed. 

### A note on Fleming's left-hand rule

You might also encounter Fleming's left-hand rule in textbooks or class notes.

It is another way of finding the direction of magnetic force on a current-carrying conductor.

The finger assignments are different:

- First finger: magnetic field.
- Second finger: conventional current.
- Thumb: force.

This uses your left hand rather than the right-hand cross-product arrangement used in this article.

Both methods give the same physical answer when applied correctly. The mistake is mixing the finger positions from one convention with the other.

For consistency, we'll continue using the right-hand cross-product rule.

## 4. Conventional current: the direction that keeps the rules consistent

There is one more detail that affects all three rules: the direction of current.

Conventional current is defined as the direction in which positive charge would move.

In a metal wire, however, the mobile charge carriers are electrons. They drift in the opposite direction to conventional current.

[IMAGE: Side-by-side illustration of a metal conductor showing conventional current arrows pointing right and electron drift arrows pointing left. Include a simple battery connection, with the positive terminal on the left and negative terminal on the right, and clearly distinguish the two directions.]

For the three rules in this article, use conventional current:

| Situation | Direction to use |
|---|---|
| Field around a straight wire | Conventional current |
| Field inside a solenoid | Conventional current through the coils |
| Force on a current-carrying wire | Conventional current |
| Force on an individual electron | Actual electron velocity and negative charge |

Why is the last row different?

For an individual moving charge, the magnetic force is described by:

\[
\vec F = q\vec v \times \vec B
\]

Here:

- \(q\) is the particle's electric charge in coulombs (C).
- \(\vec v\) is its actual velocity in metres per second (m s\(^{-1}\)).
- \(\vec B\) is the magnetic flux density in tesla (T).

The right-hand cross-product rule gives the force direction for a positive charge when your fingers represent velocity and magnetic field.

An electron has negative charge, so its force points in the opposite direction.

For a current-carrying metal wire, you don't need to reverse the answer again. The conventional-current direction already accounts for the negative charge carriers.

## 5. Two worked examples: applying the force rule

The best way to check that you understand the distinction is to use the rules in situations where the current and field point in different directions.

### Worked example: Find the force on a perpendicular wire

A straight wire has a length of \(0.15\ \text{m}\) inside a uniform magnetic field of \(0.40\ \text{T}\). It carries a conventional current of \(2.4\ \text{A}\) to the right. The external magnetic field points into the page.

Calculate the magnetic force on the wire, including its direction.

[IMAGE: Straight horizontal wire carrying conventional current to the right, shown inside a uniform magnetic field represented by evenly spaced crosses indicating a field into the page. Show the active wire length of 0.15 m and leave the force direction undrawn.]

**Step 1**: Identify the quantity being calculated.

We need the magnetic force on a current-carrying conductor, so we use the cross-product rule and the force magnitude equation.

**Step 2**: Identify the angle.

The current points horizontally to the right, while the magnetic field points into the page.

These directions are perpendicular, so:

\[
\theta=90^\circ
\]

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

\[
F=BIL\sin\theta
\]

Substitute:

\[
\begin{aligned}
F&=(0.40)(2.4)(0.15)\sin90^\circ\\
&=0.144\ \text{N}
\end{aligned}
\]

**Step 4**: Find the direction.

Point your right index finger to the right, along the conventional current.

Point your middle finger into the page, along the magnetic field.

Your thumb points upwards.

Therefore, the magnetic force is upwards.

**Final result:**

\[
\boxed{F=0.144\ \text{N upwards}}
\]

The wire experiences an upward force because the external magnetic field and conventional current are perpendicular.

### Worked example: An angled conductor in a magnetic field

A straight conductor of length \(0.20\ \text{m}\) carries a conventional current of \(4.0\ \text{A}\) to the right. It is placed in a uniform magnetic field of \(0.50\ \text{T}\), directed \(30^\circ\) above the current direction in the plane of the page.

Calculate the magnitude and direction of the magnetic force. Then determine what happens if the current is reversed while the magnetic field stays unchanged.

