How to Calculate Solenoid Turns for a Target Magnetic Field

Two air-core solenoids carry the same current of \(1.0\ \text{A}\). Solenoid A has 300 turns spread over \(0.15\ \text{m}\). Solenoid B has 600 turns spread over \(0.30\ \text{m}\).

Which produces the stronger magnetic field at its centre?

You might predict solenoid B because it has twice as many turns. But both coils have exactly 2000 turns per metre, so the ideal solenoid model predicts that their magnetic fields are equal.

This is the important idea behind calculating the number of turns needed for a target magnetic field: the number of turns matters in relation to the length of the solenoid and the current flowing through it.

The calculation you need is:

\[ \boxed{N=\frac{BL}{\mu_0 I}} \]

It tells you how many turns to wind when you already know the magnetic field you want, the solenoid’s length, and its operating current.

01Calculating the number of solenoid turns

Start with the magnetic field equation

For a long, tightly wound air-core solenoid, the magnetic flux density near its centre is approximately:

\[ B=\frac{\mu_0 NI}{L} \]

To calculate the number of turns, rearrange this equation to make \(N\) the subject:

\[ \boxed{N=\frac{BL}{\mu_0 I}} \]

This is the main equation for solving target-field solenoid problems.

Each symbol represents a physical quantity:

For HSC-style calculations, use:

\[ \mu_0\approx4\pi\times10^{-7} \ \text{T m A}^{-1} \]

Notice that \(L\) is the physical length of the coil, not the total length of copper wire used to wind it.

Why the number of turns per metre matters

A solenoid produces its magnetic field through the combined contributions of its individual current-carrying loops.

Adding turns strengthens the field, provided the other relevant quantities stay constant. However, spreading those turns over a greater length reduces their concentration.

The useful quantity is therefore the number of turns per unit length:

\[ n=\frac{N}{L} \]

Here, \(n\) is measured in turns per metre (\(\text{m}^{-1}\)).

The magnetic field equation can now be written as:

\[ B=\mu_0 nI \]

This gives three useful predictions:

• Increasing \(N\) while keeping \(L\) and \(I\) constant increases the field.
• Increasing \(L\) while keeping \(N\) and \(I\) constant decreases the field.
• Increasing \(I\) while keeping \(N\) and \(L\) constant increases the field.

For your design calculation, this means you cannot select a turn count without considering the length and current.

A quick method for finding the required turns

Use this sequence when approaching an exam question.

1. Identify the target magnetic flux density, \(B\).
2. Identify the solenoid’s axial length, \(L\).
3. Identify the current flowing through the solenoid, \(I\).
4. Convert all quantities into SI units.
5. Substitute into \(N=BL/(\mu_0 I)\).
6. Round the result appropriately and check that the resulting field meets the target.

One particularly common mistake is leaving the magnetic field in millitesla.

Remember:

\[ 1\ \text{mT}=1\times10^{-3}\ \text{T} \]

For example, a target field of \(3.5\ \text{mT}\) must be entered as \(0.0035\ \text{T}\).

When can you use this equation?

The equation assumes that the solenoid:

• Is sufficiently long compared with its diameter.
• Has turns distributed reasonably evenly along its length.
• Carries a steady current through every turn.
• Has an air core or a non-magnetic core with approximately the permeability of free space.
• Is being considered at a point well away from its ends.

The magnetic field near the centre is approximately uniform under these conditions.

A short coil has more significant end effects. A ferromagnetic core also changes the relationship between current, turns, and magnetic field. Its permeability may vary with field strength and magnetic history, so the simple air-core calculation cannot automatically be applied.

For the underlying field pattern and the derivation of the solenoid equation, see Magnetic Field Inside a Solenoid: HSC Physics Explained.

02Testing the relationship experimentally

Suppose you want to check whether increasing the number of turns actually increases the magnetic field as predicted.

You could investigate this using a solenoid, a DC power supply, an ammeter, and a Hall probe.

