How to Find Magnetic Force on an Angled Wire | HSC Physics

A straight wire carries current through a magnetic field, but it isn’t perpendicular to the field. Does it still experience the full magnetic force?

Suppose the wire is tilted at \(30^\circ\) to the magnetic field. You might be tempted to use the cosine of the angle, or perhaps calculate the force as if the wire were perpendicular to the field. Neither is correct.

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

\[ F=BIL\sin\theta \]

The crucial detail is that \(\theta\) is the angle between the direction of conventional current and the magnetic field \(B\).

If the wire makes an angle of \(30^\circ\) with the field, you use \(\sin30^\circ\). The resulting force is half the maximum force the same wire would experience if it were perpendicular to the field.

Let’s work through how to identify the correct angle, calculate the force, and determine its direction.

01Understanding the angle in the magnetic force equation

Imagine a straight wire placed between the north and south poles of a magnet. The magnetic field travels from north to south through the space between the poles.

Now run an electric current through the wire, with the wire tilted relative to the magnetic field.

The wire experiences a magnetic force. This is the motor effect.

The important point is that the force depends on how the current-carrying wire is oriented relative to the magnetic field, not simply on how much wire is present.

Why does the equation contain sine?

Think about what happens when you rotate the wire.

• If the wire is perpendicular to the magnetic field, the magnetic force is at its maximum.
• If you tilt the wire, the force becomes smaller.
• If the wire becomes parallel to the magnetic field, the force falls to zero.

Why?

Only the component of the conductor’s length perpendicular to the magnetic field contributes to the magnetic force.

If the actual length of the wire is \(L\), its perpendicular component is:

\[ L_{\perp}=L\sin\theta \]

Substituting this into the maximum-force relationship gives:

\[ F=BI(L\sin\theta) \]

Therefore:

\[ \boxed{F=BIL\sin\theta} \]

The sine term accounts for the angle of the conductor relative to the field.

What do the symbols mean?

Be careful not to confuse \(B\) and \(F\).

• \(B\) describes the magnetic field at the conductor’s position. It is measured in tesla.
• \(F\) is the force experienced by the conductor. It is measured in newtons.

They are different physical quantities, and they point in different directions.

What happens at different angles?

The following table shows how the angle changes the force when \(B\), \(I\), and \(L\) remain constant.

Notice that a conductor perpendicular to the magnetic field experiences the maximum force because its entire length contributes to the perpendicular component.

A conductor parallel to the field experiences no force because there is no perpendicular component.

This gives you a useful prediction before you even reach for your calculator.

02How to identify the correct angle

This is where many otherwise straightforward HSC calculations go wrong.

A question might give you an angle between the wire and the vertical, while the magnetic field runs horizontally. You cannot automatically substitute the supplied angle into the equation.

You need the angle between the current and the magnetic field.

When the angle is given directly

Suppose a wire carries current at \(35^\circ\) to a magnetic field.

The angle required is simply:

\[ \theta=35^\circ \]

Use:

\[ F=BIL\sin35^\circ \]

When the angle is given relative to the perpendicular

Suppose the magnetic field is horizontal, but the wire makes an angle of \(25^\circ\) with the vertical.

The vertical direction is perpendicular to the horizontal magnetic field.

Therefore, the angle between the wire and the magnetic field is:

\[ \theta=90^\circ-25^\circ \]

\[ \theta=65^\circ \]

You must use \(\sin65^\circ\), not \(\sin25^\circ\).

Equivalently, you could use:

\[ \sin65^\circ=\cos25^\circ \]

Both approaches give the same result.

A quick decision table

One more important detail: the angle is defined using the direction of conventional current, not electron flow.

For calculating the magnitude, angles of \(30^\circ\) and \(150^\circ\) give the same sine. However, reversing the current reverses the magnetic force direction. We will examine that distinction later.

03A reliable method for calculating magnetic force

When you encounter an angled conductor in an HSC Physics question, follow this process.

Step 1: Identify the magnetic field.

Find the direction and magnitude of \(B\). Make sure you are using the field acting on the conductor, rather than confusing it with the force.

Step 2: Identify the current direction.

Use conventional current to establish the direction along the wire.

Step 3: Find the angle.

Determine the angle between the current direction and the magnetic field. Convert any angle measured from the perpendicular if necessary.

Step 4: Check the length.

Use the actual length of conductor inside the magnetic field, in metres. Do not automatically use the total length of a wire if part of it lies outside the field.

