How to Find Current from a Magnetic Balance Experiment

A digital balance sits underneath a horseshoe magnet. A straight wire passes through the gap between the magnet’s poles, supported by a separate stand. When current flows through the wire, the balance reading changes, even though the wire never touches the balance.

How can that change be used to find the current?

The method is to convert the change in balance reading into a magnetic force, then rearrange the motor-effect equation to calculate the current:

\[ \boxed{I=\frac{|\Delta m|g}{BL\sin\theta}} \]

For a conductor perpendicular to the magnetic field, this simplifies to:

\[ \boxed{I=\frac{|\Delta m|g}{BL}} \]

Here, \(B\) is the magnetic flux density in the gap, \(L\) is the length of conductor inside the magnetic field, and \(\Delta m\) is the change in the balance’s mass-equivalent reading.

The important distinction is that the balance measures a force indirectly. It does not measure the magnetic field strength or the current directly.

01What happens in a magnetic balance experiment?

Start with the apparatus.

A permanent magnet sits on an electronic balance. A straight conducting wire is held independently, with a section of the wire passing through the gap between the north and south poles.

The wire is connected to a power supply so that a current can flow through it.

There are three important parts:

• The permanent magnet produces a magnetic field in the gap.
• The conductor carries an electric current through that field.
• The electronic balance measures the change in the apparent weight of the magnet.

Why does the balance reading change?

Before switching on the current, record the balance reading. Call this \(m_0\).

Now switch on the current and record the new reading, \(m_1\).

The difference is:

\[ \Delta m=m_1-m_0 \]

Why does this happen?

A current-carrying conductor experiences a magnetic force when it is placed in an external magnetic field. This is called the motor effect.

However, Newton’s third law also tells us that the conductor exerts an equal and opposite force on the magnet.

The magnet is sitting on the balance, so this reaction force changes the force that the balance must support.

The balance therefore detects the magnetic interaction indirectly.

If the magnet experiences an additional downward force, the balance reading increases. If the magnet experiences an upward force, the reading decreases.

The magnitude of the force is:

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

where:

• \(F\) is the magnetic force in newtons (N),
• \(\Delta m\) is the change in balance reading in kilograms (kg),
• \(g\) is gravitational acceleration, approximately \(9.80\ \text{m s}^{-2}\).

Important: The balance displays a mass-equivalent reading, not a force in newtons. You must multiply the change in mass by gravitational acceleration.

For example, a balance change of \(2.5\ \text{g}\) corresponds to:

\[ F=0.0025\times9.80 \]

\[ F=0.0245\ \text{N} \]

It does not correspond to a force of \(2.5\ \text{N}\).

Predict what happens when the current is reversed

Suppose the current is flowing in one direction and the balance reading increases.

What happens if you reverse the current?

The magnetic force on the conductor reverses direction. The reaction force on the magnet reverses as well, so the balance reading moves in the opposite direction.

If the magnitude of the current stays the same, the magnitude of the force also stays the same.

This gives us a useful experimental check: reversing the current should reverse the balance change.

02Understand the magnetic force equation

The magnetic force on a straight current-carrying conductor in a uniform magnetic field is:

\[ F=BIL\sin\theta \]

This is the equation you need to connect the balance reading to the unknown current.

Each symbol has a specific meaning.

The difference between B and F

This distinction is particularly important in magnetic balance questions.

\(B\) represents the magnetic flux density. It describes the strength of the external magnetic field in the gap.

\(F\) represents the magnetic force acting on the current-carrying conductor.

They are not interchangeable.

• \(B\) is measured in tesla (T).
• \(F\) is measured in newtons (N).

The magnetic field is produced by the permanent magnet. The force develops when the current-carrying conductor interacts with that field.

Why does the length of conductor matter?

Look carefully at the conductor.

Not all of the wire experiences the same magnetic interaction. The relevant length is the section actually inside the magnetic field.

Suppose the entire wire is 30 cm long, but only 8 cm passes through the uniform magnetic field.

You must use:

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

You must not use \(0.30\ \text{m}\).

The remaining wire is outside the relevant field region, so it does not contribute to the force in this simplified calculation.

What does the angle mean?

The angle \(\theta\) is measured between the direction of conventional current and the magnetic field.

There are three important cases.

This last case is worth remembering.

A conductor can carry a substantial current and still experience no magnetic force if its current is parallel to the magnetic field.

A zero balance change therefore does not automatically mean that the current is zero.

03Rearrange the equation to find current

We now have two expressions for the magnetic force:

From the balance reading:

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

From the motor-effect equation:

\[ F=BIL\sin\theta \]

Because both expressions describe the magnitude of the same force, we can equate them:

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

Rearranging to make current the subject:

\[ \boxed{I=\frac{|\Delta m|g}{BL\sin\theta}} \]

For the common arrangement where the conductor is perpendicular to the magnetic field:

\[ \sin90^\circ=1 \]

Therefore:

\[ \boxed{I=\frac{|\Delta m|g}{BL}} \]

This is the main equation to remember when finding current from a magnetic balance experiment.