Setting up the experiment

Connect the positive terminal of a low-voltage DC supply to a switch, then a current-limiting resistor, an ammeter, and the solenoid. Connect the other end of the solenoid back to the negative terminal.

The ammeter measures the current flowing through the coil.

Place a Hall probe along the solenoid’s central axis, approximately halfway between its ends. Connect the probe to a teslameter so that it measures the magnetic flux density at this position.

The probe must be aligned with the magnetic field. Its position should remain unchanged between measurements.

Before taking readings, zero the Hall probe with the solenoid switched off.

Then switch on the circuit, record the current and measure the magnetic field at the centre.

What should the experiment show?

The ideal solenoid model predicts that:

\[ B\propto\frac{N}{L} \]

provided the current remains constant.

Consider three hypothetical coils carrying a current of \(1.00\ \text{A}\).

The predicted central magnetic fields are:

These are illustrative values calculated from the ideal solenoid equation, not experimental measurements.

Notice what happens.

Coils A and B have different total turn counts but identical turn densities. Their predicted magnetic fields are equal.

Coil C has the same length as A but twice the number of turns. Its predicted magnetic field is twice as strong.

What can go wrong with the experiment?

An initial comparison might simply use two coils with different numbers of turns and assume that any change in magnetic field is caused by the turn count.

That experiment does not isolate the effect of \(N\) if the coil lengths also change.

The improved design keeps the length and current constant while changing the number of turns. Alternatively, calculate \(N/L\) for each coil and compare the predicted fields.

There is another important experimental issue: adding turns usually increases the resistance of the wire. If the supply voltage stays constant, the current may decrease.

You must therefore measure the actual current rather than assume it remains unchanged.

Real measurements may also differ from the ideal predictions because of end effects, probe alignment, surrounding magnetic fields, and heating of the wire.

03Finding the magnetic field direction

Calculating the required number of turns determines the magnitude of the magnetic field. It does not tell you which way the field points.

You need a separate direction rule.

For a solenoid, use the right-hand grip rule.

Curl the fingers of your right hand in the direction of conventional current flowing around the wire turns. Your extended thumb points in the direction of the magnetic field inside the solenoid.

It also points towards the solenoid’s north pole.

If you are looking directly at one end of the solenoid, there is an even quicker way to identify its pole.

Why conventional current matters

In metal wires, electrons actually drift in the opposite direction to conventional current.

However, standard HSC direction rules use conventional current.

If a diagram shows electron flow, reverse its direction before applying the right-hand grip rule.

Also, do not confuse the right-hand grip rule with a force rule.

They answer different questions:

• Right-hand grip rule: Which direction does the magnetic field produced by a current-carrying coil point?
• Fleming’s left-hand rule: Which direction does a force act on a current-carrying conductor in an external magnetic field?
• Right-hand cross-product rule: Which direction does the magnetic force on a positive moving charge point?

For a negative charge, the magnetic force direction is reversed.

The key distinction is that a solenoid’s current produces a magnetic field, while a force rule predicts how an object responds to an existing magnetic field.

For more on the difference between magnetic fields and magnetic forces, see Electric Fields vs Magnetic Fields: HSC Physics Explained.

04Worked examples

05Questions and solutions

Question 1

A long air-core solenoid has a length of \(0.35\ \text{m}\) and carries a current of \(1.4\ \text{A}\).

Calculate the minimum whole number of turns needed to produce a magnetic flux density of at least \(2.0\ \text{mT}\) at its centre.

Solution 1

The minimum number of turns required is 398 turns.

Start by converting the magnetic field into tesla:

\[ B=2.0\times10^{-3}\ \text{T} \]

Use:

\[ N=\frac{BL}{\mu_0 I} \]

Substitute:

\[ N= \frac{(2.0\times10^{-3})(0.35)} {(4\pi\times10^{-7})(1.4)} \]

\[ N=397.9 \]

Because the target is a minimum, round upwards:

\[ \boxed{N=398\text{ turns}} \]

Check the resulting field:

\[ B=\frac{(4\pi\times10^{-7})(398)(1.4)} {0.35} \]

\[ \boxed{B\approx2.001\ \text{mT}} \]

This is slightly above the required field.

