How to Find Magnetic Field Strength from Motor-Effect Data

A wire sits between two magnet poles. You increase the current from \(1.0\ \text{A}\) to \(2.0\ \text{A}\), and the measured magnetic force doubles from \(0.024\ \text{N}\) to \(0.048\ \text{N}\).

Has the magnetic field doubled?

No. If the magnets and wire position haven’t changed, the magnetic field is approximately constant. It is the current that has doubled.

This proportional relationship gives us a way to calculate the magnetic field strength from experimental data. Plot magnetic force \(F\) against current \(I\), calculate the gradient, and use the conductor length to determine \(B\).

For a straight conductor perpendicular to a uniform magnetic field:

\[
F=BIL
\]

Rearranging gives:

\[
\boxed{B=\frac{\text{gradient of }F\text{ against }I}{L}}
\]

The important detail is that the gradient is not \(B\) by itself. It is \(BL\). Getting that distinction right is the key to solving motor-effect graph questions.

01Understand what the experiment measures

Before plotting anything, identify the three physical quantities involved.

• \(B\) is the magnetic field strength, more precisely called magnetic flux density, measured in tesla (\(\text{T}\)).
• \(I\) is the conventional current through the conductor, measured in amperes (\(\text{A}\)).
• \(F\) is the magnetic force acting on the conductor, measured in newtons (\(\text{N}\)).

The length of conductor inside the magnetic field is also important. We represent it using \(L\), measured in metres.

Which part of the wire counts?

Imagine a straight wire passing through the gap between two magnet poles.

Only the section of conductor inside the magnetic field contributes to the motor-effect force in our simplified uniform-field model. The rest of the wire carries current, but it is outside the region where we are measuring the magnetic interaction.

This is why \(L\) means the active length inside the field, not the total length of wire connected to the power supply.

The equation behind the graph

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

\[
F=BIL\sin\theta
\]

Here, \(\theta\) is the angle between the conventional current direction and the magnetic field direction.

There are two particularly useful cases:

For a perpendicular conductor, \(\sin90^\circ=1\), so the equation simplifies to:

\[
F=BIL
\]

Notice that \(B\) and \(F\) are completely different quantities.

The magnetic field \(B\) describes the magnetic environment around the conductor. The force \(F\) describes the interaction experienced by the conductor.

The same magnetic field can produce different forces if the current or conductor length changes.

If you need to refresh what magnetic field strength represents, see Magnetic Field Models: From Field Lines to Field Strength.

02Set up the experiment so the data is meaningful

The aim is to investigate how magnetic force changes as current changes.

A typical arrangement uses:

• Two permanent magnets with facing pole pieces, producing an approximately uniform magnetic field.
• A straight conductor positioned in the gap between the poles.
• A DC power supply connected to the conductor through an ammeter and variable resistor.
• A force sensor arranged to measure the magnetic force on the conductor.

The circuit components are connected in series so the ammeter measures the current passing through the conductor.

The variable resistor allows the current to be changed without moving the conductor or magnets.

The force sensor measures the magnetic force. Before switching on the current, record or zero the initial reading.

What must stay constant?

To investigate the relationship between force and current, change only the current.

Keep the following constant:

• Magnetic field strength \(B\).
• Active conductor length \(L\).
• Angle \(\theta\) between the conductor and magnetic field.
• Position of the conductor within the magnetic field.

The current must be measured using the ammeter at each setting. Do not assume that changing the power supply voltage produces a particular current.

The conductor and magnets should also remain stationary during measurements.

Predict the graph

Suppose the current doubles while everything else stays constant.

What should happen to the magnetic force?

The force should double.

This follows directly from:

\[
F=BIL
\]

If \(B\) and \(L\) are constant, then:

\[
F\propto I
\]

We therefore expect a straight-line graph of magnetic force against current.

If the experimental data follows this pattern, it supports the motor-effect relationship.

What could go wrong in the first design?

Imagine measuring the force using a long wire that extends well beyond the magnet gap.

Some of the connecting wire may also sit inside the magnetic field. Those sections can experience magnetic forces too, meaning the sensor measures more than the contribution from the intended active segment.

A better arrangement keeps the active conductor clearly defined and routes the connecting leads away from the strong-field region.

Similarly, if the wire moves while the current changes, its position or angle may change the measured force independently of the current.