Step 5: Substitute into the equation.

\[ F=BIL\sin\theta \]

Check that:

• \(B\) is in tesla.
• \(I\) is in amperes.
• \(L\) is in metres.
• Your calculator is in degree mode if the angle is given in degrees.

Step 6: Determine the force direction.

The equation calculates the magnitude. You need a separate direction rule to determine whether the force acts into or out of the page, or in another perpendicular direction.

04Common mistakes when finding magnetic force

There are a few traps worth recognising before attempting exam questions.

A particularly important comparison

Consider two conductors with the same \(B\), \(I\), and \(L\).

• Conductor A makes an angle of \(30^\circ\) with the magnetic field.
• Conductor B makes an angle of \(30^\circ\) with the direction perpendicular to the magnetic field.

For conductor A:

\[ F_A=BIL\sin30^\circ \]

\[ F_A=0.5BIL \]

For conductor B:

\[ \theta=90^\circ-30^\circ=60^\circ \]

Therefore:

\[ F_B=BIL\sin60^\circ \]

\[ F_B=0.866BIL \]

Conductor B experiences a greater force, even though both questions mention an angle of \(30^\circ\).

The difference comes entirely from what the angle is measured relative to.

05What if the conductor is not entirely inside the magnetic field?

The equation also requires you to pay attention to the length of conductor actually experiencing the field.

Suppose a wire is \(0.75\ \mathrm{m}\) long, but only \(0.45\ \mathrm{m}\) lies within the region of a uniform magnetic field.

If the remaining \(0.30\ \mathrm{m}\) is outside the field, it does not experience the same magnetic force.

For the calculation, use:

\[ L=0.45\ \mathrm{m} \]

not \(0.75\ \mathrm{m}\).

This assumes the field outside the marked region is negligible and that the field inside is uniform.

If the magnetic field changes significantly along the conductor, you cannot automatically use one value of \(B\) for the entire segment. You may need to consider separate sections.

This is particularly important when a conductor is bent or passes through different magnetic field regions.

For this article, we will focus on straight conductors in uniform fields for the standard calculations, then use a more challenging bent-wire problem to explore what happens when the field or conductor geometry changes.

06Questions and solutions

The following questions progress from a straightforward force calculation to problems involving current reversal, partial field exposure, and multiple conductor segments.

Question 1

A straight conductor of length \(0.40\ \mathrm{m}\) is placed in a uniform magnetic field of \(0.30\ \mathrm{T}\), directed east.

A conventional current of \(2.0\ \mathrm{A}\) flows through the conductor at \(30^\circ\) north of east.

Calculate:

a. The magnitude of the magnetic force.

b. The direction of the force.

Solution 1

The magnetic force is \(0.12\ \mathrm{N}\), directed into the page.

Step 1: Identify the angle.

The current flows \(30^\circ\) north of east, and the magnetic field points east.

Therefore:

\[ \theta=30^\circ \]

Step 2: Apply the equation.

\[ F=BIL\sin\theta \]

Substituting:

\[ \begin{aligned} F&=(0.30)(2.0)(0.40)\sin30^\circ\\ &=(0.30)(2.0)(0.40)(0.5)\\ &=0.12\ \mathrm{N} \end{aligned} \]

Step 3: Determine the direction.

The current has an eastward and northward component, while the magnetic field points east.

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

Therefore, the conductor experiences a force of \(0.12\ \mathrm{N}\) into the page.

The important point is that only the component of the current-carrying conductor perpendicular to the field contributes to the force.

Question 2

A straight conductor of length \(0.60\ \mathrm{m}\) carries a current of \(4.0\ \mathrm{A}\) in a uniform magnetic field of \(0.45\ \mathrm{T}\).

The magnetic field points horizontally east. The conductor is tilted at \(25^\circ\) to the vertical, and conventional current flows from the lower left towards the upper right.

Calculate:

a. The magnitude of the magnetic force.

b. The maximum possible magnetic force if the conductor were rotated to become perpendicular to the magnetic field, without changing \(B\), \(I\), or \(L\).

c. Explain why using \(\sin25^\circ\) would give an incorrect result.

Solution 2

The conductor experiences a force of \(0.98\ \mathrm{N}\). Its maximum possible force is \(1.08\ \mathrm{N}\).

Step 1: Calculate the angle between the current and magnetic field.

The wire is \(25^\circ\) from the vertical, while the field is horizontal.