Rounding down to 397 turns would not satisfy the minimum target.

Question 2

A long air-core solenoid has 720 turns distributed over a length of \(0.36\ \text{m}\). It carries a current of \(1.20\ \text{A}\).

The solenoid is redesigned with a length of \(0.48\ \text{m}\). The new power supply can provide a current of \(1.50\ \text{A}\).

The designer wants to maintain exactly the same magnetic field at the centre.

Calculate:

(a) The number of turns required in the redesigned solenoid.

(b) The percentage change in magnetic field if the designer instead keeps the original 720 turns.

Solution 2

(a) The redesigned solenoid requires 768 turns.

(b) If it retains 720 turns, its magnetic field decreases by 6.25%.

Part (a): Calculate the new turn count.

The original and redesigned solenoids must produce the same magnetic field.

Therefore:

\[ B_A=B_B \]

Using the solenoid equation:

\[ \frac{\mu_0 N_A I_A}{L_A} = \frac{\mu_0 N_B I_B}{L_B} \]

The permeability cancels:

\[ \frac{N_A I_A}{L_A} = \frac{N_B I_B}{L_B} \]

Rearrange:

\[ N_B=N_A\frac{L_B}{L_A} \frac{I_A}{I_B} \]

Substitute:

\[ N_B= 720\times\frac{0.48}{0.36} \times\frac{1.20}{1.50} \]

\[ N_B=720\times1.333\times0.800 \]

\[ \boxed{N_B=768\text{ turns}} \]

The new solenoid needs more turns because it is longer.

However, its greater current helps compensate for the increased length, so the number of turns does not need to increase by the same percentage as the length.

Part (b): Calculate the field change if the turn count stays at 720.

Compare the new field with the original field:

\[ \frac{B_B}{B_A} = \frac{N_B I_B/L_B}{N_A I_A/L_A} \]

Both turn counts are now 720, so:

\[ \frac{B_B}{B_A} = \frac{1.50/0.48}{1.20/0.36} \]

\[ \frac{B_B}{B_A}=0.9375 \]

The redesigned field is 93.75% of the original field.

Therefore:

\[ \text{Percentage decrease} =(1-0.9375)\times100 \]

\[ \boxed{6.25\%\text{ decrease}} \]

The tempting approach is to increase the turns by the same proportion as the length. That ignores the increased current.

Both quantities affect the field, so they must be considered together.

Question 3

Two long air-core solenoids are used in an experiment to investigate how turn density affects magnetic field strength.

Solenoid P has:

• 360 turns
• Length of \(0.18\ \text{m}\)
• Current of \(0.80\ \text{A}\)

Solenoid Q has:

• 600 turns
• Length of \(0.30\ \text{m}\)
• Current of \(0.80\ \text{A}\)

A Hall probe is positioned at the centre of each solenoid.

(a) Calculate the predicted magnetic field at the centre of each solenoid.

(b) Explain why the solenoid with 600 turns does not produce a stronger field initially.

(c) Solenoid Q is then compressed uniformly until its length is \(0.24\ \text{m}\). Its current and number of turns remain unchanged. Calculate its new magnetic field and the percentage increase in field strength.

Solution 3

(a) Both original solenoids produce approximately \(2.01\ \text{mT}\).

(b) They have equal turn densities, so their magnetic fields are equal despite their different total turn counts.

(c) After compression, solenoid Q produces approximately \(2.51\ \text{mT}\), an increase of 25%.

Part (a): Calculate the field in solenoid P.