A reliable calculation sequence

When you encounter one of these questions, follow this order:

1. Find the difference between the balance readings.
2. Convert the mass difference into kilograms.
3. Calculate the magnetic force using \(F=|\Delta m|g\).
4. Identify the magnetic flux density \(B\).
5. Identify the active conductor length \(L\), in metres.
6. Determine the angle \(\theta\) between the current and magnetic field.
7. Substitute into the rearranged equation and calculate \(I\).

A useful check is to make sure your final current is in amperes.

The units also make sense:

\[ 1\ \text{T}=\frac{1\ \text{N}}{1\ \text{A}\times1\ \text{m}} \]

Therefore, dividing force by magnetic flux density and conductor length gives current in amperes.

04Worked examples

05Common mistakes and experimental limitations

There are several ways to get the wrong current even when you remember the correct equation.

What if the magnetic field is not known?

You cannot calculate the current from the balance reading alone.

You also need the magnetic flux density \(B\).

It must be supplied in the question or measured independently, for example using a calibrated magnetic field sensor.

Be careful not to confuse this with the equation for the magnetic field around a long straight current-carrying wire:

\[ B=\frac{\mu_0 I}{2\pi r} \]

That equation describes the field produced by a wire itself. In a magnetic balance experiment, \(B\) normally refers to the external field produced by the permanent magnet.

For that separate situation, see Magnetic Field Around a Current-Carrying Wire: HSC Physics.

What if the magnetic force is not vertical?

An electronic balance measures the vertical component of the force acting on it.

The standard magnetic balance arrangement is designed so that the magnetic force is vertical. This allows us to use the full force magnitude:

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

If the force is at an angle to the vertical, the balance measures only its vertical component.

You cannot automatically equate the total magnetic force with the balance-derived force in that situation.

How can you improve the experiment?

A single current-on reading can be affected by a small balance offset or drift.

A useful improvement is to take readings with the current flowing in both directions.

Because reversing the current reverses the magnetic force, the two readings move in opposite directions. Their difference can be used to eliminate a constant balance offset.

If the readings are \(m_+\) and \(m_-\), the magnitude of the mass-equivalent magnetic change is:

\[ |\Delta m|=\frac{|m_+-m_-|}{2} \]

This assumes the current has the same magnitude in both directions and the balance offset remains constant.

Other useful precautions include:

• Ensure the conductor does not touch the magnet.
• Keep the active conductor segment straight.
• Use a uniform magnetic field over the active length.
• Keep the current steady while taking readings.
• Avoid excessive current that could heat the wire.
• Measure the active length carefully.
• Repeat readings and calculate an average.

These precautions help ensure that the balance change is caused by the magnetic interaction rather than an unrelated disturbance.

06Questions and solutions

The following questions progress from a standard current calculation to more demanding experimental and conceptual problems.

Question 1

A straight conductor is placed perpendicular to a uniform magnetic field between the poles of a permanent magnet.

The balance initially reads \(182.40\ \text{g}\). When the current is switched on, the reading changes to \(180.44\ \text{g}\).

The magnetic flux density is \(0.30\ \text{T}\), and the active conductor length is \(8.0\ \text{cm}\).

Calculate the magnitude of the current.

Solution 1

The current is approximately \(0.80\ \text{A}\).

First, calculate the change in balance reading:

\[ \Delta m=180.44-182.40 \]

\[ \Delta m=-1.96\ \text{g} \]

Taking the magnitude and converting to kilograms:

\[ |\Delta m|=0.00196\ \text{kg} \]

Calculate the magnetic force:

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

\[ F=0.00196\times9.80 \]

\[ F=0.019208\ \text{N} \]

Convert the conductor length:

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

Since the conductor is perpendicular to the magnetic field:

\[ \sin90^\circ=1 \]

Rearrange and substitute:

\[ \begin{aligned} I&=\frac{F}{BL}\\ &=\frac{0.019208}{0.30\times0.080}\\ &=0.8003\ \text{A} \end{aligned} \]

Therefore:

\[ \boxed{I=0.80\ \text{A}} \]

The important step was converting the balance change into kilograms before calculating the force.

Question 2

A conductor of active length \(0.12\ \text{m}\) is placed in a uniform magnetic field of \(0.36\ \text{T}\).

The angle between the conventional current direction and the magnetic field is \(30^\circ\).

The balance reading changes by \(2.20\ \text{g}\).

The conductor and magnetic field both lie in the horizontal plane, so the magnetic force is vertical.

Calculate the current.

Explain what error would occur if a student incorrectly assumed that the conductor was perpendicular to the magnetic field.

