How to Calculate the Length of Wire for a Magnetic Force

A wire carries a current through a magnetic field, and you need it to experience a particular force. How much wire must actually sit inside the field?

The calculation is usually straightforward. The important part is choosing the correct length: **you need the length of the conductor inside the magnetic field, not the total length of wire connected to the circuit.**

For a straight wire, the magnetic force is

\[
F=BIL\sin\theta
\]

Rearrange this to find the required length:

\[
\boxed{L=\frac{F}{BI\sin\theta}}
\]

Here:

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

If the wire is perpendicular to the field, \(\theta=90^\circ\), so \(\sin90^\circ=1\). The calculation becomes

\[
\boxed{L=\frac{F}{BI}}
\]

The distinction between \(B\) and \(F\) is important. \(B\) describes the magnetic field itself. \(F\) is the force experienced by the current-carrying wire because it is in that field.

## 1. Identify the length that experiences the force

Imagine a long wire connected to a power supply. It passes through the gap between the poles of a magnet, then continues out the other side.

Only the section inside the magnetic field contributes to the motor-effect force, assuming the field outside the gap is negligible. The sections connected to the power supply still carry current, but they do not contribute appreciably to the force being calculated.

[IMAGE: Annotated physical arrangement of a long straight conducting wire passing horizontally through the uniform gap between the north and south poles of a horseshoe magnet. Show the wire extending well beyond both sides of the magnet, with only the central segment inside the field marked L. Label the magnetic field B across the pole gap, conventional current I along the wire, the angle theta between current and field, and the magnetic force F perpendicular to both. Use clear perspective so the active segment and external wire are easy to distinguish.]

Think of \(L\) as the active length of conductor. It is the portion of the wire exposed to the magnetic field, not the distance from one end of the wire to the other.

This is the first decision to make in every question: locate the field region, then identify the conductor segment within it.

02Account for the angle between the wire and the field

The perpendicular arrangement produces the greatest possible force for a given wire length, current, and magnetic flux density.

If the wire is tilted, only its component perpendicular to the magnetic field contributes to the force. The effective length is

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

This is why the full equation includes the sine term.

There are three cases worth recognising:

Be careful if the question gives an angle measured from the perpendicular rather than from the magnetic field. If that angle is \(\phi\), use

\[ L=\frac{F}{BI\cos\phi} \]

For a more detailed explanation of identifying the correct angle, see How to Find Magnetic Force on an Angled Wire.

A reliable method for finding wire length

Use this sequence when approaching an exam question:

1. Identify the magnetic flux density \(B\), in tesla.
2. Identify the current \(I\), in amperes.
3. Find the length of conductor actually inside the field.
4. Determine the angle between conventional current and the magnetic field.
5. Rearrange \(F=BIL\sin\theta\) to make \(L\) the subject.
6. Substitute the values using SI units.
7. Check whether the result makes physical sense.

The units provide a useful check. Since \(1\ \mathrm{T}=1\ \mathrm{N\\,A^{-1}\\,m^{-1}}\), the expression \(F/(BI)\) has units of metres.

Also remember that the rearranged equation calculates the magnitude of the force. If a question asks for its direction, use the conventional current and magnetic field directions with Fleming’s left-hand rule. Reversing the current changes the force direction, but not the length required for the same force magnitude.

03Worked calculations

04How can an experiment verify the required length?

The relationship can be investigated using a magnetic balance experiment.

Imagine a permanent magnet sitting on an electronic balance. Its north and south poles face each other across a gap. A straight conducting wire, supported independently by a clamp stand, passes through this gap.

The wire is connected to a power supply, with an ammeter in series to measure the current. The conductor segment inside the gap experiences the magnetic force.

What should happen when the current is switched on?

The wire experiences a magnetic force. By Newton’s third law, the wire also exerts an equal and opposite force on the magnet.

Because the magnet rests on the balance, this reaction force changes the balance reading.

