What Happens to a Solenoid’s Field When Its Length Doubles?

A solenoid has 600 turns spread over a length of 30 cm. You pull the turns apart until the solenoid is 60 cm long. You haven’t added or removed any wire, and the current stays exactly the same.

What happens to the magnetic field inside it?

The magnetic field halves.

The reason is that the same number of turns now occupies twice the length. There are half as many turns per metre, so the magnetic contributions from the windings produce half the original field in the ideal-solenoid model.

But there’s an important condition hidden in that answer: the number of turns and the current must remain constant. Change either of those, and the result can be different.

01Why doubling the length halves the magnetic field

Start by picturing a solenoid as a long coil of wire. Each turn carries conventional current and contributes to the magnetic field inside the coil.

When the turns are closely packed, many turns contribute to the field within a given length. Spread those same turns over a longer distance, and the turn density decreases.

Think of fence posts along a straight track. If you keep the same number of posts but double the track length, the posts become half as closely spaced.

The same density idea applies to a solenoid. The analogy is about spacing, though: every turn still contributes to the magnetic field, including turns some distance away.

The equation behind the change

For a long, ideal, air-core solenoid, the magnetic flux density inside the coil is

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

where:

• \(B\) is the magnetic flux density, measured in tesla (T).
• \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7}\ \text{T m A}^{-1}\).
• \(N\) is the total number of turns.
• \(I\) is the conventional current through the solenoid, measured in amperes (A).
• \(L\) is the length of the solenoid, measured in metres (m).

The important part is the fraction:

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

This is the number of turns per metre, also called the turn density.

We can therefore write

\[ B = \mu_0 nI \]

where \(n=N/L\).

The equation tells us that the magnetic field depends on two things:

1. How many turns are packed into each metre.
2. How much current flows through those turns.

Total turns alone do not determine the field.

What happens mathematically when the length doubles?

Suppose the original solenoid has:

• \(N\) turns
• Length \(L\)
• Current \(I\)
• Magnetic field \(B_1\)

Its original field is

\[ B_1 = \frac{\mu_0 NI}{L} \]

Now double the length while keeping the number of turns and current unchanged.

The new length is \(2L\), so

\[ B_2 = \frac{\mu_0 NI}{2L} \]

Comparing this with the original equation,

\[ B_2 = \frac{1}{2}B_1 \]

Therefore,

\[ \boxed{B_2 = 0.5B_1} \]

The new magnetic field is 50% of the original field.

Notice that we haven’t changed the direction of the current. The magnetic field still points in the same direction. Only its magnitude has decreased.

02The condition that changes the answer

Doubling the length does not automatically halve the magnetic field.

You must check what happens to the number of turns and the current at the same time.

Consider these four situations.

The second situation is particularly important.

Imagine a solenoid that originally has 500 turns over 25 cm. Its turn density is

\[ n_1 = \frac{500}{0.25} = 2000\ \text{turns/m} \]

Now increase its length to 50 cm and add another 500 turns.

The new turn density is

\[ n_2 = \frac{1000}{0.50} = 2000\ \text{turns/m} \]

The turn density has not changed.

If the current also remains constant, the magnetic field stays the same.

This is the distinction worth remembering:

• Increasing the total number of turns increases the field only if the turn density or current also increases.
• Increasing the length decreases the field only if the other relevant quantities do not compensate for that increase.

A useful comparison formula

Rather than calculating two magnetic fields separately, you can compare them directly.

Starting with

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

the ratio of the new field to the original field is

\[ \boxed{ \frac{B_2}{B_1} = \frac{N_2}{N_1} \frac{I_2}{I_1} \frac{L_1}{L_2} } \]

This is particularly useful when an exam question changes several quantities at once.

If the length doubles, the final fraction becomes

\[ \frac{L_1}{L_2} = \frac{1}{2} \]

You then account for any changes in turns and current.

03How could we test whether the field really halves?

The equation makes a prediction. An experiment can test it.

The question is straightforward:

Does stretching a solenoid to twice its original length, while keeping its total turns and current constant, reduce the magnetic field at its centre to approximately half?

Setting up the experiment

A suitable apparatus would contain:

• A long air-core solenoid with a known number of turns.
• An adjustable, non-magnetic former that allows the turns to be spread along the axis.
• A regulated DC power supply.
• An ammeter connected in series with the solenoid.
• A Hall probe connected to a magnetic-field meter.
• A ruler for measuring the solenoid’s length.