The improved method is to:

1. Mark the active length of conductor.
2. Position it centrally in the uniform-field region.
3. Keep its orientation fixed.
4. Zero the force sensor with no current.
5. Change the current using the variable resistor.
6. Record the actual current and corresponding force.
7. Repeat readings to reduce random measurement errors.

The experiment now tests one relationship rather than several changing conditions at once.

03Plot magnetic force against current

This is the central graphing technique.

For the standard motor-effect experiment, use:

• Horizontal axis: current \(I\), in amperes.
• Vertical axis: magnetic force \(F\), in newtons.

Do not swap these axes if you want to use the gradient relationship developed below.

Consider the following illustrative experimental data.

The active conductor length is \(0.080\ \text{m}\), and the conductor is perpendicular to the magnetic field.

Illustrative data (not actual experimental measurements):

Why does the gradient matter?

The gradient of a straight-line graph is:

\[
m=\frac{\Delta y}{\Delta x}
\]

For our force-current graph:

\[
m=\frac{\Delta F}{\Delta I}
\]

Now return to the motor-effect equation:

\[
F=BIL
\]

Rearrange it into a straight-line form:

\[
F=(BL)I
\]

Compare this with the general straight-line equation:

\[
y=mx+c
\]

We can identify the corresponding quantities:

Therefore:

\[
\boxed{\text{Gradient}=BL}
\]

And rearranging:

\[
\boxed{B=\frac{m}{L}}
\]

This is the equation you need to remember.

The gradient tells you how much the magnetic force increases for each additional ampere of current. Dividing by the active conductor length isolates the magnetic field strength.

04What if the conductor is not perpendicular to the field?

So far, we have assumed the conductor is at \(90^\circ\) to the magnetic field.

But what happens if the wire is angled?

Imagine a conductor positioned at \(30^\circ\) to the field. It is tempting to use the full conductor length in the equation and ignore the angle.

That would overestimate the force.

Only the component of the conductor perpendicular to the magnetic field contributes to the motor-effect force.

The effective perpendicular length is:

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

Therefore, the full equation is:

\[
F=BIL\sin\theta
\]

The graphing method does not change.

You still plot \(F\) against \(I\), provided the conductor’s angle and active length remain constant.

However, the gradient becomes:

\[
\boxed{m=BL\sin\theta}
\]

Rearranging gives:

\[
\boxed{B=\frac{m}{L\sin\theta}}
\]

05Choosing the correct graph

The graph you need depends on which experimental variable changes.

You do not always have to plot force against current.

The key is to rearrange the motor-effect equation into a straight-line relationship.

Starting with:

\[
F=BIL\sin\theta
\]

we can obtain three useful graphing methods.

For most introductory motor-effect data questions, you will be asked to vary current while keeping the conductor length and angle constant.

That is why plotting \(F\) against \(I\) is such a useful method.

A quick units check

Units are an excellent way to catch mistakes.

The gradient of a force-current graph has units:

\[
\frac{\text{N}}{\text{A}}
\]

But magnetic field strength has units of tesla.

Since:

\[
1\ \text{T}=
1\frac{\text{N}}{\text{A}\cdot\text{m}}
\]

we must divide the gradient by a length measured in metres.

This gives:

\[
\frac{\text{N/A}}{\text{m}}=\text{T}
\]

If your calculated magnetic field strength has units of \(\text{N/A}\), you have stopped one step too early.

06Experimental limitations and common graphing mistakes

A straight-line graph is useful, but you still need to understand what could affect its gradient.

Mistake 1: Using the total wire length

Only the active conductor length inside the magnetic field should be used in the equation.

If a \(20\ \text{cm}\) wire has only \(8\ \text{cm}\) inside the uniform-field region, use:

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

Using the full \(0.20\ \text{m}\) would underestimate the field strength.

Mistake 2: Treating the gradient as B

For a graph of \(F\) against \(I\):

\[
m=BL\sin\theta
\]

The gradient contains both the field strength and the conductor geometry.

You must divide by \(L\sin\theta\) to obtain \(B\).

Mistake 3: Forgetting to convert units

The conductor length must be in metres.

For example:

\[
8.0\ \text{cm}=0.080\ \text{m}
\]

Likewise, convert force readings from millinewtons to newtons before using them in calculations, unless you explicitly convert the gradient first.