Solution 2

The current is approximately \(1.00\ \text{A}\). Assuming the conductor is perpendicular would underestimate the current.

First, convert the balance change into kilograms:

\[ \Delta m=0.00220\ \text{kg} \]

Calculate the magnetic force:

\[ F=0.00220\times9.80 \]

\[ F=0.02156\ \text{N} \]

Now use the full magnetic force equation:

\[ I=\frac{F}{BL\sin\theta} \]

Substitute:

\[ I= \frac{0.02156} {0.36\times0.12\times\sin30^\circ} \]

Since:

\[ \sin30^\circ=0.5 \]

\[ I=\frac{0.02156}{0.0216} \]

\[ \boxed{I\approx1.00\ \text{A}} \]

Now consider the incorrect perpendicular assumption.

If a student used \(\sin90^\circ=1\), they would calculate:

\[ I=\frac{0.02156}{0.36\times0.12} \]

\[ I\approx0.50\ \text{A} \]

This is only half the correct result.

The reason is that the conductor is at \(30^\circ\) to the field, so its magnetic force is only half the force it would experience if it were perpendicular.

A good habit is to identify the angle before substituting anything into the equation.

Question 3

A student is using a magnetic balance experiment to find an unknown current.

The magnetic flux density is \(0.25\ \text{T}\), and the active conductor length is \(0.080\ \text{m}\). The conductor is perpendicular to the magnetic field.

The balance has a small unknown constant offset, so a reliable zero-current reading is unavailable.

The student records:

Assume the current has the same magnitude in both directions.

Calculate the current, explaining how reversing the current allows the student to account for the unknown balance offset.

Solution 3

The current is \(0.98\ \text{A}\).

The key is to recognise that reversing the current reverses the magnetic force, producing equal and opposite balance changes.

The difference between the two readings is:

\[ 254.6-250.6=4.0\ \text{g} \]

Because the magnetic force reverses, this difference represents twice the magnitude of the mass-equivalent magnetic change.

Therefore:

\[ |\Delta m|=\frac{4.0}{2} \]

\[ |\Delta m|=2.0\ \text{g} \]

Convert to kilograms:

\[ |\Delta m|=0.0020\ \text{kg} \]

Calculate the force:

\[ F=0.0020\times9.80 \]

\[ F=0.0196\ \text{N} \]

Now calculate the current:

\[ I=\frac{F}{BL} \]

\[ I=\frac{0.0196}{0.25\times0.080} \]

\[ \boxed{I=0.98\ \text{A}} \]

The unknown constant balance offset cancels because the readings are taken on opposite sides of the same midpoint.

A tempting mistake is to use the full \(4.0\ \text{g}\) difference as the mass change for one direction. That would double the calculated force and current.

The half-difference method works because the two magnetic forces are equal in magnitude and opposite in direction.

Question 4

An experimenter investigates how the magnetic force changes with the length of conductor inside a uniform magnetic field.

The current and magnetic flux density remain constant throughout the experiment.

The magnetic flux density is \(0.40\ \text{T}\), and the conductor is perpendicular to the field.

The experiment produces the following data:

The balance readings contain a small constant positive offset.

(a) Determine the current using the gradient of the mass-change-versus-length relationship.

(b) Explain why the graph does not pass through the origin.

(c) Explain why calculating the current using only the first data point would give a less reliable result.

Solution 4

(a) The current is approximately \(0.49\ \text{A}\).

The relationship between magnetic force and active conductor length is:

\[ F=BIL \]

Since \(B\) and \(I\) are constant:

\[ F\propto L \]

The gradient of the force-versus-length graph is therefore:

\[ \text{Gradient}=BI \]

However, the experiment records mass-equivalent changes rather than force directly.

Since:

\[ F=\Delta m g \]

we can calculate the gradient using the mass data.

Use the first and last data points:

\[ \text{Gradient} = \frac{3.4-1.0}{0.16-0.04} \]

\[ =\frac{2.4\ \text{g}}{0.12\ \text{m}} \]

\[ =20\ \text{g m}^{-1} \]

Convert to kilograms per metre:

\[ \text{Gradient}=0.020\ \text{kg m}^{-1} \]

Convert this to a force gradient:

\[ \frac{\Delta F}{\Delta L} =0.020\times9.80 \]

\[ =0.196\ \text{N m}^{-1} \]

Since:

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

we obtain:

\[ I=\frac{0.196}{0.40} \]

\[ \boxed{I=0.49\ \text{A}} \]

(b) The graph does not pass through the origin because of the constant balance offset.

The magnetic force should be zero when the active conductor length is zero.

However, the balance has a small positive offset, which adds approximately \(0.20\ \text{g}\) to each recorded value.

This shifts the graph upwards without changing its gradient.