[IMAGE: Horizontal conductor with current arrow pointing right, and a uniform magnetic field arrow pointing diagonally upwards and to the right at 30 degrees to the wire. Show the angle between the two vectors and leave the force direction undrawn.]

**Step 1**: Identify the angle between the current and field.

The magnetic field is \(30^\circ\) above the current direction.

Therefore:

\[
\theta=30^\circ
\]

**Step 2**: Apply the force equation.

\[
F=BIL\sin\theta
\]

Substitute:

\[
\begin{aligned}
F&=(0.50)(4.0)(0.20)\sin30^\circ\\
&=(0.50)(4.0)(0.20)(0.5)\\
&=0.20\ \text{N}
\end{aligned}
\]

**Step 3**: Determine the force direction.

Point your right index finger to the right.

Point your middle finger diagonally upwards and to the right.

Your thumb points out of the page.

Therefore, the force acts out of the page.

**Step 4**: Reverse the current.

Reversing the current reverses the first direction in the cross product, while the magnetic field stays unchanged.

The force must therefore reverse.

Its new direction is into the page, but its magnitude remains \(0.20\ \text{N}\).

**Final result:**

\[
\boxed{F=0.20\ \text{N out of the page}}
\]

After reversing the current:

\[
\boxed{F=0.20\ \text{N into the page}}
\]

Notice that the force magnitude depends on the sine of the angle, not simply the size of the magnetic field. A conductor parallel to the field would experience no magnetic force, even if the field were strong.

## 6. Which right-hand rule should you use?

Before attempting an HSC question, identify exactly what the question wants you to determine.

This short comparison is worth remembering.

| What the question asks | What to do |
|---|---|
| Direction of the magnetic field around a straight wire | Point your right thumb along conventional current; curl your fingers. |
| Direction of the field inside a solenoid | Curl your right fingers with conventional current; use your thumb. |
| North and south poles of a solenoid | Look at one end and use anticlockwise = north, clockwise = south. |
| Direction of force on a current-carrying wire | Point your index finger along conventional current, middle finger along external field, and thumb along force. |
| Magnitude of force on a wire | Use \(F=BIL\sin\theta\). |
| Direction of force on an individual electron | Use its actual velocity and reverse the positive-charge result. |

There are three particularly common traps.

**Trap 1: Using the grip rule to find force.**

The grip rule finds the field produced by a current. It does not directly tell you the force experienced by that conductor.

**Trap 2: Using the wire's own magnetic field in the force equation.**

When calculating the force on a wire in a magnet's field, use the external magnetic field acting on the wire. Do not substitute the circular field produced by that same wire.

**Trap 3: Using electron flow instead of conventional current.**

For the current-based rules, use conventional current. Only use actual electron velocity when analysing the force on an individual electron.

## Questions and solutions

The following questions move from a straightforward motor-effect calculation to problems that combine the grip rule, the cross-product rule, and the distinction between force and torque.

### Question 1

A straight conductor carries a conventional current of \(3.0\ \text{A}\) out of the page. It is placed in a uniform magnetic field of \(0.25\ \text{T}\), directed horizontally to the right.

The section of wire inside the field is \(0.10\ \text{m}\) long.

Calculate the magnetic force on the wire and state its direction.

[IMAGE: End-on view of a straight wire represented by a dot, indicating conventional current out of the page. Show a uniform magnetic field directed to the right with evenly spaced arrows and label the wire's active length as 0.10 m in an accompanying perspective view. Do not show the force direction.]

### Solution 1

The magnetic force is \(0.075\ \text{N}\), directed upwards.

The current is out of the page, while the external magnetic field points right. These directions are perpendicular.

**Step 1**: Identify the given values.

\[
\begin{aligned}
I&=3.0\ \text{A}\\
B&=0.25\ \text{T}\\
L&=0.10\ \text{m}\\
\theta&=90^\circ
\end{aligned}
\]

**Step 2**: Calculate the force magnitude.

\[
F=BIL\sin\theta
\]

\[
\begin{aligned}
F&=(0.25)(3.0)(0.10)\sin90^\circ\\
&=0.075\ \text{N}
\end{aligned}
\]

**Step 3**: Determine the direction.

Point your index finger out of the page and your middle finger to the right.