Therefore:

\[ \theta=90^\circ-25^\circ \]

\[ \theta=65^\circ \]

Step 2: Calculate the magnetic force.

\[ F=BIL\sin\theta \]

Substituting:

\[ \begin{aligned} F&=(0.45)(4.0)(0.60)\sin65^\circ\\ &=(1.08)(0.9063)\\ &=0.9788\ \mathrm{N} \end{aligned} \]

Therefore:

\[ F=0.98\ \mathrm{N} \]

Step 3: Calculate the maximum possible force.

The maximum occurs when the conductor is perpendicular to the magnetic field.

This means:

\[ \theta=90^\circ \]

Since \(\sin90^\circ=1\):

\[ F_{\max}=BIL \]

\[ \begin{aligned} F_{\max}&=(0.45)(4.0)(0.60)\\ &=1.08\ \mathrm{N} \end{aligned} \]

Step 4: Explain the angle misconception.

Using \(\sin25^\circ\) would incorrectly treat the angle between the conductor and the vertical as the angle between the conductor and the magnetic field.

The vertical is perpendicular to the horizontal magnetic field.

The correct angle is \(65^\circ\), not \(25^\circ\).

The calculated force is approximately 91% of the maximum possible force.

Question 3

A straight conductor of length \(0.50\ \mathrm{m}\) carries a current of \(3.0\ \mathrm{A}\) in a uniform magnetic field of \(0.40\ \mathrm{T}\), directed east.

The conductor is positioned at \(40^\circ\) north of east.

Initially, conventional current flows from the lower left towards the upper right.

The current is then reversed without moving the conductor or changing the magnetic field.

Determine:

a. The magnitude and direction of the magnetic force before the current is reversed.

b. The magnitude and direction of the magnetic force after the current is reversed.

c. Explain why the magnitude remains unchanged even though the angle used in the equation changes.

Solution 3

The force has a magnitude of \(0.39\ \mathrm{N}\) in both situations, but its direction reverses.

Step 1: Calculate the initial force.

Initially, the current flows northeast, making an angle of \(40^\circ\) with the magnetic field.

Therefore:

\[ \theta_1=40^\circ \]

Using:

\[ F=BIL\sin\theta \]

\[ \begin{aligned} F_1&=(0.40)(3.0)(0.50)\sin40^\circ\\ &=(0.60)(0.6428)\\ &=0.3857\ \mathrm{N} \end{aligned} \]

Therefore:

\[ F_1=0.39\ \mathrm{N} \]

Using the right-hand cross-product rule, the initial force is into the page.

Step 2: Reverse the current.

When the current reverses, it flows southwest.

The angle between the reversed current direction and the eastward magnetic field is now:

\[ \theta_2=140^\circ \]

Substitute this angle:

\[ \begin{aligned} F_2&=(0.40)(3.0)(0.50)\sin140^\circ\\ &=(0.60)(0.6428)\\ &=0.3857\ \mathrm{N} \end{aligned} \]

Therefore:

\[ F_2=0.39\ \mathrm{N} \]

The force now acts out of the page.

Step 3: Explain why the magnitude is unchanged.

The key mathematical relationship is:

\[ \sin140^\circ=\sin40^\circ \]

Therefore, the magnitude calculated using the force equation is identical.

However, the direction of the current has reversed. Since magnetic force depends on the cross product of current direction and magnetic field direction, reversing the current reverses the force.

This is an important distinction:

• Reversing current does not change the force magnitude if all other quantities remain constant.
• Reversing current does change the force direction.

Never rely on the sine calculation alone to determine the direction.

Question 4

A straight conductor has a total length of \(0.75\ \mathrm{m}\). It passes diagonally through a region of uniform magnetic field.

Only \(0.45\ \mathrm{m}\) of the conductor lies inside the field. The remaining section is outside the field, where the magnetic field is negligible.

The conductor carries a current of \(2.5\ \mathrm{A}\), and the magnetic flux density is \(0.60\ \mathrm{T}\).

The angle between the conventional current and the magnetic field is \(55^\circ\).

Calculate the magnetic force on the conductor.

Explain why using the full \(0.75\ \mathrm{m}\) length would produce an incorrect result.

Solution 4

The magnetic force on the conductor is \(0.55\ \mathrm{N}\), directed into the page.

Step 1: Identify the length of conductor inside the field.

The full wire is \(0.75\ \mathrm{m}\) long, but only \(0.45\ \mathrm{m}\) is exposed to the magnetic field.