Mistake 4: Forcing the line through the origin

The ideal motor-effect relationship predicts zero magnetic force at zero current.

However, real instruments may have a zero offset.

If the force sensor records a small non-zero reading before current is switched on, a best-fit line may have a non-zero intercept.

You can either subtract the initial reading from every measurement or fit a straight line that allows for the intercept.

Do not automatically force the line through the origin. Doing so can distort the gradient.

Mistake 5: Ignoring the uniform-field assumption

The simple equation assumes the active conductor experiences a uniform magnetic field.

If the conductor moves towards the edge of the magnet gap, the field may change.

The measured force may then change because of both current and magnetic field strength.

This is why the conductor should remain in the central region of the gap throughout the experiment.

What if a balance gives mass instead of force?

Some motor-effect experiments measure the change in a digital balance reading rather than using a force sensor directly.

A balance measures mass, not force.

To convert the change in mass into a force magnitude, use:

\[
F=\Delta m g
\]

where:

• \(\Delta m\) is the change in mass in kilograms.
• \(g=9.8\ \text{m s}^{-2}\) is the gravitational acceleration.

For example, if the balance reading changes by \(2.4\ \text{g}\):

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

Therefore:

\[
F=0.0024\times9.8
\]

\[
F=0.0235\ \text{N}
\]

The balance may measure the reaction force on the magnet assembly rather than the force on the wire itself.

By Newton’s third law, these forces have equal magnitudes and opposite directions.

For calculating magnetic field strength, the magnitude is what matters.

07Questions and solutions

The following questions build from a straightforward gradient calculation to experiments where you need to decide what the graph actually represents.

Question 1

A straight conductor has an active length of \(0.075\ \text{m}\) inside a uniform magnetic field. It is perpendicular to the field.

A graph of magnetic force \(F\) against current \(I\) has a gradient of:

\[
m=0.0180\ \text{N A}^{-1}
\]

Calculate the magnetic field strength.

Solution 1

The magnetic field strength is \(0.240\ \text{T}\).

Because the conductor is perpendicular to the magnetic field:

\[
F=BIL
\]

The gradient of the force-current graph is:

\[
m=BL
\]

Rearranging:

\[
B=\frac{m}{L}
\]

Substitute the values:

\[
B=\frac{0.0180}{0.075}
\]

\[
\boxed{B=0.240\ \text{T}}
\]

The result is in tesla because the gradient is divided by a length measured in metres.

The important step is recognising that the gradient represents \(BL\), not \(B\) alone.

Question 2

A conductor is positioned perpendicular to a uniform magnetic field. Its active length inside the field is \(0.040\ \text{m}\).

The magnet assembly is placed on a digital balance, while the conductor is supported independently. When current flows, the balance reading changes because of the reaction force acting on the magnets.

The following illustrative readings are obtained:

Using a graph of balance reading against current:

(a) Determine the gradient.

(b) Calculate the magnetic field strength.

(c) Explain why the force acting on the magnet assembly is opposite to the force acting on the conductor.

Solution 2

The magnetic field strength is approximately \(0.049\ \text{T}\).

(a) The gradient of the mass-current graph is approximately:

\[
m_{\Delta m}=
0.20\ \text{g A}^{-1}
\]

Convert grams to kilograms:

\[
m_{\Delta m}=
2.0\times10^{-4}\ \text{kg A}^{-1}
\]

(b) The balance measures a change in mass, so we must convert the gradient into a force-current gradient.

Using:

\[
F=\Delta m g
\]

the force-current gradient is:

\[
m_F=gm_{\Delta m}
\]

Substitute:

\[
m_F=9.8(2.0\times10^{-4})
\]

\[
m_F=1.96\times10^{-3}\ \text{N A}^{-1}
\]

For a perpendicular conductor:

\[
B=\frac{m_F}{L}
\]

Substitute:

\[
B=\frac{1.96\times10^{-3}}{0.040}
\]

\[
\boxed{B=0.049\ \text{T}}
\]

(c) The magnetic field exerts a force on the current-carrying conductor.

The conductor simultaneously exerts an equal-magnitude force in the opposite direction on the magnets.

This is an action-reaction pair described by Newton’s third law.

The balance detects the force on the magnets, not directly the force on the wire. Their magnitudes are equal, so the same magnetic field calculation applies.

Question 3

A straight conductor has a total length of \(14\ \text{cm}\). It is positioned perpendicular to a uniform magnetic field.

