What Happens When You Reverse Both Current and Magnetic Field?

A straight conductor sits in a magnetic field. Conventional current runs upwards, while the external magnetic field points to the right. The motor-effect force points into the page.

Now reverse the current and the entire magnetic field at the same time. Does the force reverse too?

No. The force remains in the same direction, with the same magnitude, provided the current magnitude, magnetic field magnitude, conductor length, and angle remain unchanged.

This is because reversing both the current and magnetic field produces two directional reversals that cancel each other. The force on the conductor is unchanged.

01First, separate the current, magnetic field, and force

Three quantities matter when analysing the motor effect.

• Current (\(I\)): The conventional current flowing through the conductor, measured in amperes (A).
• Magnetic flux density (\(B\)): The external magnetic field at the conductor, measured in teslas (T). Its direction is from the north pole towards the south pole.
• Force (\(F\)): The magnetic force acting on the current-carrying conductor, measured in newtons (N).

The important distinction is that \(B\) represents the magnetic field, not the force.

The magnetic field and current determine the force. Reversing either one changes the force direction, while reversing both leaves the original force direction unchanged.

For this explanation, assume a straight conductor in a uniform magnetic field. The conductor remains stationary, and its active length inside the field does not change.

02How could we test what happens?

Imagine investigating this with a straight wire placed between the pole pieces of an electromagnet.

The wire runs horizontally through the gap. Its ends connect to a DC power supply, with an ammeter connected in series to measure the current.

The electromagnet produces a magnetic field perpendicular to the wire. A separate power supply controls the electromagnet, allowing its field direction to be reversed independently of the current in the wire.

A force sensor measures the vertical force on the conductor.

Start with the current flowing to the right and the magnetic field pointing into the page.

The motor-effect force acts upwards.

Now consider three trials:

1. Reverse the current but leave the magnetic field unchanged.
2. Reverse the magnetic field but leave the current unchanged.
3. Reverse both the current and magnetic field.

What should the force sensor show?

For a fair comparison, keep the magnitudes of the current and magnetic field constant. Also keep the conductor in exactly the same position.

The following illustrative readings show the expected pattern. They are not measurements from an actual experiment.

Take upwards as the positive force direction.

Notice that Trial B reverses the force, but Trial C restores the original force direction.

This is the experimental pattern we want to explain.

A force of \(0.20\text{ N}\) is relatively small. To picture its scale, it is approximately equal to the weight of a \(20.4\text{ g}\) mass near Earth’s surface:

\[ m=\frac{F}{g} \]

\[ m=\frac{0.20}{9.8} =0.0204\text{ kg} \]

This is why a sensitive force sensor is useful.

Improving the experiment

A simple demonstration using a permanent magnet can introduce an experimental problem.

Physically turning the magnet around might change the distance between the pole pieces or disturb the conductor’s alignment. That could change the magnetic field strength, making the comparison unfair.

A better design uses an electromagnet. Reversing its coil current reverses the magnetic field without physically moving the pole pieces.

A Hall probe can check that the field magnitude remains approximately constant, while the ammeter confirms that the conductor current has the required magnitude.

This separates the effect of reversing the directions from unwanted changes in the experimental setup.

03Why reversing both directions leaves the force unchanged

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

\[ F=BIL\sin\theta \]

where:

• \(F\) is the magnetic force in newtons (N).
• \(B\) is the magnetic flux density in teslas (T).
• \(I\) is the conventional current in amperes (A).
• \(L\) is the length of conductor inside the magnetic field, in metres (m).
• \(\theta\) is the angle between the conventional current direction and the magnetic field.

This equation calculates the force magnitude. It does not tell you whether the force points into or out of the page.

For that, use Fleming’s left-hand rule:

• First finger: magnetic field direction.
• Second finger: conventional current direction.
• Thumb: force direction.

Keep your thumb, first finger, and second finger perpendicular to each other.

What happens when both directions reverse?

Return to the original arrangement:

• Current points upwards.
• Magnetic field points right.
• Force points into the page.

Now reverse the current. The second finger changes direction, so the thumb points out of the page.

Instead, imagine reversing only the magnetic field. The first finger changes direction, and the thumb again points out of the page.

But what if you reverse both?

The first and second fingers both change direction. The thumb returns to its original direction.