Use:

\[ B=\frac{\mu_0 NI}{L} \]

For solenoid P:

\[ B_P= \frac{(4\pi\times10^{-7})(360)(0.80)} {0.18} \]

\[ B_P=2.011\times10^{-3}\ \text{T} \]

Therefore:

\[ \boxed{B_P=2.01\ \text{mT}} \]

Now calculate the field in solenoid Q:

\[ B_Q= \frac{(4\pi\times10^{-7})(600)(0.80)} {0.30} \]

\[ B_Q=2.011\times10^{-3}\ \text{T} \]

Therefore:

\[ \boxed{B_Q=2.01\ \text{mT}} \]

Part (b): Explain why the fields are equal.

Calculate the turn density of each solenoid.

For P:

\[ n_P=\frac{360}{0.18} =2000\ \text{m}^{-1} \]

For Q:

\[ n_Q=\frac{600}{0.30} =2000\ \text{m}^{-1} \]

Both solenoids have the same turn density and carry the same current.

Since:

\[ B=\mu_0 nI \]

their magnetic fields must be equal under the ideal solenoid model.

Having more total turns does not automatically produce a stronger magnetic field.

Part (c): Calculate the field after compression.

The new length is:

\[ L=0.24\ \text{m} \]

The number of turns and current are unchanged.

Substitute:

\[ B_{\text{new}}= \frac{(4\pi\times10^{-7})(600)(0.80)} {0.24} \]

\[ B_{\text{new}}=2.513\times10^{-3}\ \text{T} \]

Therefore:

\[ \boxed{B_{\text{new}}=2.51\ \text{mT}} \]

Calculate the percentage increase:

\[ \text{Percentage increase} = \frac{B_{\text{new}}-B_{\text{initial}}} {B_{\text{initial}}}\times100 \]

Using the unrounded values:

\[ =\frac{2.513-2.011}{2.011}\times100 \]

\[ \boxed{25.0\%} \]

The field increases because the same 600 turns are now distributed over a shorter length.

The turn density has increased from 2000 to 2500 turns per metre.

Question 4

A long air-core solenoid has 640 turns distributed over a length of \(0.32\ \text{m}\). It carries a conventional current of \(1.25\ \text{A}\).

An observer looks directly at the right-hand end of the solenoid and sees the conventional current circulating clockwise.

(a) Calculate the magnetic flux density at the centre.

(b) Identify the north and south poles of the solenoid and state the direction of the magnetic field inside it.

(c) In a separate motor-effect experiment, a straight wire carries conventional current vertically upwards through a uniform magnetic field directed into the page. Determine the direction of the force on the wire.

Explain why the right-hand grip rule should not be used for part (c).

Solution 4

(a) The magnetic flux density is approximately \(3.14\ \text{mT}\).

(b) The right-hand end is south, the left-hand end is north, and the internal magnetic field points from right to left.

(c) The force on the separate straight wire acts towards the left.

Part (a): Calculate the field magnitude.

Use:

\[ B=\frac{\mu_0 NI}{L} \]

Substitute:

\[ B= \frac{(4\pi\times10^{-7})(640)(1.25)} {0.32} \]

\[ B=3.142\times10^{-3}\ \text{T} \]

Therefore:

\[ \boxed{B=3.14\ \text{mT}} \]

Part (b): Determine the poles and field direction.

The observer sees clockwise conventional current when looking at the right-hand end.

The end-on right-hand rule states:

• Anticlockwise current means north.
• Clockwise current means south.

Therefore, the right-hand end is the south pole.

The opposite end must be north.

Inside a solenoid, magnetic field lines travel from south to north. Therefore, the field points from the right-hand end towards the left-hand end.

Part (c): Determine the force direction.

The right-hand grip rule is not appropriate here because we are no longer finding the magnetic field produced by a solenoid.

We are finding the force on a current-carrying wire placed in an existing magnetic field.

Use Fleming’s left-hand rule:

• First finger: magnetic field, into the page.
• Second finger: conventional current, upwards.
• Thumb: force, towards the left.