The balance does not directly measure magnetic force. It measures a change in the force it supports, displayed as a mass-equivalent reading.

The experiment proceeds as follows:

1. Switch off the current and record the initial balance reading, \(m_0\).
2. Switch on the current and allow it to become steady.
3. Record the new balance reading, \(m_1\).
4. Calculate the change in reading, \(\Delta m=m_1-m_0\).
5. Convert the mass change into a force using \(F=|\Delta m|g\).

Here, \(\Delta m\) is measured in kilograms and \(g\) is gravitational acceleration, approximately \(9.8\ \mathrm{m\\,s^{-2}}\).

We can then combine the balance measurement with the motor-effect equation:

\[ |\Delta m|g=BIL\sin\theta \]

Rearranging gives

\[ \boxed{L=\frac{|\Delta m|g}{BI\sin\theta}} \]

This allows the active conductor length to be calculated from the change in balance reading.

A small experimental calculation

Suppose the balance reading increases by \(3.0\ \mathrm{g}\) when the current is switched on.

The illustrative experimental conditions are:

• Magnetic flux density: \(0.30\ \mathrm{T}\).
• Current: \(2.0\ \mathrm{A}\).
• Wire perpendicular to the field.

Find the length of conductor inside the field.

Step 1: Convert the mass change into kilograms.

\[ \Delta m=3.0\ \mathrm{g}=0.0030\ \mathrm{kg} \]

Step 2: Calculate the magnetic force.

\[ \begin{aligned} F&=|\Delta m|g\\\ &=0.0030\times9.8\\\ &=0.0294\ \mathrm{N} \end{aligned} \]

Step 3: Rearrange the force equation and calculate the length.

\[ \begin{aligned} L&=\frac{F}{BI}\\\ &=\frac{0.0294}{0.30\times2.0}\\\ &=0.049\ \mathrm{m} \end{aligned} \]

Therefore,

\[ \boxed{L=4.9\ \mathrm{cm}} \]

A balance change of just 3.0 g corresponds to a magnetic force of approximately \(0.029\ \mathrm{N}\), produced by 4.9 cm of active conductor under these conditions.

For a fuller explanation of the balance readings, including how reversing the current changes the measurement, see How to Find Current from a Magnetic Balance Experiment.

What should happen if the active length increases?

We can also use the apparatus to test the relationship between force and conductor length.

Keep the magnetic flux density, current, and angle constant. Change only the length of conductor inside the uniform field.

For example, take:

\[ B=0.30\ \mathrm{T},\quad I=2.0\ \mathrm{A},\quad \theta=90^\circ \]

The following values are illustrative ideal predictions, not recorded experimental results.

The pattern is clear: doubling the active length doubles the force.

The straight-line relationship is what we expect from

\[ F=BIL \]

when \(B\) and \(I\) remain constant.

The gradient of the force-versus-length graph is \(BI\) for a perpendicular conductor. For an angled conductor, the gradient is \(BI\sin\theta\).

Why the experimental setup matters

A first attempt might measure the entire wire between the power supply terminals and use that measurement as \(L\). This is a flawed design because much of the wire may sit outside the magnetic field.

The improved approach is to identify the boundaries of the uniform field and measure only the conductor segment between them. Keep the active segment away from the pole edges, where the field may become less uniform.

You should also keep the current and angle constant while changing the active length. Otherwise, a change in force could be caused by more than one variable.

In a real experiment, stray fields, fringing at the magnet’s edges, heating, and measurement uncertainty can affect the results. The HSC calculation generally assumes a steady current and a uniform magnetic field over the active conductor segment.

05Common mistakes when calculating wire length

One final physical check is particularly useful.

For fixed \(B\), \(I\), and \(\theta\), increasing the required force must increase the required length. Increasing the magnetic flux density or current must decrease the required length.

If your answer does the opposite, revisit your rearrangement.

05Questions and solutions

Question 1

A straight wire is placed perpendicular to a uniform magnetic field of flux density \(0.35\ \mathrm{T}\). It carries a current of \(2.0\ \mathrm{A}\).