The force therefore points into the page again.

This is the key relationship:

The table assumes the same current magnitude, field magnitude, conductor length, and geometry, with a non-zero initial force.

There is a mathematical way to understand this result, too.

The directional relationship can be represented using a vector cross product:

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

Here, \(\vec L\) is a length vector pointing along the conventional current, and \(\vec B\) is the magnetic field vector.

Reversing both directions gives:

\[ \vec F'=I[(-\vec L)\times(-\vec B)] \]

The two negative signs cancel:

\[ \vec F'=I(\vec L\times\vec B) \]

Therefore:

\[ \boxed{\vec F'=\vec F} \]

This means the entire force vector remains unchanged, not just its magnitude.

You do not need to calculate a cross product to answer an HSC question. Fleming’s left-hand rule gives you the same directional result.

04Does this still work when the wire is at an angle?

Yes. The double-reversal principle is not limited to a conductor perpendicular to the magnetic field.

Remember:

\[ F=BIL\sin\theta \]

If both current and magnetic field reverse, the angle between their directions remains unchanged.

For example, an initial angle of \(35^\circ\) remains \(35^\circ\) after both directions reverse.

The sine term therefore stays the same, and the force direction is also unchanged.

There is an interesting difference if you reverse only one direction.

The angle changes from \(35^\circ\) to \(145^\circ\). However:

\[ \sin145^\circ=\sin35^\circ \]

The force magnitude remains unchanged, even though its direction reverses.

This is an easy detail to miss. A larger angle does not necessarily mean a larger motor-effect force.

For a fuller explanation of the angle dependence, see Maximum and Zero Motor-Effect Force in HSC Physics.

05When does the double-reversal rule need more care?

There are two important limitations.

1. The magnitudes must also remain unchanged

Reversing both directions preserves the force direction. It does not automatically preserve the force magnitude if the current or magnetic field strength changes.

For example, doubling the current while keeping everything else constant doubles the force magnitude.

Likewise, halving the magnetic field strength halves the force magnitude.

The relationship is:

\[ F\propto BI \]

when the conductor length and angle remain constant.

2. You must reverse the entire external magnetic field

This is a more subtle point.

Imagine a conductor placed in a region where the magnetic field comes from two sources:

• A permanent magnet.
• An electromagnet.

The magnetic field \(B\) in the motor-effect equation is the resultant external field at the conductor.

Reversing the electromagnet does not automatically reverse the permanent magnet’s field.

The two contributions must be added to determine the actual field experienced by the conductor.

This distinction becomes important in more complicated questions involving multiple magnetic field sources.

06A real-world contrast: The loudspeaker

A loudspeaker provides a useful example of what happens when only the current reverses.

Its voice coil sits inside the magnetic field of a permanent magnet.

When the audio current reverses, the magnetic field remains approximately constant, but the force on the voice coil reverses.

This allows the cone to move backwards and forwards.

If both the current and the entire magnetic field reversed together, the magnetic force would instead retain its original direction.

You can explore this application in How a Loudspeaker Turns Current into Sound: The Motor Effect.

07Questions and solutions

The following questions move from a straightforward double reversal to more complicated situations involving angled conductors, changing magnitudes, multiple wire segments, and competing magnetic fields.

Question 1

A straight conductor has an active length of \(0.18\text{ m}\) inside a uniform magnetic field of \(0.40\text{ T}\).

It carries a conventional current of \(2.0\text{ A}\) upwards, perpendicular to the magnetic field, which points to the right.

(a) Calculate the magnitude of the motor-effect force.

(b) Both the current and the entire magnetic field are reversed. State the new force magnitude and direction.

Solution 1

The force is \(0.14\text{ N}\) into the page, and it remains \(0.14\text{ N}\) into the page after both reversals.

Part (a)

Use:

\[ F=BIL\sin\theta \]

Since the conductor is perpendicular to the field:

\[ \theta=90^\circ \]

Substitute:

\[ \begin{aligned} F&=(0.40)(2.0)(0.18)\sin90^\circ\\ &=(0.40)(2.0)(0.18)\\ &=0.144\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F=0.14\text{ N}} \]

Part (b)

Initially, the current points upwards and the magnetic field points right.

Fleming’s left-hand rule gives a force into the page.

After both directions reverse:

• The current points downwards.
• The magnetic field points left.