Therefore:

\[ \boxed{\text{Force acts to the left}} \]

The distinction is important. The solenoid grip rule determines a magnetic field, while Fleming’s left-hand rule determines a magnetic force.

Question 5

A student is designing a long air-core solenoid with a fixed axial length of \(0.20\ \text{m}\).

The solenoid is connected in series with a \(1.0\ \Omega\) resistor to an ideal \(6.0\ \text{V}\) DC supply.

The resistance contributed by each wire turn is \(0.020\ \Omega\). Assume that:

• Every additional turn has the same mean circumference.
• The same wire is used in both designs.
• The coil’s length and diameter remain effectively unchanged.
• The resistor remains at \(1.0\ \Omega\).
• Heating and changes in wire resistance are negligible.
• The supply voltage remains constant at \(6.0\ \text{V}\).

The student compares two designs:

• Design A: 200 turns.
• Design B: 400 turns.

(a) Calculate the total circuit resistance and current for each design.

(b) Calculate the magnetic field at the centre of each solenoid.

(c) Determine the percentage increase in magnetic field when the number of turns doubles.

(d) The student removes the \(1.0\ \Omega\) series resistor and repeats the comparison using the same \(6.0\ \text{V}\) supply. Predict whether doubling the turns now increases, decreases, or leaves the magnetic field unchanged. Justify your answer mathematically.

(e) Instead, suppose a current-regulated supply maintains a current of \(1.20\ \text{A}\) for both original designs, with the \(1.0\ \Omega\) series resistor still present. Calculate the magnetic field for the 400-turn design and the voltage required from the supply.

Solution 5

(a) The circuit currents are \(1.20\ \text{A}\) for the 200-turn design and \(0.667\ \text{A}\) for the 400-turn design.

(b) Their magnetic fields are approximately \(1.51\ \text{mT}\) and \(1.68\ \text{mT}\), respectively.

(c) Doubling the turns increases the magnetic field by approximately 11.1%, not 100%.

(d) Without the series resistor, the magnetic field remains unchanged when the number of turns doubles.

(e) With a regulated current of \(1.20\ \text{A}\), the 400-turn solenoid produces approximately \(3.02\ \text{mT}\) and requires a supply voltage of \(10.8\ \text{V}\).

This question tests an important limitation of the simple proportionality \(B\propto N\): it only holds when the current and length remain constant.

Part (a): Calculate the current in each design.

Start with Design A.

The coil resistance is:

\[ R_{\text{coil}}=200(0.020) \]

\[ R_{\text{coil}}=4.0\ \Omega \]

The \(1.0\ \Omega\) resistor is connected in series, so:

\[ R_{\text{total}}=4.0+1.0 \]

\[ R_{\text{total}}=5.0\ \Omega \]

Using Ohm’s law:

\[ I=\frac{V}{R} \]

\[ I_A=\frac{6.0}{5.0} \]

\[ \boxed{I_A=1.20\ \text{A}} \]

Now consider Design B.

Its coil resistance is:

\[ R_{\text{coil}}=400(0.020) \]

\[ R_{\text{coil}}=8.0\ \Omega \]

The total resistance becomes:

\[ R_{\text{total}}=8.0+1.0 \]

\[ R_{\text{total}}=9.0\ \Omega \]

Therefore:

\[ I_B=\frac{6.0}{9.0} \]

\[ \boxed{I_B=0.667\ \text{A}} \]

Doubling the number of turns has increased the coil resistance and reduced the current.

Part (b): Calculate the magnetic field for each design.

For Design A:

\[ B_A=\frac{\mu_0 N_A I_A}{L} \]

Substitute:

\[ B_A= \frac{(4\pi\times10^{-7})(200)(1.20)} {0.20} \]

\[ \boxed{B_A=1.51\ \text{mT}} \]

For Design B:

\[ B_B=\frac{\mu_0 N_B I_B}{L} \]

Substitute:

\[ B_B= \frac{(4\pi\times10^{-7})(400)(0.667)} {0.20} \]

\[ \boxed{B_B=1.68\ \text{mT}} \]

Part (c): Calculate the percentage increase.