Therefore:

\[ L=0.45\ \mathrm{m} \]

Step 2: Identify the remaining quantities.

\[ \begin{aligned} B&=0.60\ \mathrm{T}\\ I&=2.5\ \mathrm{A}\\ L&=0.45\ \mathrm{m}\\ \theta&=55^\circ \end{aligned} \]

Step 3: Calculate the magnetic force.

\[ F=BIL\sin\theta \]

Substituting:

\[ \begin{aligned} F&=(0.60)(2.5)(0.45)\sin55^\circ\\ &=(0.675)(0.8192)\\ &=0.553\ \mathrm{N} \end{aligned} \]

Therefore:

\[ F=0.55\ \mathrm{N} \]

Step 4: Explain why the total wire length is not used.

The magnetic force acts on the section of conductor exposed to the magnetic field.

The remaining \(0.30\ \mathrm{m}\) is outside the field, so it contributes no significant magnetic force under the stated assumptions.

If you incorrectly used \(0.75\ \mathrm{m}\), you would calculate:

\[ F=(0.60)(2.5)(0.75)\sin55^\circ \]

\[ F=0.92\ \mathrm{N} \]

This overestimates the force because it assumes the entire conductor experiences the magnetic field.

The actual force is \(0.55\ \mathrm{N}\) into the page.

Question 5

A conductor is bent into an L shape. It consists of two straight sections joined at point C.

• Section AC is horizontal and has a length of \(0.40\ \mathrm{m}\).
• Section CD is vertical and has a length of \(0.30\ \mathrm{m}\).
• Conventional current of \(2.0\ \mathrm{A}\) flows from A to C, then from C to D.

The entire conductor lies inside a uniform magnetic field of \(0.50\ \mathrm{T}\), directed into the page.

Consider the following questions:

a. Calculate the magnetic force on each straight section, including its direction.

b. Calculate the magnitude and direction of the resultant magnetic force on the entire conductor.

c. A student argues that the total wire length is \(0.70\ \mathrm{m}\), so the total magnetic force must be \(BIL=0.70\ \mathrm{N}\). Explain precisely why this reasoning is incorrect.

d. Suppose the L-shaped conductor is replaced by a smooth curved conductor connecting the same endpoints A and D. The entire new conductor remains inside the same uniform magnetic field. Would its resultant magnetic force change?

e. Now suppose the magnetic field on the vertical section is reduced to \(0.25\ \mathrm{T}\), while the horizontal section remains in a \(0.50\ \mathrm{T}\) field. Calculate the new resultant force. Explain why the conclusion in part (d) no longer necessarily applies.

Solution 5

In the uniform field, the two sections experience forces of \(0.40\ \mathrm{N}\) north and \(0.30\ \mathrm{N}\) west. Their resultant is \(0.50\ \mathrm{N}\), directed \(53.1^\circ\) north of west.

The crucial idea is that the forces on the two sections point in different directions. You must add them as vectors, not simply add their magnitudes.

Step 1: Calculate the force on the horizontal section AC.

The current flows east, while the magnetic field points into the page.

These directions are perpendicular, so:

\[ \theta=90^\circ \]

The magnetic force is:

\[ F_{AC}=BIL\sin90^\circ \]

Substituting:

\[ \begin{aligned} F_{AC}&=(0.50)(2.0)(0.40)(1)\\ &=0.40\ \mathrm{N} \end{aligned} \]

Using the right-hand cross-product rule, the force acts north.

Step 2: Calculate the force on the vertical section CD.

The current now flows north, while the magnetic field still points into the page.

These directions are also perpendicular.

Therefore:

\[ F_{CD}=BIL\sin90^\circ \]

Substituting:

\[ \begin{aligned} F_{CD}&=(0.50)(2.0)(0.30)(1)\\ &=0.30\ \mathrm{N} \end{aligned} \]

The right-hand cross-product rule gives a force towards the west.

Step 3: Calculate the resultant force.

We now have two forces:

• \(0.40\ \mathrm{N}\) north.
• \(0.30\ \mathrm{N}\) west.

These forces are perpendicular to each other.

Therefore, use Pythagoras to find the magnitude of their resultant:

\[ F_{\text{net}}=\sqrt{F_{AC}^2+F_{CD}^2} \]

Substituting:

\[ \begin{aligned} F_{\text{net}}&=\sqrt{0.40^2+0.30^2}\\ &=\sqrt{0.16+0.09}\\ &=\sqrt{0.25}\\ &=0.50\ \mathrm{N} \end{aligned} \]

The direction can be found using trigonometry.