The two reversals cancel, leaving the force in its original direction.

The magnitudes of \(B\), \(I\), and \(L\) have not changed, so the force magnitude is also unchanged.

\[ \boxed{F=0.14\text{ N into the page}} \]

A common mistake is to assume that two reversed inputs must produce a reversed force. The directional relationship is a combination of the two inputs, not a response to either one independently.

Question 2

A straight conductor has an active length of \(0.30\text{ m}\) inside a uniform magnetic field of \(0.80\text{ T}\).

The conductor carries a conventional current of \(3.5\text{ A}\), directed \(35^\circ\) above the rightward magnetic field.

The current is reversed, but the magnetic field remains unchanged.

(a) Calculate the original force magnitude.

(b) Calculate the new force magnitude.

(c) Explain what happens to the force direction and why the magnitude does not change.

Solution 2

The force magnitude is approximately \(0.48\text{ N}\) in both arrangements, but reversing the current alone reverses the force direction.

Part (a)

The initial angle is:

\[ \theta=35^\circ \]

Using:

\[ F=BIL\sin\theta \]

Substitute:

\[ \begin{aligned} F&=(0.80)(3.5)(0.30)\sin35^\circ\\ &=(0.84)(0.574)\\ &=0.482\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F=0.48\text{ N}} \]

With the current directed upwards and to the right, and the magnetic field directed right, the force is into the page.

Part (b)

Reversing the current changes the angle between the current and magnetic field to:

\[ \theta'=180^\circ-35^\circ \]

\[ \theta'=145^\circ \]

The new force magnitude is:

\[ F'=BIL\sin145^\circ \]

Since:

\[ \sin145^\circ=\sin35^\circ \]

we obtain:

\[ \boxed{F'=0.48\text{ N}} \]

Part (c)

The force now points out of the page.

The tempting mistake is to assume that \(145^\circ\) produces a larger force than \(35^\circ\), because the angle is larger.

However, the sine of the two angles is identical.

The magnitude therefore remains unchanged, while Fleming’s left-hand rule shows that the force direction reverses.

Question 3

A straight conductor has an active length of \(0.25\text{ m}\) and is positioned at \(60^\circ\) to a uniform magnetic field.

Initially:

• Magnetic flux density is \(0.40\text{ T}\).
• Current is \(3.0\text{ A}\).
• The force points into the page.

The current and magnetic field are then both reversed. However, the new magnetic flux density is \(0.50\text{ T}\), and the new current is \(2.0\text{ A}\).

The conductor’s geometry remains unchanged.

(a) Calculate the initial force magnitude.

(b) Calculate the final force magnitude.

(c) Determine the final force direction and explain whether the force vector is unchanged.

Solution 3

The final force remains directed into the page, but its magnitude decreases from \(0.26\text{ N}\) to \(0.22\text{ N}\). The force vector is therefore not unchanged.

Part (a)

Initially:

\[ B_1=0.40\text{ T} \]

\[ I_1=3.0\text{ A} \]

\[ L=0.25\text{ m} \]

\[ \theta=60^\circ \]

Using:

\[ F_1=B_1I_1L\sin60^\circ \]

Substitute:

\[ \begin{aligned} F_1&=(0.40)(3.0)(0.25)(0.866)\\ &=0.2598\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F_1=0.26\text{ N}} \]

Part (b)

After both directions reverse:

\[ B_2=0.50\text{ T} \]

\[ I_2=2.0\text{ A} \]

The angle remains \(60^\circ\).

Therefore:

\[ F_2=B_2I_2L\sin60^\circ \]

Substitute:

\[ \begin{aligned} F_2&=(0.50)(2.0)(0.25)(0.866)\\ &=0.2165\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F_2=0.22\text{ N}} \]

Part (c)

The final force points into the page, the same direction as the initial force.

However, its magnitude has decreased.

The reason is that the product of magnetic flux density and current has changed:

Initially:

\[ B_1I_1=(0.40)(3.0)=1.20 \]

Finally:

\[ B_2I_2=(0.50)(2.0)=1.00 \]

The product is smaller in the second arrangement.

This question exposes an important distinction:

• Reversing both directions preserves the force direction.
• Preserving the force magnitude also requires the relevant magnitudes and geometry to remain unchanged.