Use:

\[ \text{Percentage increase} = \frac{B_B-B_A}{B_A}\times100 \]

Substituting the unrounded values:

\[ =\frac{1.6755-1.5080}{1.5080}\times100 \]

\[ \boxed{11.1\%} \]

The field increases, but it does not double.

Why?

The number of turns doubles, but the current decreases from \(1.20\ \text{A}\) to \(0.667\ \text{A}\).

The fixed series resistor means that the total circuit resistance does not double. It increases from \(5.0\ \Omega\) to \(9.0\ \Omega\).

Consequently, the product \(NI\), which determines the magnetic field for a fixed length, increases by only a factor of \(10/9\).

Part (d): Remove the series resistor.

Without the external resistor, the circuit resistance is simply the resistance of the solenoid.

For Design A:

\[ I_A=\frac{6.0}{4.0} \]

\[ I_A=1.50\ \text{A} \]

The product of turns and current is:

\[ N_AI_A=200(1.50) \]

\[ N_AI_A=300\ \text{A-turns} \]

For Design B:

\[ I_B=\frac{6.0}{8.0} \]

\[ I_B=0.75\ \text{A} \]

Therefore:

\[ N_BI_B=400(0.75) \]

\[ N_BI_B=300\ \text{A-turns} \]

Both designs now have exactly the same value of \(NI\).

Since the length is also unchanged:

\[ B=\frac{\mu_0 NI}{L} \]

their magnetic fields must be equal.

Calculating the common field:

\[ B=\frac{(4\pi\times10^{-7})(300)} {0.20} \]

\[ \boxed{B=1.885\ \text{mT}} \]

This is the surprising result.

Doubling the number of turns doubles the coil resistance. With a fixed voltage and no additional series resistance, the current halves.

The increase in \(N\) is exactly cancelled by the decrease in \(I\).

Part (e): Use a current-regulated supply.

Now the current is maintained at \(1.20\ \text{A}\) for both designs.

For the 400-turn solenoid:

\[ B=\frac{\mu_0 NI}{L} \]

\[ B=\frac{(4\pi\times10^{-7})(400)(1.20)} {0.20} \]

\[ \boxed{B=3.02\ \text{mT}} \]

Because the current is now held constant, doubling the turns doubles the magnetic field.

However, the supply must provide enough voltage to maintain that current.

The total resistance of the 400-turn circuit is:

\[ R_{\text{total}}=8.0+1.0 \]

\[ R_{\text{total}}=9.0\ \Omega \]

Using Ohm’s law:

\[ V=IR \]

\[ V=(1.20)(9.0) \]

\[ \boxed{V=10.8\ \text{V}} \]

The original \(6.0\ \text{V}\) supply cannot maintain the required \(1.20\ \text{A}\) through this circuit.

The current-regulated design needs a supply capable of providing at least \(10.8\ \text{V}\) under the stated assumptions.

The deeper lesson is that the solenoid equation describes the relationship between field, turns, length, and current. It does not independently determine what current the electrical circuit will actually deliver.

06What to check after calculating the turns

Finding \(N\) is only the first part of designing a solenoid.

Before accepting your answer, check three things:

• Electrical conditions: Is the stated current actually maintained, or will increasing the wire length increase resistance and reduce current?
• Physical construction: Can the required turns fit over the available winding length without unrealistic spacing?
• Magnetic assumptions: Is the solenoid sufficiently long, evenly wound, and measured near its centre for the ideal equation to be reliable?

Most importantly, keep the two direction problems separate.

Use the right-hand grip rule to determine the magnetic field produced by the solenoid. Use a force rule when investigating how a separate current-carrying conductor or moving charge responds to that field.

That distinction becomes particularly useful when you move from calculating magnetic fields to analysing the motor effect and electromagnetic induction.