Calculate the length of conductor inside the field required to produce a magnetic force of \(0.14\ \mathrm{N}\).

Solution 1

The required conductor length is \(0.20\ \mathrm{m}\).

Because the wire is perpendicular to the field, we can use \(F=BIL\).

Step 1: Rearrange the equation.

\[ L=\frac{F}{BI} \]

Step 2: Substitute the values.

\[ \begin{aligned} L&=\frac{0.14}{0.35\times2.0}\\\ &=\frac{0.14}{0.70}\\\ &=0.20\ \mathrm{m} \end{aligned} \]

Step 3: Interpret the result.

The conductor needs 20 cm inside the magnetic field. Any additional wire outside the field does not need to be included.

Question 2

A straight conductor passes through a uniform magnetic field. Its direction is \(25^\circ\) away from the direction perpendicular to the magnetic field.

The conductor carries a current of \(2.5\ \mathrm{A}\), and the magnetic flux density is \(0.32\ \mathrm{T}\).

Calculate the length of conductor required to experience a magnetic force of \(0.30\ \mathrm{N}\).

Solution 2

The required length is \(0.414\ \mathrm{m}\).

The important step is converting the supplied angle. The equation needs the angle between the current and the magnetic field, not the angle between the wire and the perpendicular direction.

Step 1: Calculate the correct angle.

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

\[ \theta=65^\circ \]

Step 2: Rearrange the general equation.

\[ L=\frac{F}{BI\sin\theta} \]

Step 3: Substitute the values.

\[ \begin{aligned} L&=\frac{0.30}{0.32\times2.5\times\sin65^\circ}\\\ &=\frac{0.30}{0.725}\\\ &=0.414\ \mathrm{m} \end{aligned} \]

Step 4: Explain the angle trap.

A tempting mistake is to substitute \(25^\circ\) directly into the sine term. That would give approximately \(0.887\ \mathrm{m}\), which is incorrect.

The supplied angle is measured from the perpendicular to the field. The correct angle between the current and field is \(65^\circ\).

Question 3

A straight conducting wire has a total length of \(0.75\ \mathrm{m}\). Only \(0.22\ \mathrm{m}\) of the wire lies inside a uniform magnetic field.

The active section makes an angle of \(35^\circ\) with the magnetic field. The field has a flux density of \(0.60\ \mathrm{T}\), and the current is \(1.8\ \mathrm{A}\).

(a) Calculate the magnetic force acting on the wire.

(b) The current is reduced to \(1.2\ \mathrm{A}\). The required magnetic force is now twice the original force. Calculate the new active conductor length required, assuming the magnetic field and angle remain unchanged.

(c) Is this new length physically possible if the total wire length remains \(0.75\ \mathrm{m}\)? Explain.

Solution 3

(a) The magnetic force is approximately \(0.136\ \mathrm{N}\).

(b) The new active length must be \(0.660\ \mathrm{m}\).

(c) Yes. The required active length is less than the total available wire length.

Step 1: Identify the active length.

The total wire is \(0.75\ \mathrm{m}\), but only \(0.22\ \mathrm{m}\) is inside the field.

Therefore,

\[ L=0.22\ \mathrm{m} \]

Step 2: Calculate the original force.

Use

\[ F=BIL\sin\theta \]

Substitute:

\[ \begin{aligned} F&=0.60\times1.8\times0.22\times\sin35^\circ\\\ &=0.1363\ \mathrm{N} \end{aligned} \]

Therefore,

\[ \boxed{F=0.136\ \mathrm{N}} \]

Step 3: Calculate the new target force.

The required force is twice the original:

\[ F_{\text{new}}=2\times0.1363 \]

\[ F_{\text{new}}=0.2726\ \mathrm{N} \]

Step 4: Calculate the new active length.