A double reversal does not guarantee an identical force if other conditions have changed.

Question 4

An L-shaped conductor lies in the plane of the page.

Current enters at the lower-left end, travels:

• \(0.20\text{ m}\) horizontally to the right.
• Then \(0.15\text{ m}\) vertically upwards.

Both straight sections are completely inside a uniform magnetic field of \(0.40\text{ T}\), directed into the page.

The current is \(3.0\text{ A}\). Assume the bend is very short and its contribution to the total force is negligible.

(a) Calculate the magnetic force on each straight section.

(b) Determine the magnitude and direction of the resultant force on the conductor.

(c) If both the current and the entire magnetic field are reversed, determine the new resultant force.

Solution 4

The resultant force is \(0.30\text{ N}\), directed \(36.9^\circ\) west of north. Reversing both the current and magnetic field leaves this resultant completely unchanged.

Part (a)

Both straight sections are perpendicular to the magnetic field.

Therefore:

\[ F=BIL \]

For the horizontal section:

\[ \begin{aligned} F_H&=(0.40)(3.0)(0.20)\\ &=0.24\text{ N} \end{aligned} \]

Using Fleming’s left-hand rule, the force acts upwards.

For the vertical section:

\[ \begin{aligned} F_V&=(0.40)(3.0)(0.15)\\ &=0.18\text{ N} \end{aligned} \]

The force acts towards the left.

Part (b)

The two forces are perpendicular to each other.

They cannot be added as ordinary scalar magnitudes. Instead, use vector addition.

The horizontal force is \(0.18\text{ N}\) to the left, and the vertical force is \(0.24\text{ N}\) upwards.

The resultant magnitude is:

\[ F_R=\sqrt{F_H^2+F_V^2} \]

Substitute:

\[ \begin{aligned} F_R&=\sqrt{0.24^2+0.18^2}\\ &=\sqrt{0.0576+0.0324}\\ &=\sqrt{0.0900}\\ &=0.30\text{ N} \end{aligned} \]

To find the direction, measure the angle west of north:

\[ \tan\theta=\frac{0.18}{0.24} \]

\[ \theta=\tan^{-1}(0.75) \]

\[ \theta=36.9^\circ \]

Therefore:

\[ \boxed{F_R=0.30\text{ N at }36.9^\circ \text{ west of north}} \]

Part (c)

Reversing the current reverses the current direction in both conductor sections.

Reversing the magnetic field changes its direction for both sections.

For each section, both directional inputs reverse, so each magnetic force remains unchanged.

Consequently, the two individual forces retain their original directions and magnitudes.

Their resultant is also unchanged:

\[ \boxed{F_R=0.30\text{ N at }36.9^\circ \text{ west of north}} \]

A tempting mistake is to assume that the forces on the two sections cancel because the current changes direction at the bend.

They do not. The forces are perpendicular, not opposite. They must be added as vectors.

Question 5

A straight vertical conductor has an active length of \(0.25\text{ m}\) and carries a conventional current of \(2.0\text{ A}\) upwards.

It sits in a magnetic field produced by two separate sources:

• A permanent magnet produces a magnetic field of \(0.30\text{ T}\) to the right.
• An electromagnet produces a magnetic field of \(0.20\text{ T}\), also to the right.

Both fields are perpendicular to the conductor and point in the same direction.

The current in the conductor is reversed. At the same time, the current in the electromagnet is reversed, but the permanent magnet is left unchanged.

Assume both magnetic field contributions are uniform and collinear at the conductor.

(a) Calculate the initial resultant magnetic field and the initial force magnitude and direction.

(b) Calculate the new resultant magnetic field and the new force magnitude and direction after both currents are reversed.

(c) Now suppose the electromagnet initially produces \(0.30\text{ T}\) instead of \(0.20\text{ T}\). What happens to the final force?

(d) Repeat the reasoning if the electromagnet initially produces \(0.40\text{ T}\). Explain why the final force direction differs between these cases.

Solution 5

For the original \(0.20\text{ T}\) electromagnet, the force changes from \(0.25\text{ N}\) into the page to \(0.05\text{ N}\) out of the page.

The reason is that reversing the electromagnet does not reverse the entire magnetic field. The permanent magnet continues producing a rightward field.

This is the most important constraint in the question.