Your thumb points upwards.

[IMAGE: Right-hand cross-product solution for a conductor carrying current out of the page in a magnetic field pointing right. Show the resulting force arrow upwards, with all three directions labelled.]

Therefore:

\[
\boxed{F=0.075\ \text{N upwards}}
\]

The force is perpendicular to both the conventional current and the external magnetic field, as required by the cross-product relationship.

### Question 2

A long, straight vertical wire carries a conventional current of \(4.0\ \text{A}\) upwards.

Point \(P\) is located \(3.0\ \text{cm}\) directly to the right of the wire. The wire is also placed in an external uniform magnetic field of \(0.30\ \text{T}\), directed horizontally to the left.

The length of the wire inside the external field is \(0.20\ \text{m}\).

Determine:

**(a)** The direction of the magnetic field produced by the wire at point \(P\).

**(b)** The magnitude and direction of the magnetic force on the wire.

Explain why the magnetic field at \(P\) is not the field direction you use to calculate the force on the wire.

[IMAGE: Top-down and side-view combination of a vertical current-carrying wire, with conventional current upwards in the side view and point P positioned to its right in the horizontal cross-section. Show the external magnetic field directed left. Label the wire length 0.20 m and distance to P as 3.0 cm. Do not draw the wire's own magnetic field or the force.]

### Solution 2

The magnetic field produced by the wire at \(P\) points into the page. The magnetic force on the wire is \(0.24\ \text{N}\), directed out of the page.

The important detail is that these directions come from two different magnetic interactions.

**(a) Magnetic field at point P**

**Step 1**: Apply the right-hand grip rule.

Point your right thumb upwards, in the direction of conventional current.

Your fingers curl around the wire.

At a point directly to the right of the wire, the magnetic field points into the page.

Therefore:

\[
\boxed{B_{\text{wire at P}}\text{ is into the page}}
\]

**(b) Magnetic force on the wire**

**Step 1**: Identify the external magnetic field.

The external field points to the left.

This is the field supplied by the surrounding magnetic arrangement, not the circular field produced by the wire itself.

**Step 2**: Calculate the force magnitude.

The current and external field are perpendicular.

\[
F=BIL\sin90^\circ
\]

\[
\begin{aligned}
F&=(0.30)(4.0)(0.20)(1)\\
&=0.24\ \text{N}
\end{aligned}
\]

**Step 3**: Find the force direction.

Point your index finger upwards, along the conventional current.

Point your middle finger left, along the external field.

Your thumb points out of the page.

Therefore:

\[
\boxed{F=0.24\ \text{N out of the page}}
\]

The tempting mistake is to use the magnetic field at \(P\), which points into the page, as the field in the force calculation.

But that field is produced by the wire itself. The force equation uses the external magnetic field acting on the conductor.

This question tests whether you can distinguish a magnetic field produced by a current from a magnetic field that causes a force on that current.

### Question 3

A long horizontal solenoid is viewed directly from its left-hand end. The conventional current appears anticlockwise.

The magnetic flux density inside the solenoid is \(2.0\ \text{mT}\).

A short straight conductor is positioned at the centre of the solenoid. Its active length is \(0.080\ \text{m}\), and it carries a conventional current of \(1.5\ \text{A}\) vertically upwards in the plane of the page.

Determine:

**(a)** The north and south poles of the solenoid.

**(b)** The direction of the magnetic field inside the solenoid.

**(c)** The direction and magnitude of the magnetic force on the straight conductor.

**(d)** What happens to the force direction if the current through the solenoid is reversed but the current through the straight conductor stays unchanged?

[IMAGE: A horizontal solenoid shown in perspective with its left end facing the observer. Show anticlockwise conventional current around the left end and a straight test conductor positioned at the centre, with its current directed vertically upwards. Label the internal field magnitude as 2.0 mT and the active test-wire length as 0.080 m. Do not show the solenoid poles, internal field direction or force direction.]