The current has changed to \(1.2\ \mathrm{A}\), so we must use the new current rather than the original value.

\[ L=\frac{F_{\text{new}}}{BI_{\text{new}}\sin\theta} \]

Substitute:

\[ \begin{aligned} L&=\frac{0.2726}{0.60\times1.2\times\sin35^\circ}\\\ &=\frac{0.2726}{0.4130}\\\ &=0.660\ \mathrm{m} \end{aligned} \]

Step 5: Check whether the result is possible.

The total available wire length is \(0.75\ \mathrm{m}\).

The required active length is \(0.660\ \mathrm{m}\).

Since

\[ 0.660<0.750 \]

the arrangement is physically possible, assuming the field region is large enough to contain that active length.

The tempting approach is to double the original length because the target force has doubled. However, the current has also decreased.

The new current is only two-thirds of its original value. We therefore need three times the original active length to produce twice the force.

This is a useful reminder that length and current must be considered together.

Question 4

A permanent magnet sits on an electronic balance. A straight conducting wire passes through the gap between its poles and is supported independently.

The wire is perpendicular to the magnetic field.

The following readings are recorded:

The magnetic flux density is \(0.28\ \mathrm{T}\), and the current is \(1.5\ \mathrm{A}\).

(a) Calculate the length of conductor inside the magnetic field.

(b) The current is then reversed without changing its magnitude. Predict the new balance reading, assuming the experimental conditions remain unchanged.

Solution 4

(a) The active conductor length is \(0.056\ \mathrm{m}\).

(b) The new balance reading is \(317.6\ \mathrm{g}\).

Step 1: Calculate the change in balance reading.

\[ \Delta m=322.4-320.0 \]

\[ \Delta m=2.4\ \mathrm{g} \]

Convert to kilograms:

\[ \Delta m=0.0024\ \mathrm{kg} \]

Step 2: Convert the mass change into magnetic force.

\[ F=|\Delta m|g \]

\[ \begin{aligned} F&=0.0024\times9.8\\\ &=0.02352\ \mathrm{N} \end{aligned} \]

Step 3: Calculate the active conductor length.

The wire is perpendicular to the field, so

\[ L=\frac{F}{BI} \]

Substitute:

\[ \begin{aligned} L&=\frac{0.02352}{0.28\times1.5}\\\ &=\frac{0.02352}{0.42}\\\ &=0.056\ \mathrm{m} \end{aligned} \]

Therefore,

\[ \boxed{L=5.6\ \mathrm{cm}} \]

Step 4: Predict the effect of reversing the current.

Reversing the current reverses the magnetic force on the wire.

By Newton’s third law, the force exerted by the wire on the magnet also reverses.

The original balance reading increased by \(2.4\ \mathrm{g}\). With the current reversed, it should decrease by the same amount.

\[ m_{\text{new}}=320.0-2.4 \]

\[ \boxed{m_{\text{new}}=317.6\ \mathrm{g}} \]

The tempting prediction is that the balance reading remains at \(322.4\ \mathrm{g}\) because the current magnitude has not changed.

That overlooks the direction of the force. The magnitude stays the same, but the direction reverses, so the balance reading changes in the opposite direction.

Question 5

A single conducting wire is bent into a three-sided rectangular shape, with its left side open.

The wire lies entirely inside a uniform magnetic field of flux density \(0.40\ \mathrm{T}\). The field points vertically upwards in the plane of the page.

The wire has three sections:

• A lower horizontal section of length \(0.18\ \mathrm{m}\).
• A right-hand vertical connecting section of length \(0.12\ \mathrm{m}\).
• An upper horizontal section of length \(0.10\ \mathrm{m}\).

A conventional current of \(2.0\ \mathrm{A}\) enters through the lower-left end, travels right along the lower section, upwards through the connecting section, and left along the upper section.

The entire wire is within the uniform field.

(a) Calculate the net magnetic force on the entire wire, including its magnitude and direction.

(b) The target net force is \(0.10\ \mathrm{N}\) out of the page. If the upper horizontal section remains \(0.10\ \mathrm{m}\), what must the lower horizontal section’s length be?