The experiment varies the length of conductor inside the magnetic field while maintaining a constant current of \(2.0\ \text{A}\).

The following readings are obtained:

(a) Identify the appropriate graph to plot.

(b) Calculate its gradient in SI units.

(c) Determine the magnetic field strength.

(d) Explain why the total conductor length of \(14\ \text{cm}\) is not used.

Solution 3

The magnetic field strength is \(0.30\ \text{T}\).

(a) Since the active conductor length changes while current and magnetic field remain constant, plot magnetic force \(F\) against active length \(L\).

The horizontal axis should represent \(L\) in metres, and the vertical axis should represent \(F\) in newtons.

(b) Convert the lengths into metres:

\[
2.0\ \text{cm}=0.020\ \text{m}
\]

\[
8.0\ \text{cm}=0.080\ \text{m}
\]

Calculate the gradient:

\[
m=\frac{\Delta F}{\Delta L}
\]

Using the first and last readings:

\[
m=\frac{0.048-0.012}{0.080-0.020}
\]

\[
m=\frac{0.036}{0.060}
\]

\[
m=0.60\ \text{N m}^{-1}
\]

(c) The motor-effect equation is:

\[
F=BIL
\]

Rearrange:

\[
B=\frac{F}{IL}
\]

For a graph of \(F\) against \(L\):

\[
m=BI
\]

Therefore:

\[
B=\frac{m}{I}
\]

Substitute:

\[
B=\frac{0.60}{2.0}
\]

\[
\boxed{B=0.30\ \text{T}}
\]

(d) The remaining conductor is outside the uniform magnetic-field region.

It does not contribute to the force in the simplified model being used.

The relevant length is the portion of the wire inside the field, not the total length of wire connected to the circuit.

This question demonstrates why identifying the independent variable is important before choosing a graph.

Question 4

A straight conductor of active length \(0.100\ \text{m}\) is positioned at an angle of \(60^\circ\) to a uniform magnetic field.

A graph of magnetic force magnitude \(F\) against current magnitude \(I\) produces a gradient of:

\[
m=0.0260\ \text{N A}^{-1}
\]

(a) Calculate the magnetic field strength.

(b) A student suggests using \(\cos60^\circ\) instead of \(\sin60^\circ\). Explain why this is incorrect.

(c) The experiment is repeated with the current reversed. The horizontal axis still records positive current magnitudes, while the vertical axis records signed force relative to the original force direction. Determine the new gradient.

Solution 4

The magnetic field strength is approximately \(0.300\ \text{T}\), and reversing the current changes the signed gradient to \(-0.0260\ \text{N A}^{-1}\).

(a) The conductor is angled, so we must use:

\[
F=BIL\sin\theta
\]

The gradient of the force-current graph is:

\[
m=BL\sin\theta
\]

Rearranging:

\[
B=\frac{m}{L\sin\theta}
\]

Substitute:

\[
B=\frac{0.0260}{0.100\sin60^\circ}
\]

Using:

\[
\sin60^\circ\approx0.866
\]

we obtain:

\[
B=\frac{0.0260}{0.0866}
\]

\[
\boxed{B\approx0.300\ \text{T}}
\]

(b) Using cosine would select the component of the conductor parallel to the magnetic field.

That component does not contribute to the motor-effect force.

The force depends on the perpendicular component:

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

This is why sine appears in the equation.

(c) Reversing the current reverses the direction of the magnetic force.

The horizontal axis still records positive current magnitudes, but the signed force is now negative relative to the original direction.

Therefore, the new graph has a negative gradient:

\[
\boxed{m=-0.0260\ \text{N A}^{-1}}
\]

The magnitude of the gradient is unchanged.

The magnetic field strength is still \(0.300\ \text{T}\). Reversing the current changes the force direction, not the magnetic field strength.

Question 5

A straight conductor passes through two adjacent uniform magnetic-field regions, labelled A and B.

The conventional current travels in the same direction through both regions. However, the magnetic fields point in opposite directions.

The active conductor lengths are:

• Region A: \(0.040\ \text{m}\).
• Region B: \(0.060\ \text{m}\).

Both sections of conductor are perpendicular to their respective magnetic fields.

A force sensor measures the net force on the conductor. The positive force direction is defined as the direction of the force produced by region A.

