Inductance and Energy Storage
1. Inductance in Solenoids
Learning outcomes
- I can define inductance.
- I can explain how solenoids behave as inductors.
- I can identify factors affecting inductance.
- I can describe how coil geometry influences inductance.
- I can solve simple inductance problems.
Inductance in Solenoids
Inductance is the ability of a conductor or circuit to produce an induced EMF when the current through it changes.
A coil of wire has much greater inductance than a straight wire because the magnetic fields produced by its individual turns reinforce one another. A long coil containing many closely spaced turns is called a solenoid.
Solenoids are therefore natural inductors.
When current through a solenoid changes:
changing current → changing magnetic field → changing magnetic flux → self-induced EMF
The self-induced EMF acts to oppose the change in current according to Lenz's law.
What Is Inductance?
Inductance measures how strongly a circuit responds to a changing current by producing a self-induced EMF.
The symbol for inductance is:
L
The SI unit is the:
henry (H)
The relationship between induced EMF and changing current is:
ε = −L ΔI/Δt
where:
- ε = self-induced EMF in volts (V)
- L = inductance in henries (H)
- ΔI = change in current in amperes (A)
- Δt = time interval in seconds (s)
For magnitude calculations:
|ε| = L|ΔI/Δt|
Meaning of One Henry
A circuit has an inductance of:
1 H
if a current changing at:
1 A/s
produces an induced EMF with magnitude:
1 V
Therefore:
1 H = 1 V·s/A
Large inductance means a given rate of current change produces a relatively large self-induced EMF.
Why a Solenoid Has Inductance
A solenoid consists of many turns of wire.
When current flows through each turn, it produces a magnetic field.
Inside the solenoid, the fields from the individual turns combine.
This creates a relatively strong magnetic field along the solenoid's axis.
When the current changes, this magnetic field changes.
The changing field changes the magnetic flux through the turns of the solenoid.
Faraday's law then produces a self-induced EMF.
Magnetic Field Inside a Solenoid
For an ideal long solenoid:
B = μNI/l
where:
- B = magnetic field strength
- μ = magnetic permeability of the material inside the solenoid
- N = number of turns
- I = current
- l = length of the solenoid
The quantity:
N/l
represents the number of turns per unit length.
The equation shows that the magnetic field becomes stronger when:
- current increases
- number of turns increases
- magnetic permeability increases
For the ideal model, increasing the solenoid length while keeping N and I fixed spreads the turns farther apart and decreases the field.
Magnetic Flux Through a Turn
Magnetic flux through one turn can be written:
Φ = BA
when the magnetic field is perpendicular to the area of the loop.
For a solenoid:
B = μNI/l
Therefore:
Φ = (μNI/l)A
or:
Φ = μNIA/l
The flux through each turn depends on:
- current
- number of turns
- cross-sectional area
- solenoid length
- magnetic permeability
Flux Linkage
A solenoid contains many turns.
If the magnetic flux through one turn is Φ, the total flux linkage is:
NΦ
Substituting the expression for flux:
NΦ = N(μNIA/l)
Therefore:
NΦ = μN²AI/l
Inductance can be defined using flux linkage:
L = NΦ/I
Substituting:
L = μN²A/l
This is the important inductance equation for an ideal long solenoid.
Inductance Equation for a Solenoid
For an ideal long solenoid:
L = μN²A/l
where:
- L = inductance in henries (H)
- μ = magnetic permeability in H/m
- N = number of turns
- A = cross-sectional area in m²
- l = length of the solenoid in m
This equation reveals how the geometry and material of a solenoid affect its inductance.
Factors Affecting Inductance
The inductance of an ideal solenoid depends mainly on:
- number of turns, N
- cross-sectional area, A
- solenoid length, l
- magnetic permeability, μ
The relationship is:
L = μN²A/l
Therefore:
L ∝ N²
L ∝ A
L ∝ μ
L ∝ 1/l
These relationships are extremely useful for comparing solenoids without doing a complete calculation.
Effect of Number of Turns
Inductance is proportional to the square of the number of turns:
L ∝ N²
This means that changing the number of turns has a particularly large effect.
If N doubles:
N → 2N
then:
L → (2²)L
Therefore:
L becomes 4 times larger
If N triples:
L becomes 9 times larger
This is one of the most important relationships for solenoids.
Why the Number of Turns Is Squared
The N² relationship occurs for two reasons.
First, increasing N increases the magnetic field:
B ∝ N
Second, increasing N also increases the number of turns linked by that magnetic flux.
Therefore:
one factor of N strengthens the field
and:
another factor of N increases flux linkage
Together:
L ∝ N²
Example: Changing the Number of Turns
Solenoid A has inductance:
L₁ = 0.20 H
Its number of turns is doubled while all other properties remain unchanged.
Since:
L ∝ N²
then:
L₂/L₁ = (N₂/N₁)²
L₂/L₁ = 2²
L₂ = 4(0.20)
L₂ = 0.80 H
Effect of Cross-Sectional Area
Inductance is directly proportional to cross-sectional area:
L ∝ A
A larger cross-sectional area allows more magnetic flux to pass through each turn.
Therefore:
larger area → greater flux linkage → greater inductance
If the area doubles while everything else remains unchanged:
inductance doubles
Diameter and Area
For a circular solenoid:
A = πr²
Therefore, changing the radius has a squared effect on area.
Suppose the radius doubles:
r → 2r
Then:
A → π(2r)²
A → 4πr²
Therefore:
area becomes 4 times larger
and, if all other factors remain constant:
inductance becomes 4 times larger
Example: Changing Radius
A solenoid has inductance:
L₁ = 0.15 H
Its radius is doubled while all other factors remain unchanged.