(c) Once this target has been achieved, both horizontal sections are increased in length by \(0.05\ \mathrm{m}\). Does the net magnetic force change? Explain.

Solution 5

(a) The net force is \(0.064\ \mathrm{N}\) out of the page.

(b) The lower horizontal section must be \(0.225\ \mathrm{m}\) long.

(c) The net force remains \(0.10\ \mathrm{N}\) out of the page.

This problem has an important twist: we cannot simply add the lengths of all three sections and substitute the total into the straight-wire equation.

The force is a vector, so we must calculate the force on each section and then combine the results.

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

The current flows to the right, perpendicular to the upward magnetic field.

Therefore, the lower section experiences a force out of the page.

\[ \begin{aligned} F_{\text{lower}}&=BIL_{\text{lower}}\\\ &=0.40\times2.0\times0.18\\\ &=0.144\ \mathrm{N} \end{aligned} \]

Step 2: Calculate the force on the upper horizontal section.

The current flows to the left, so its force acts in the opposite direction.

\[ \begin{aligned} F_{\text{upper}}&=BIL_{\text{upper}}\\\ &=0.40\times2.0\times0.10\\\ &=0.080\ \mathrm{N} \end{aligned} \]

This force acts into the page.

Step 3: Consider the vertical connecting section.

The current flows upwards, parallel to the magnetic field.

Therefore,

\[ F_{\text{vertical}}=0 \]

The connecting section experiences no magnetic force in this ideal arrangement.

Step 4: Calculate the net force.

The lower and upper forces act in opposite directions.

We must subtract their magnitudes:

\[ \begin{aligned} F_{\text{net}}&=F_{\text{lower}}-F_{\text{upper}}\\\ &=0.144-0.080\\\ &=0.064\ \mathrm{N} \end{aligned} \]

The lower section produces the larger force, so the net force is out of the page.

\[ \boxed{F_{\text{net}}=0.064\ \mathrm{N}\text{ out of the page}} \]

Step 5: Find the lower section length needed for the target force.

The required net force is \(0.10\ \mathrm{N}\) out of the page.

Let the unknown lower section length be \(L_{\text{lower}}\).

The net force is

\[ F_{\text{net}}=BI(L_{\text{lower}}-L_{\text{upper}}) \]

Rearrange:

\[ L_{\text{lower}}= L_{\text{upper}}+\frac{F_{\text{net}}}{BI} \]

Substitute:

\[ \begin{aligned} L_{\text{lower}} &=0.10+\frac{0.10}{0.40\times2.0}\\\ &=0.10+\frac{0.10}{0.80}\\\ &=0.10+0.125\\\ &=0.225\ \mathrm{m} \end{aligned} \]

Therefore,

\[ \boxed{L_{\text{lower}}=0.225\ \mathrm{m}} \]

Step 6: Determine what happens when both horizontal sections increase equally.

The lower section becomes

\[ 0.225+0.05=0.275\ \mathrm{m} \]

The upper section becomes

\[ 0.10+0.05=0.15\ \mathrm{m} \]

The new net force is

\[ \begin{aligned} F_{\text{net}} &=BI(0.275-0.15)\\\ &=0.40\times2.0\times0.125\\\ &=0.10\ \mathrm{N} \end{aligned} \]

The net force has not changed.

Why? Both sections produce larger forces, but they act in opposite directions. Increasing both lengths by the same amount increases the opposing forces equally, leaving their difference unchanged.

This is the key conceptual point: the total length of a bent conductor is not necessarily the length that determines its net magnetic force.

For a single straight section, use \(F=BIL\sin\theta\). For a conductor made of several sections, calculate the force on each section and add the forces as vectors.

The next useful application is to use the calculated magnetic force in a suspended-wire problem, where the magnetic force can balance the wire’s weight. That introduces the connection between the motor effect, gravitational force, and equilibrium.