Part (a)

Initially, both magnetic fields point right.

Therefore, the resultant magnetic field is the sum of their magnitudes:

\[ B_{\text{total}}=B_P+B_E \]

\[ B_{\text{total}}=0.30+0.20 \]

\[ B_{\text{total}}=0.50\text{ T} \]

The current points upwards, and the resultant field points right.

Using:

\[ F=BIL \]

because the conductor is perpendicular to the field:

\[ \begin{aligned} F&=(0.50)(2.0)(0.25)\\ &=0.25\text{ N} \end{aligned} \]

Using Fleming’s left-hand rule, the force points into the page.

Therefore:

\[ \boxed{F=0.25\text{ N into the page}} \]

Part (b)

Now reverse the current in the conductor and the current in the electromagnet.

The permanent magnet still produces a rightward field of \(0.30\text{ T}\).

However, the electromagnet now produces a leftward field of \(0.20\text{ T}\).

The fields oppose each other, so the resultant is:

\[ B_{\text{total}}=0.30-0.20 \]

\[ B_{\text{total}}=0.10\text{ T} \]

The resultant magnetic field still points right because the permanent magnet’s contribution is larger.

The conductor current, however, now points downwards.

Calculate the new force:

\[ \begin{aligned} F'&=B_{\text{total}}IL\\ &=(0.10)(2.0)(0.25)\\ &=0.050\text{ N} \end{aligned} \]

Since the current has reversed but the resultant magnetic field has not, the force direction reverses.

Therefore:

\[ \boxed{F'=0.050\text{ N out of the page}} \]

The tempting answer is that both directions were reversed, so the force must remain unchanged.

The problem is that only one contribution to the magnetic field was reversed. The total field was not.

Part (c)

Now the electromagnet initially produces:

\[ B_E=0.30\text{ T} \]

The initial resultant field is:

\[ B_{\text{initial}}=0.30+0.30 \]

\[ B_{\text{initial}}=0.60\text{ T} \]

The initial force is:

\[ F=(0.60)(2.0)(0.25) \]

\[ F=0.30\text{ N} \]

It points into the page.

After reversing the electromagnet, its field points left with a magnitude of \(0.30\text{ T}\).

The permanent magnet still produces \(0.30\text{ T}\) to the right.

Therefore:

\[ B_{\text{final}}=0.30-0.30=0 \]

The two magnetic fields cancel completely.

With no resultant external magnetic field:

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

There is no motor-effect force, even though the conductor still carries current.

Part (d)

Finally, suppose the electromagnet initially produces:

\[ B_E=0.40\text{ T} \]

Initially, the two fields add:

\[ B_{\text{initial}}=0.30+0.40 \]

\[ B_{\text{initial}}=0.70\text{ T} \]

The initial force is:

\[ F=(0.70)(2.0)(0.25) \]

\[ F=0.35\text{ N} \]

It points into the page.

After reversing the electromagnet, the fields oppose each other:

\[ B_{\text{final}}=0.30-0.40 \]

\[ B_{\text{final}}=-0.10\text{ T} \]

The negative sign indicates that the resultant field now points left.

The current in the conductor also points downwards.

Therefore, both the current and the resultant magnetic field have reversed.

The final force is:

\[ \begin{aligned} F'&=(0.10)(2.0)(0.25)\\ &=0.050\text{ N} \end{aligned} \]

The force points into the page, the same direction as initially.

\[ \boxed{F'=0.050\text{ N into the page}} \]

The three cases can be summarised as follows:

The deeper lesson is that you must reverse the actual magnetic field experienced by the conductor, not simply reverse the current in one device that produces part of that field.

08The key decision for HSC Physics

When a question asks what happens after reversing the current and magnetic field, follow this order:

1. Identify the conventional current direction.
2. Identify the direction of the external resultant magnetic field.
3. Use \(F=BIL\sin\theta\) to calculate the force magnitude.
4. Use Fleming’s left-hand rule to determine its direction.
5. Check whether the entire magnetic field has reversed, and whether its magnitude and the current magnitude have stayed constant.

For a single uniform magnetic field, reversing both current and field leaves the motor-effect force unchanged.

The next useful step is applying this reasoning to multiple conductor sections and current-carrying coils, where the individual magnetic forces must be combined to determine the resultant force or torque.