The electrical circuit is arranged so that current flows from the positive terminal of the power supply, through the ammeter, through the solenoid, and back to the negative terminal.

The Hall probe sits at the centre of the solenoid, with its sensitive axis aligned with the solenoid’s central axis.

What should we predict?

Keep the following quantities constant:

• Total number of turns, \(N\).
• Current, \(I\).
• Coil diameter.
• Core material.
• Hall probe orientation.

Only change the solenoid’s axial length.

If the original length is \(L\), the second length will be \(2L\).

The ideal-solenoid equation predicts

\[ B_2 = \frac{B_1}{2} \]

So the magnetic-field meter should record approximately half the original reading.

Why a simple compass experiment is not enough

You could place a compass near the solenoid and compare its deflection before and after stretching the coil.

However, this is not a particularly reliable way to measure the change in field strength.

A compass responds to the combined magnetic field from the solenoid and its surroundings, including Earth’s magnetic field. Its deflection angle is also not directly proportional to the solenoid’s field strength.

A better design uses a calibrated Hall probe at the centre of the coil.

We can improve the measurement further by recording the background magnetic field with the solenoid switched off, then correcting the readings for that background.

The current must also be monitored throughout the experiment. If it changes, we cannot attribute the entire change in magnetic field to the solenoid’s length.

Illustrative experimental results

Suppose we use a 600-turn air-core solenoid carrying a constant current of 0.80 A.

We measure the magnetic field at its centre at two different lengths.

The following values are illustrative, not measurements from an actual experiment.

The predicted field approximately halves.

The illustrative measurements also show a reduction of almost 50%.

What could make the measurements differ from the prediction?

The equation assumes an ideal, long solenoid.

A real solenoid has ends. Near those ends, the magnetic field spreads out rather than remaining perfectly uniform.

If the coil is too short compared with its diameter, the central field may not follow the simple inverse-length relationship exactly.

Other sources of experimental error include:

• Small changes in current during the measurement.
• Hall probe misalignment.
• Changes in the probe’s position relative to the centre.
• Nearby magnets or electrical equipment affecting the background field.
• Changes in the temperature of the wire.

The experiment should therefore use a sufficiently long coil, maintain a constant current, and measure at the centre.

The conclusion is that the ideal model predicts an exact halving, while a real experiment should give an approximately halved reading under well-controlled conditions.

04Does doubling the length change the field direction?

No. Not if the conventional current direction remains unchanged.

This is where students sometimes mix up two different magnetic-field rules.

A solenoid produces its own magnetic field. To determine that field’s direction, use the right-hand grip rule.

The right-hand grip rule

Curl the fingers of your right hand around the solenoid in the direction of conventional current.

Your thumb points:

• Along the magnetic field inside the solenoid.
• Towards the solenoid’s north pole.

You can also determine the north and south poles by looking directly at one end.

Remember that these rules use conventional current, which is the direction positive charge would move.

In a metal wire, electrons actually drift in the opposite direction. If a question gives you electron flow, reverse that direction before applying the right-hand grip rule.

For more practice distinguishing these directions, see Electron Flow vs Conventional Current in HSC Magnetic Questions.

Don’t confuse a field direction rule with a force direction rule

The right-hand grip rule determines the magnetic field produced by a solenoid.

Fleming’s left-hand rule, on the other hand, determines the direction of the force on a current-carrying conductor placed in an external magnetic field.

These are different jobs.

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

\[ F = BIL\sin\theta \]

where \(B\) is the external magnetic flux density, \(I\) is the conventional current, \(L\) is the conductor length inside the field, and \(\theta\) is the angle between the current and the field.

If the solenoid’s field halves while everything else about the conductor stays constant, the magnetic force on a perpendicular conductor also halves.

But you must determine the solenoid’s field direction first. Do not use the motor-effect force rule to work out the solenoid’s own magnetic field.

For a broader explanation of the field pattern and right-hand grip rule, see Magnetic Field Inside a Solenoid: HSC Physics Explained.

05Worked examples

06Questions and solutions

Try each question before reading its solution. The questions progress from a standard calculation to situations where several physical effects must be considered together.

Question 1

A long air-core solenoid contains 500 turns over a length of 0.25 m. It carries a constant current of 0.60 A.

(a) Calculate the magnetic flux density inside the solenoid.

(b) The solenoid is stretched until its length doubles. The number of turns and current remain unchanged. Calculate the new magnetic flux density.

(c) When viewed directly from the right-hand end, the conventional current flows anticlockwise around the turns. State the direction of the magnetic field inside the solenoid.