Let \(\phi\) be the angle north of west.

\[ \tan\phi=\frac{0.40}{0.30} \]

\[ \phi=\tan^{-1}(1.333) \]

\[ \phi=53.1^\circ \]

Therefore, the resultant magnetic force is:

\[ F_{\text{net}}=0.50\ \mathrm{N} \]

at \(53.1^\circ\) north of west.

Step 4: Explain why the total wire length does not give the resultant force.

The actual length of the conductor is:

\[ L_{\text{total}}=0.40+0.30 \]

\[ L_{\text{total}}=0.70\ \mathrm{m} \]

However, substituting this into \(F=BIL\) would incorrectly assume that the magnetic force acts in one direction along the entire conductor.

The two sections experience forces in perpendicular directions.

Their magnitudes cannot simply be added:

\[ 0.40+0.30=0.70\ \mathrm{N} \]

Instead, the forces must be combined using vector addition:

\[ F_{\text{net}}=\sqrt{0.40^2+0.30^2} \]

This gives \(0.50\ \mathrm{N}\), not \(0.70\ \mathrm{N}\).

The actual path length is not enough to determine the resultant force when the conductor changes direction.

Step 5: Consider replacing the L-shaped wire with a smooth curve.

The resultant force would remain \(0.50\ \mathrm{N}\), with the same direction, provided the entire new conductor remains in the same uniform magnetic field and the current flows between the same endpoints.

Why?

The magnetic force on each small section depends on its direction. However, in a uniform magnetic field, the vector contributions from all the sections add to the same overall result as a straight conductor connecting the endpoints.

The displacement from A to D is:

\[ AD=\sqrt{0.40^2+0.30^2} \]

\[ AD=0.50\ \mathrm{m} \]

A straight conductor joining A and D would therefore have a length of \(0.50\ \mathrm{m}\).

It would carry current at \(36.9^\circ\) north of east, perpendicular to the magnetic field directed into the page.

Its force magnitude would be:

\[ \begin{aligned} F&=BIL\sin90^\circ\\ &=(0.50)(2.0)(0.50)\\ &=0.50\ \mathrm{N} \end{aligned} \]

The direction would also be northwest.

This is a useful result: in a uniform magnetic field, the resultant magnetic force on a conductor depends on its endpoints, not the precise shape of the path connecting them.

Step 6: Recalculate when the magnetic field is no longer uniform.

Now the horizontal section experiences a field of \(0.50\ \mathrm{T}\), while the vertical section experiences a field of \(0.25\ \mathrm{T}\).

The horizontal section is unchanged:

\[ F_{AC}=0.40\ \mathrm{N}\text{ north} \]

The vertical section now experiences:

\[ \begin{aligned} F_{CD}&=BIL\\ &=(0.25)(2.0)(0.30)\\ &=0.15\ \mathrm{N} \end{aligned} \]

This force acts west.

The new resultant magnitude is:

\[ \begin{aligned} F_{\text{net}}&=\sqrt{0.40^2+0.15^2}\\ &=\sqrt{0.1825}\\ &=0.427\ \mathrm{N} \end{aligned} \]

Therefore:

\[ F_{\text{net}}=0.43\ \mathrm{N} \]

The direction is:

\[ \begin{aligned} \tan\phi&=\frac{0.40}{0.15}\\ \phi&=\tan^{-1}(2.667)\\ &=69.4^\circ \end{aligned} \]

The resultant is \(69.4^\circ\) north of west.

The force has changed from \(0.50\ \mathrm{N}\) to \(0.43\ \mathrm{N}\), and its direction has also changed.

This happens because the magnetic field is no longer uniform. The two sections experience different field strengths, so their forces are weighted differently.

The endpoints alone are no longer sufficient to determine the resultant force.

This is the important limitation: the path-independent result applies to a conductor entirely within a uniform magnetic field, not an arbitrary field that changes along its length.

07Where this concept leads next

Finding the magnetic force on an angled wire comes down to three decisions:

• Identify the magnetic field \(B\) acting on the conductor.
• Find the angle between the conventional current and the magnetic field.
• Calculate the force magnitude and determine its direction separately.

Once you can handle an individual conductor segment, the next step is understanding what happens when several segments experience magnetic forces at once.

This leads naturally to the torque on a current-carrying coil, where forces on opposite sides of a conductor create a turning effect.

If you want to explore how the magnetic field itself is produced by another current-carrying wire, read Magnetic Field Around a Current-Carrying Wire: HSC Physics.

For the related problem of calculating the force per unit length between two parallel current-carrying wires, see Force Between Parallel Current-Carrying Wires: HSC Physics.