The offset is a measurement issue, not evidence that a magnetic force exists when no conductor is inside the field.

(c) Using the first data point is less reliable because the offset has a proportionally larger effect on smaller measurements.

For example, the first reading is \(1.0\ \text{g}\), but approximately \(0.20\ \text{g}\) comes from the offset.

Using that single reading would produce:

\[ I=\frac{0.0010\times9.80}{0.40\times0.04} \]

\[ I=0.6125\ \text{A} \]

This is noticeably different from the gradient result.

Using several measurements to calculate the gradient reduces the influence of a constant offset and makes better use of the experimental data.

Question 5

A particularly interesting magnetic balance arrangement uses two active conductor segments instead of one.

A single continuous insulated wire is folded into a hairpin shape. Two straight, parallel sections of the wire pass through separate magnetic gaps.

Both magnet assemblies rest on the same electronic balance. The wire is held by an independent support.

The magnetic fields in both gaps point horizontally from left to right. However, because the wire doubles back on itself, the conventional currents in the two active sections travel in opposite directions.

Both sections are perpendicular to their respective magnetic fields. Their magnetic forces therefore act vertically in opposite directions.

The experimental information is:

The balance records:

The balance has an unknown constant offset.

(a) Calculate the current flowing through the wire.

(b) Explain why adding the two magnetic force magnitudes would give an incorrect result.

(c) Suppose the magnetic flux density and active length of the two sections were adjusted so that \(B_A L_A=B_B L_B\). What would the balance read when the current is switched on? Would this prove that there is no current?

Solution 5

(a) The current is approximately \(1.93\ \text{A}\).

The important detail is that the two active conductor sections experience magnetic forces in opposite directions.

We must calculate the net force, not add the two force magnitudes.

First, use the two balance readings to remove the unknown constant offset.

The difference between the readings is:

\[ 306.9-293.1=13.8\ \text{g} \]

Since reversing the current reverses the net force:

\[ |\Delta m|=\frac{13.8}{2} \]

\[ |\Delta m|=6.9\ \text{g} \]

Convert to kilograms:

\[ |\Delta m|=0.0069\ \text{kg} \]

The magnitude of the net magnetic force is:

\[ F_{\text{net}}=|\Delta m|g \]

\[ F_{\text{net}}=0.0069\times9.80 \]

\[ F_{\text{net}}=0.06762\ \text{N} \]

Now consider the two conductor sections separately.

For section A:

\[ F_A=B_AIL_A \]

\[ F_A=0.40I\times0.15 \]

\[ F_A=0.060I \]

For section B:

\[ F_B=B_BIL_B \]

\[ F_B=0.25I\times0.10 \]

\[ F_B=0.025I \]

The forces act in opposite directions, so the net force magnitude is:

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

Therefore:

\[ F_{\text{net}}=0.060I-0.025I \]

\[ F_{\text{net}}=0.035I \]

Substitute the experimentally determined net force:

\[ 0.06762=0.035I \]

Rearranging:

\[ I=\frac{0.06762}{0.035} \]

\[ \boxed{I=1.93\ \text{A}} \]

The current through the continuous wire is approximately \(1.93\ \text{A}\).

(b) Adding the two force magnitudes would be incorrect because the forces act in opposite directions.

A tempting approach is to calculate:

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

This would be correct only if the two forces acted in the same direction.

Here, the wire doubles back on itself. The current directions in the two active sections are opposite, while the magnetic fields point in the same direction.

Consequently, the magnetic forces oppose one another.

The balance measures their net effect.

This is a vector addition problem, not simply an addition of force magnitudes.

(c) The balance would show no change, even though current is flowing.

If:

\[ B_A L_A=B_B L_B \]

then:

\[ F_A=F_B \]

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

Therefore:

\[ F_{\text{net}}=F_A-F_B=0 \]

The balance would register no net change.

However, this does not mean that the current is zero. Both sections can experience substantial magnetic forces that cancel when considered together.

This is the important limitation of using a magnetic balance: it measures the net force on the supported magnet assembly, not every individual magnetic force acting within the apparatus.

If the forces cancel perfectly, the balance alone cannot reveal the current.

An independent measurement or a modified experimental arrangement would be needed.

06Where to go next

Finding current from a magnetic balance experiment brings together three important ideas:

• The balance converts a change in apparent mass into a force.
• The motor-effect equation connects that force to magnetic flux density, current, conductor length, and angle.
• Newton’s third law explains why a balance supporting the magnet can measure the interaction with a conductor it does not support.

The next useful step is to become confident predicting the direction of the magnetic force using Fleming’s left-hand rule.

Being able to calculate the magnitude is only half the job. In more demanding HSC Physics questions, you may also need to explain why reversing the current reverses the force, why changing the conductor’s orientation changes its magnitude, and why multiple sections of a conductor can produce forces that partially or completely cancel.