Solution 1

(a) The original magnetic flux density is \(1.51\times10^{-3}\ \text{T}\).

For an ideal solenoid,

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

Substitute:

\[ \begin{aligned} B_1 &=\frac{(4\pi\times10^{-7})(500)(0.60)} {0.25}\\ &=1.51\times10^{-3}\ \text{T} \end{aligned} \]

Therefore,

\[ \boxed{B_1=1.51\ \text{mT}} \]

(b) The new magnetic flux density is \(7.54\times10^{-4}\ \text{T}\).

The new length is

\[ L_2=2(0.25)=0.50\ \text{m} \]

Since the turns and current remain unchanged, the field halves.

\[ B_2=\frac{B_1}{2} \]

\[ B_2=\frac{1.51\times10^{-3}}{2} \]

Therefore,

\[ \boxed{B_2=7.54\times10^{-4}\ \text{T}} \]

(c) The magnetic field points towards the right-hand end of the solenoid.

The conventional current appears anticlockwise when viewed from the right-hand end.

Using the right-hand grip rule, curl your fingers anticlockwise. Your thumb points towards the observer.

The right-hand end is therefore the north pole, and the internal magnetic field points towards it.

The important distinction is that stretching the coil changes the field’s magnitude, not its direction.

Question 2

A long air-core solenoid has 720 turns over a length of 0.24 m and carries a current of 0.40 A.

The solenoid is stretched to twice its original length. During this process, a current controller increases the current to 0.60 A without changing its direction.

(a) Determine the ratio of the final magnetic field to the initial magnetic field.

(b) Calculate the percentage change in magnetic field strength.

(c) Explain why the increase in current does not completely compensate for the increased length.

Solution 2

(a) The final magnetic field is 0.75 times the initial field.

Use the ratio equation:

\[ \frac{B_2}{B_1} = \frac{N_2}{N_1} \frac{I_2}{I_1} \frac{L_1}{L_2} \]

The number of turns is unchanged, so

\[ \frac{N_2}{N_1}=1 \]

The current ratio is

\[ \frac{I_2}{I_1} = \frac{0.60}{0.40} =1.5 \]

The length ratio is

\[ \frac{L_1}{L_2} = \frac{0.24}{0.48} =0.5 \]

Therefore,

\[ \begin{aligned} \frac{B_2}{B_1} &=1\times1.5\times0.5\\ &=0.75 \end{aligned} \]

Hence,

\[ \boxed{B_2=0.75B_1} \]

(b) The magnetic field decreases by 25%.

\[ \begin{aligned} \text{Percentage decrease} &=(1-0.75)\times100\\ &=25\% \end{aligned} \]

Therefore,

\[ \boxed{25\%\text{ decrease}} \]

(c) The current increases by 50%, but the turn density halves.

The increased current makes each turn’s contribution stronger. However, spreading the same number of turns over twice the length reduces the turn density by 50%.

These changes combine multiplicatively:

\[ 1.5\times0.5=0.75 \]

The field therefore becomes 75% of its original value rather than returning to its original strength.

The tempting prediction is that a 50% increase in current should compensate for a 50% reduction in turn density. It does not, because the changes act as multiplication factors, not quantities that can simply be added or subtracted.

Question 3

Two long air-core solenoids, A and B, carry different currents.

Their properties are:

Both solenoids have the same diameter, and their internal fields are measured near their centres.

A student argues that solenoid B must produce the stronger magnetic field because it has 400 additional turns.

Determine whether this prediction is correct. Calculate the magnetic field in each solenoid and compare the results.

Solution 3

Solenoid A produces the stronger magnetic field. Its field is twice that of solenoid B.

The student’s prediction is incorrect because total turns alone do not determine the field. We must consider both turn density and current.

For solenoid A,

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

Substituting,

\[ \begin{aligned} B_A &=\frac{(4\pi\times10^{-7})(500)(1.20)} {0.25}\\ &=3.02\times10^{-3}\ \text{T} \end{aligned} \]

Therefore,

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

For solenoid B,

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

Substituting,

\[ \begin{aligned} B_B &=\frac{(4\pi\times10^{-7})(900)(0.80)} {0.60}\\ &=1.51\times10^{-3}\ \text{T} \end{aligned} \]

Therefore,

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

Comparing the fields,

\[ \frac{B_A}{B_B} = \frac{3.02}{1.51} =2 \]

Thus,

\[ \boxed{B_A=2B_B} \]

Solenoid B has more total turns, but its turns are spread over a much greater length. It also carries a smaller current.