### Solution 3

The left-hand end is north, the right-hand end is south, and the internal field points left. The test conductor experiences a force of \(2.4\times10^{-4}\ \text{N}\) out of the page.

**(a) Identify the poles**

**Step 1**: Use the end-on solenoid rule.

Anticlockwise conventional current, viewed from the left-hand end, means that end is north.

Therefore:

- Left-hand end: north.
- Right-hand end: south.

**(b) Find the internal magnetic field direction**

**Step 2**: Use the magnetic field convention for a solenoid.

Inside a solenoid, the field points from south to north.

Since the north pole is on the left, the internal field points from right to left.

**(c) Calculate the force**

**Step 3**: Convert the magnetic flux density into tesla.

\[
2.0\ \text{mT}=2.0\times10^{-3}\ \text{T}
\]

**Step 4**: Identify the directions.

The conventional current in the test conductor points upwards.

The magnetic field points left.

The two directions are perpendicular.

Using the right-hand cross-product rule, the force points out of the page.

**Step 5**: Calculate the magnitude.

\[
F=BIL\sin90^\circ
\]

\[
\begin{aligned}
F&=(2.0\times10^{-3})(1.5)(0.080)\\
&=2.4\times10^{-4}\ \text{N}
\end{aligned}
\]

Therefore:

\[
\boxed{F=2.4\times10^{-4}\ \text{N out of the page}}
\]

[IMAGE: Completed explanatory diagram of the solenoid and central test conductor. Label the left end N, right end S, internal magnetic field pointing left, conventional current in the test wire pointing upwards, and magnetic force pointing out of the page.]

**(d) Reverse the solenoid current**

Reversing the solenoid current reverses its magnetic field.

The test conductor's current remains upwards, so the first direction in the cross product is unchanged while the second direction reverses.

The magnetic force therefore reverses and now points into the page.

Its magnitude remains unchanged because the magnetic field strength and test-wire current have the same magnitudes as before.

This is an example of using the solenoid grip rule first, followed by the force rule. The order matters because you need to establish the external field direction before finding the force.

### Question 4

An electron beam travels horizontally to the right at a speed of \(3.0\times10^6\ \text{m s}^{-1}\) through a uniform magnetic field of \(0.20\ \text{T}\), directed vertically upwards in the plane of the page.

**(a)** Determine the direction and magnitude of the magnetic force on each electron.

**(b)** A metal wire has electrons drifting to the right through it in the same magnetic field. State the direction of conventional current in the wire.

**(c)** Determine the direction of the magnetic force on the current-carrying wire.

Explain why the direction procedure for the individual electron is different from the procedure for the entire wire.

Use the magnitude of the electron's charge as \(1.60\times10^{-19}\ \text{C}\).

[IMAGE: Two-part illustration. Left: electron beam travelling horizontally right through a magnetic field pointing vertically upwards, with electron symbols along the beam. Right: metal conductor showing electron drift right in the same upward magnetic field. Do not show either magnetic force or the conventional current direction in the wire.]

### Solution 4

The force on each electron is \(9.6\times10^{-14}\ \text{N}\) into the page. In the metal wire, conventional current points left, and the magnetic force on the wire also points into the page.

**(a) Force on an individual electron**

**Step 1**: Identify the particle's velocity and charge.

The electron moves right, and its charge is negative.

The magnitude of its charge is:

\[
|q|=1.60\times10^{-19}\ \text{C}
\]

**Step 2**: Apply the right-hand cross-product rule.

For a positive charge moving right through a field pointing upwards, the cross-product direction is out of the page.

However, an electron is negatively charged.

Its force therefore points in the opposite direction:

\[
\boxed{\text{Into the page}}
\]

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

For perpendicular velocity and magnetic field:

\[
F=|q|vB
\]

Substitute:

\[
\begin{aligned}
F&=(1.60\times10^{-19})(3.0\times10^6)(0.20)\\
&=9.6\times10^{-14}\ \text{N}
\end{aligned}
\]

Therefore:

\[
\boxed{F=9.6\times10^{-14}\ \text{N into the page}}
\]

**(b) Conventional current direction**

Electrons drift to the right.