Two experiments are performed:

Experiment 1: Both magnetic-field regions are active. A graph of signed net force against positive current magnitude has a gradient of:

\[
m_{AB}=+0.0040\ \text{N A}^{-1}
\]

Experiment 2: Region B is shielded so that only region A produces a magnetic force. The graph now has a gradient of:

\[
m_A=+0.020\ \text{N A}^{-1}
\]

Assume the fields are uniform within their respective regions, the current is the same through both sections, and forces on connecting leads are negligible.

(a) Determine the magnetic field strength in region A.

(b) Determine the magnetic field strength in region B.

(c) A student argues that dividing the combined gradient by the total active length gives the magnetic field strength in both regions. Explain precisely why this conclusion is incorrect.

(d) What would a zero gradient in Experiment 1 tell you? Would it prove that both magnetic fields are zero?

Solution 5

The magnetic field strengths are:

\[
\boxed{B_A=0.50\ \text{T}}
\]

and

\[
\boxed{B_B\approx0.27\ \text{T}}
\]

The important point is that the force sensor measures the net force from both regions, not the individual magnetic forces.

(a) In Experiment 2, only region A produces a magnetic force.

The gradient is therefore:

\[
m_A=B_A L_A
\]

Rearrange:

\[
B_A=\frac{m_A}{L_A}
\]

Substitute:

\[
B_A=\frac{0.020}{0.040}
\]

\[
\boxed{B_A=0.50\ \text{T}}
\]

(b) In Experiment 1, both regions produce forces.

Because their magnetic fields point in opposite directions, their forces also point in opposite directions.

The net force is the force from A minus the force from B:

\[
F_{\text{net}}=F_A-F_B
\]

Using the motor-effect equation for each region:

\[
F_{\text{net}}=B_AIL_A-B_BIL_B
\]

Factor out the current:

\[
F_{\text{net}}=
(B_A L_A-B_B L_B)I
\]

Therefore, the gradient of the combined graph is:

\[
m_{AB}=B_A L_A-B_B L_B
\]

Substitute the known values:

\[
0.0040=(0.50)(0.040)-B_B(0.060)
\]

Simplify:

\[
0.0040=0.020-B_B(0.060)
\]

Rearrange:

\[
0.060B_B=0.016
\]

\[
B_B=\frac{0.016}{0.060}
\]

\[
\boxed{B_B=0.267\ \text{T}}
\]

The magnetic field in region B has a magnitude of approximately \(0.27\ \text{T}\), directed opposite to the field in region A.

(c) The student’s calculation would give:

\[
B_{\text{apparent}}=
\frac{0.0040}{0.040+0.060}
\]

\[
B_{\text{apparent}}=
\frac{0.0040}{0.100}
\]

\[
B_{\text{apparent}}=0.040\ \text{T}
\]

This is not the magnetic field strength in either region.

The problem is that the student has treated two opposing magnetic forces as though they act in the same direction.

The gradient represents the difference between the two force contributions per ampere:

\[
m_{AB}=B_A L_A-B_B L_B
\]

Dividing by the total length produces a signed, length-weighted average of the two fields, not either local field strength.

The distinction is important: a small net force does not necessarily mean the individual magnetic fields are weak.

(d) A zero gradient would mean:

\[
B_A L_A=B_B L_B
\]

The two magnetic forces would have equal magnitudes and opposite directions.

Their net force would be zero, even though both magnetic fields could be substantial.

For example, if both fields had a strength of \(0.50\ \text{T}\), the unequal conductor lengths would produce unequal forces. A zero gradient would instead require the field strengths to compensate for the different active lengths.

Also, if only the combined graph were available, one equation would contain two unknown field strengths. There would not be enough information to determine both fields uniquely.

The second experiment solves this problem by isolating region A.

This is the deeper lesson: a graph measures the net effect of everything contributing to the measured quantity. Before interpreting its gradient, you must know exactly which forces are being combined.

08What to do next

Once you can extract \(B\) from a force-current graph, you can use that field strength in more complicated motor-effect problems.

A natural next step is analysing a wire suspended in a magnetic field, where the magnetic force balances the wire’s weight.

That changes the task from measuring magnetic field strength experimentally to using the motor-effect equation to predict equilibrium:

\[
BIL\sin\theta=mg
\]

The graphing method tells you where \(B\) comes from. The force-balance method shows what you can do with it.