Both effects reduce its magnetic field.

The trap is treating additional turns as though they must increase the field regardless of how far apart those turns are.

Question 4

A long air-core solenoid initially produces a uniform magnetic field of \(2.4\times10^{-3}\ \text{T}\) towards the right.

The solenoid is stretched until its length doubles. Its total number of turns and current remain unchanged.

A straight wire of length 0.12 m is positioned at the centre of the solenoid. It carries a conventional current of 2.0 A vertically upwards, perpendicular to the solenoid’s magnetic field.

(a) Calculate the new magnetic flux density inside the solenoid.

(b) Calculate the magnetic force on the wire before stretching.

(c) Calculate the magnetic force on the wire after stretching.

(d) State the direction of the force in both situations and explain why the force rule is different from the rule used to determine the solenoid’s own magnetic field.

Solution 4

(a) The new magnetic flux density is \(1.2\times10^{-3}\ \text{T}\).

The number of turns and current are unchanged, while the length doubles.

Therefore,

\[ B_2=\frac{B_1}{2} \]

\[ B_2=\frac{2.4\times10^{-3}}{2} \]

Hence,

\[ \boxed{B_2=1.2\times10^{-3}\ \text{T}} \]

The field still points to the right.

(b) The initial magnetic force is \(5.76\times10^{-4}\ \text{N}\).

Use the motor-effect equation:

\[ F=BIL\sin\theta \]

The wire is perpendicular to the field, so

\[ \theta=90^\circ \]

Therefore,

\[ \begin{aligned} F_1 &=(2.4\times10^{-3})(2.0)(0.12) \sin90^\circ\\ &=5.76\times10^{-4}\ \text{N} \end{aligned} \]

Hence,

\[ \boxed{F_1=5.76\times10^{-4}\ \text{N}} \]

(c) The new magnetic force is \(2.88\times10^{-4}\ \text{N}\).

The wire’s current, length, and angle remain unchanged. Only the magnetic field strength halves.

Therefore,

\[ \begin{aligned} F_2 &=(1.2\times10^{-3})(2.0)(0.12) \sin90^\circ\\ &=2.88\times10^{-4}\ \text{N} \end{aligned} \]

Hence,

\[ \boxed{F_2=2.88\times10^{-4}\ \text{N}} \]

(d) The force is directed into the page in both situations.

There are two separate direction calculations here.

First, the solenoid’s right-hand grip rule determines its internal magnetic field. Since the solenoid’s current direction is unchanged, its field still points right.

Second, Fleming’s left-hand rule determines the force on the straight wire.

The conventional current points upwards, and the magnetic field points right. The resulting force is into the page.

You can also express this using the motor-effect vector relationship:

\[ \vec F=I\vec L\times\vec B \]

The force direction remains unchanged because neither the current direction nor the magnetic-field direction reverses.

Its magnitude halves because the magnetic field halves.

This is a useful distinction: changing the strength of a magnetic field does not necessarily change the direction of the force it produces.

Question 5

An ideal long air-core solenoid initially contains \(N\) turns over a length \(L\). It has electrical resistance \(R\) and is connected to an ideal fixed-voltage supply \(V\).

Its initial current is

\[ I_0=\frac{V}{R} \]

and its initial magnetic field is

\[ B_0=\frac{\mu_0 NI_0}{L} \]

Two technicians independently modify the solenoid so that its total length doubles.

Technician A: Re-spacing the existing winding

Technician A spreads the existing \(N\) turns uniformly over a length of \(2L\). No additional wire is added. Assume the resistance remains \(R\) to the precision of the model.

The solenoid remains connected to the original fixed-voltage supply.

Technician B: Adding a second winding section

Technician B adds an identical winding section end-to-end. The new section has the same turn density and winding orientation as the original. It is connected in series-aiding, so conventional current circulates in the same sense around every turn.

The new solenoid therefore has:

• \(2N\) total turns.
• Length \(2L\).
• Total resistance \(2R\).

It is also connected to the original fixed-voltage supply \(V\).

Answer the following:

(a) Determine the magnetic field produced by Technician A’s solenoid in terms of \(B_0\).

(b) Determine the current flowing through Technician B’s solenoid.

(c) Determine the magnetic field produced by Technician B’s solenoid in terms of \(B_0\).

(d) Technician B replaces the fixed-voltage supply with an ideal current-regulated supply that maintains the original current \(I_0\). What is the new magnetic field?