Conventional current is defined in the opposite direction to electron flow in a metal conductor.

Therefore:

\[
\boxed{\text{Conventional current points left}}
\]

**(c) Force on the metal wire**

**Step 4**: Apply the current-carrying wire rule.

The conventional current points left, while the external magnetic field points upwards.

Point your right index finger left and your middle finger upwards.

Your thumb points into the page.

Therefore, the force on the wire is into the page.

[IMAGE: Completed comparison showing the electron's velocity to the right, upward magnetic field and electron force into the page. Beside it, show the metal wire with conventional current to the left and the force on the conductor into the page.]

The subtle point is that the two calculations start with different information.

For an individual electron, use its actual velocity and negative charge.

For a current-carrying wire, use conventional current directly in the force equation. Its direction already represents the equivalent positive-charge flow.

If you reverse the electron's velocity using conventional current and then reverse the force again for negative charge, you would count the sign change twice.

### Question 5

A one-turn rectangular conducting loop is mounted on a frictionless vertical axle through its centre. It sits inside the central region of a long solenoid, where the magnetic field is uniform.

The solenoid's internal magnetic field has a magnitude of \(0.30\ \text{T}\) and points horizontally to the right.

The rectangular loop lies initially in the plane of the page, with its horizontal sides parallel to the solenoid's axis. Its dimensions are:

- Horizontal width: \(0.12\ \text{m}\).
- Vertical height: \(0.080\ \text{m}\).

The loop carries a steady conventional current of \(2.0\ \text{A}\), clockwise when viewed from the front.

Assume the loop is entirely within the uniform-field region, the current remains constant, and the axle is frictionless. Ignore fringe fields and the magnetic field produced by the loop itself when calculating the forces.

**(a)** Determine the direction and magnitude of the magnetic force on each of the four sides of the loop.

**(b)** Determine the resultant force on the loop.

**(c)** Calculate the magnitude of the torque acting on the loop about its vertical axle. Explain why the loop can rotate even though its resultant force is zero.

**(d)** Suppose the loop rotates through \(90^\circ\), so that its plane becomes perpendicular to the magnetic field. What happens to the magnetic torque?

**(e)** Return to the original orientation. Predict what happens to the force and torque directions if:

- Only the solenoid current is reversed.
- Only the loop current is reversed.
- Both currents are reversed.

[IMAGE: Question setup showing a rectangular one-turn loop in the plane of the page, positioned centrally inside a horizontal solenoid. The solenoid axis and supplied internal magnetic field direction point right. Show the vertical axle through the centre, clockwise conventional current around the loop, and dimensions of 0.12 m horizontal width and 0.080 m vertical height. Do not draw magnetic forces, torque arrows or the solenoid poles.]

### Solution 5

The loop experiences two equal and opposite forces of \(0.048\ \text{N}\) on its vertical sides. The resultant force is zero, but these forces form a couple that produces a torque of \(5.76\times10^{-3}\ \text{N m}\).

This is the important result: zero resultant force does not necessarily mean zero rotational effect.

**(a) Forces on the four sides**

**Step 1**: Identify the current direction on each side.

The loop current is clockwise when viewed from the front.

Therefore:

- Top side: current to the right.
- Right side: current downwards.
- Bottom side: current to the left.
- Left side: current upwards.

The external magnetic field points to the right.

**Step 2**: Find the force on the horizontal sides.

The top and bottom sides are parallel and antiparallel to the magnetic field.

For both sides:

\[
\theta=0^\circ\text{ or }180^\circ
\]

Therefore:

\[
F=BIL\sin\theta=0
\]

Neither horizontal side experiences a magnetic force.

**Step 3**: Find the force on the vertical sides.

The vertical sides are perpendicular to the magnetic field.

Their active length is:

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

The force magnitude on either side is:

\[
F=BIL
\]

Substitute:

\[
\begin{aligned}
F&=(0.30)(2.0)(0.080)\\
&=0.048\ \text{N}
\end{aligned}
\]

Now determine the directions.

On the left side, current points upwards and the field points right. The force points into the page.

On the right side, current points downwards and the field points right. The force points out of the page.

Therefore:

| Loop section | Magnetic force |
|---|---|
| Top | Zero |
| Bottom | Zero |
| Left | \(0.048\ \text{N}\) into the page |
| Right | \(0.048\ \text{N}\) out of the page |

[IMAGE: Completed solution diagram of the rectangular loop in the horizontal magnetic field. Show the left vertical side experiencing a force into the page, the right vertical side experiencing an equal force out of the page, and zero force on the top and bottom sides. Include the vertical axle and label the equal force magnitudes.]

**(b) Resultant force**

The two non-zero forces have equal magnitudes and opposite directions.

Therefore, they cancel:

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

The loop has no resultant translational force.

But this does not settle whether it rotates.

**(c) Calculate the torque**

**Step 4**: Identify the lever arms.

The two vertical forces act on opposite sides of the loop.

Each force acts a perpendicular distance of half the loop's width from the central axle:

\[
r=\frac{0.12}{2}=0.060\ \text{m}
\]

Both forces produce torque in the same rotational sense.

They form a couple.

**Step 5**: Calculate the total torque.

Torque from one force is:

\[
\tau=Fr
\]

Since there are two equal forces producing torque in the same direction:

\[
\tau=2F\left(\frac{w}{2}\right)
\]

Therefore:

\[
\tau=Fw
\]

Substitute:

\[
\begin{aligned}
\tau&=(0.048)(0.12)\\
&=0.00576\ \text{N m}
\end{aligned}
\]

Hence:

\[
\boxed{\tau=5.76\times10^{-3}\ \text{N m}}
\]

The loop can therefore rotate despite having zero resultant force.

The two forces cancel as a net force, but their separated lines of action create a turning effect. This is called a couple.

A useful comparison is a steering wheel. You can push one side forwards and the other backwards with equal forces. The overall force can be zero while the wheel still turns.

The analogy describes the rotational effect, not the magnetic mechanism producing it.

**(d) Rotate the loop through \(90^\circ\)**

After a quarter-turn, the plane of the loop becomes perpendicular to the magnetic field.

The loop's normal is now aligned with the field.

The magnetic torque becomes zero.

Why?

The magnetic forces on the loop no longer produce a turning couple about its axle. The loop has reached an orientation in which its magnetic moment is aligned with the external field.

The torque decreases as the loop rotates towards this orientation.

Thus:

\[
\boxed{\tau=0\ \text{N m at this orientation}}
\]

The important distinction is that the loop can experience magnetic forces on its individual sides without necessarily experiencing a net torque.

**(e) Reverse the currents**

Return to the original orientation.

The magnetic force on each side depends on the cross product of the current direction and magnetic field direction.

Consider the three possibilities.

| Change | Magnetic field | Loop current | Force and torque |
|---|---|---|---|
| Reverse solenoid current only | Reverses | Unchanged | Both force directions and torque reverse |
| Reverse loop current only | Unchanged | Reverses | Both force directions and torque reverse |
| Reverse both currents | Reverses | Reverses | Original force directions and torque are restored |

The final case is the most interesting.

Reversing both vectors leaves their cross product unchanged:

\[
(-\vec I)\times(-\vec B)
=
\vec I\times\vec B
\]

Therefore, the force on each side remains in its original direction, and the torque also remains unchanged.

The tempting prediction is that reversing both currents must reverse the force twice and therefore produce the opposite torque. But two reversals restore the original vector relationship.

This question brings all three rules together:

- The solenoid grip rule establishes the magnetic field direction.
- The current direction around the loop determines the direction of the conductor's current on each side.
- The right-hand cross-product rule determines the force on each side.
- The arrangement of those forces determines the torque.

A zero resultant force and a non-zero torque can exist at the same time. That is the central idea behind the turning effect in a simple electric motor.

## Where this leads next

The next useful step is understanding how a current-carrying coil behaves in a magnetic field and how a DC motor maintains rotation.

The rules in this article give you the tools to work out the force on each side of the coil. The next challenge is understanding how a commutator reverses the coil's current at the right point in its rotation so that the turning effect continues in the same rotational sense.