(e) Explain why the two technicians initially obtain the same magnetic field reduction, even though they change the solenoid in different ways. State whether the magnetic-field direction changes.

Solution 5

(a) Technician A produces a magnetic field of

\[ \boxed{B_A=\frac{B_0}{2}} \]

The number of turns remains \(N\), and the length doubles to \(2L\).

The resistance also remains \(R\), so the fixed-voltage supply maintains the original current:

\[ I_A=\frac{V}{R}=I_0 \]

Substitute into the solenoid equation:

\[ B_A=\frac{\mu_0 NI_0}{2L} \]

Comparing this with the original field,

\[ B_0=\frac{\mu_0 NI_0}{L} \]

gives

\[ \boxed{B_A=0.5B_0} \]

This is the familiar case: the same turns and current are spread over twice the length.

(b) Technician B’s solenoid carries half the original current.

The new resistance is \(2R\), while the supply voltage remains \(V\).

Using Ohm’s law,

\[ I_B=\frac{V}{2R} \]

Since

\[ I_0=\frac{V}{R} \]

we obtain

\[ \boxed{I_B=\frac{I_0}{2}} \]

The current halves because the resistance doubles.

(c) Technician B also produces a magnetic field of

\[ \boxed{B_B=\frac{B_0}{2}} \]

This is the less obvious result.

The total number of turns doubles, but the length also doubles. Therefore, the turn density stays constant:

\[ \frac{2N}{2L}=\frac{N}{L} \]

If we only considered turn density, we might predict that the magnetic field remains unchanged.

But the current has also halved.

Substitute the new values into the solenoid equation:

\[ B_B=\frac{\mu_0(2N)(I_0/2)}{2L} \]

Simplifying,

\[ B_B=\frac{\mu_0 NI_0}{2L} \]

Therefore,

\[ \boxed{B_B=0.5B_0} \]

The field halves because the turn density stays constant while the current halves.

(d) With an ideal current-regulated supply maintaining \(I_0\), Technician B produces the original magnetic field:

\[ \boxed{B_B=B_0} \]

The number of turns and length have both doubled, so the turn density is unchanged.

The current is now maintained at its original value rather than halving.

Therefore,

\[ \begin{aligned} B_B &=\frac{\mu_0(2N)I_0}{2L}\\ &=\frac{\mu_0 NI_0}{L}\\ &=B_0 \end{aligned} \]

The current-regulated supply must adjust its output voltage to maintain the current through the increased resistance.

(e) Both technicians initially produce half the original magnetic field, but for different reasons.

Technician A keeps the number of turns and current constant. The turn density halves, so the field halves.

Technician B keeps the turn density constant, but the current halves because the resistance doubles while the supply voltage stays fixed.

The resulting field therefore also halves.

In both cases, the magnetic-field direction remains unchanged. Technician A retains the original current direction, and Technician B’s second section is connected so that all turns reinforce the original field.

There are two tempting mistakes here:

• Assuming that doubling the solenoid’s axial length automatically doubles the resistance. It does not necessarily do so. Technician A has re-spaced the existing winding rather than adding a second length of wire.
• Assuming that keeping the turn density constant guarantees an unchanged field. It does not. The current must also remain constant.

The deeper lesson is that the magnetic field depends on both turn density and current. Neither quantity can be ignored when the solenoid is modified.

06The rule to take into your HSC exam

When a question asks what happens to the magnetic field when a solenoid’s length doubles, don’t immediately write that the field halves.

First, identify what happens to the other quantities.

For a long, ideal, air-core solenoid,

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

If the number of turns and current stay constant,

\[ \boxed{L\rightarrow2L \quad\Longrightarrow\quad B\rightarrow\frac{B}{2}} \]

If the turns also double, the turn density stays constant. If the current also changes, include that change in your calculation.

Keep the direction calculation separate from the magnitude calculation:

• Use the right-hand grip rule to find the solenoid’s magnetic-field direction.
• Use conventional current consistently.
• Use Fleming’s left-hand rule or the motor-effect vector rule to find the force on a separate current-carrying conductor.

Once you can track how \(N\), \(I\), and \(L\) affect the field independently, you can handle much more than a question about doubling length. You can predict how a solenoid behaves when it is stretched, rewound, supplied with a different current, or combined with another winding.

The next useful step is to apply that field calculation to magnetic forces and the motor effect, where both the field’s magnitude and its direction determine what happens to a conductor.