Because:
A ∝ r²
the area becomes four times larger.
Since:
L ∝ A
the new inductance is:
L₂ = 4(0.15)
L₂ = 0.60 H
Effect of Solenoid Length
For an ideal solenoid with a fixed number of turns:
L ∝ 1/l
Therefore, increasing the length reduces inductance.
If the solenoid length doubles while N, A, and μ remain constant:
inductance becomes half as large
This occurs because the turns become more spread out, weakening the magnetic field.
Important Condition About Length
The statement:
L ∝ 1/l
assumes that the number of turns remains constant.
If a longer solenoid is made by adding additional turns while keeping the same turns per metre, both N and l change.
The effect must then be analyzed using the full equation:
L = μN²A/l
Always check which quantities are being held constant.
Example: Changing Length
A solenoid has:
L₁ = 0.40 H
Its length is doubled while the number of turns, area, and core material remain unchanged.
Since:
L ∝ 1/l
then:
L₂ = L₁/2
L₂ = 0.20 H
Effect of Core Material
The magnetic permeability μ describes how a material responds to a magnetic field.
For an air-core solenoid:
μ ≈ μ₀
where:
μ₀ = 4π × 10⁻⁷ H/m
Placing a suitable magnetic material inside the solenoid can greatly increase the magnetic field and flux.
Therefore:
higher permeability → greater inductance
Relative Permeability
For many materials:
μ = μ₀μᵣ
where:
- μ₀ = permeability of free space
- μᵣ = relative permeability of the material
For air:
μᵣ ≈ 1
Some magnetic materials have much larger relative permeabilities.
Using such a core can therefore greatly increase inductance.
In real materials, however, permeability may depend on field strength, frequency, temperature, and other factors.
Air-Core Solenoids
An air-core solenoid contains no strongly magnetic core material.
Advantages can include:
- predictable inductance
- no magnetic-core saturation
- useful behaviour at high frequencies
However, its inductance may be smaller than that of a similar coil containing a high-permeability magnetic core.
Magnetic-Core Solenoids
Adding a suitable magnetic core can concentrate the magnetic field.
This increases:
- magnetic flux
- flux linkage
- inductance
Magnetic cores are therefore widely used in inductors and transformers.
Different applications use different core materials depending on the required frequency, inductance, efficiency, and power.
Coil Geometry
The geometry of a coil determines how effectively its magnetic field links its turns.
For a long ideal solenoid, the most important geometrical quantities are:
- number of turns
- length
- cross-sectional area
A coil with:
- many turns
- a large cross-sectional area
- relatively short length
will generally have greater inductance than one with:
- few turns
- small area
- greater length
assuming the same core material.
Comparing Two Solenoids
Suppose two air-core solenoids have the same length and cross-sectional area.
Solenoid A:
N = 200 turns
Solenoid B:
N = 400 turns
Since:
L ∝ N²
then:
L_B/L_A = (400/200)²
L_B/L_A = 2²
L_B/L_A = 4
Solenoid B has four times the inductance.
Worked Example: Air-Core Solenoid
An air-core solenoid has:
N = 1000 turns
A = 2.0 × 10⁻⁴ m²
l = 0.50 m
Use:
L = μ₀N²A/l
with:
μ₀ = 4π × 10⁻⁷ H/m
Substitute:
L = (4π × 10⁻⁷)(1000²)(2.0 × 10⁻⁴)/(0.50)
First:
1000² = 1.0 × 10⁶
Therefore:
L ≈ 5.0 × 10⁻⁴ H
So:
L ≈ 0.00050 H
or:
L ≈ 0.50 mH
Millihenries and Microhenries
Many practical inductors have values much smaller than one henry.
Useful units include:
1 mH = 10⁻³ H
and:
1 μH = 10⁻⁶ H
For example:
0.005 H = 5 mH
0.000020 H = 20 μH
Being comfortable converting between these units is important in inductance calculations.
Worked Example: Magnetic-Core Solenoid
A solenoid has:
N = 500
A = 4.0 × 10⁻⁴ m²
l = 0.20 m
The core has permeability:
μ = 2.0 × 10⁻⁴ H/m
Use:
L = μN²A/l
L = (2.0 × 10⁻⁴)(500²)(4.0 × 10⁻⁴)/(0.20)
Since:
500² = 250,000
then:
L = 0.10 H
Therefore:
L = 0.10 H
Worked Example: Finding the Number of Turns
A solenoid has:
L = 0.080 H
A = 5.0 × 10⁻⁴ m²
l = 0.25 m
μ = 1.0 × 10⁻⁴ H/m
Start with:
L = μN²A/l
Rearrange:
N² = Ll/(μA)
Substitute:
N² = (0.080)(0.25)/[(1.0 × 10⁻⁴)(5.0 × 10⁻⁴)]
N² = 400,000
Therefore:
N = √400,000
N ≈ 632 turns
A real coil would use a whole number of turns, so approximately 632 turns would be required under the ideal assumptions.
Worked Example: Finding Cross-Sectional Area
A solenoid has:
L = 0.050 H
N = 800
l = 0.40 m
μ = 1.0 × 10⁻⁴ H/m
Use:
L = μN²A/l
Rearrange:
A = Ll/(μN²)
Substitute:
A = (0.050)(0.40)/[(1.0 × 10⁻⁴)(800²)]
A = 0.020/64
A = 3.125 × 10⁻⁴ m²
Approximately:
A = 3.1 × 10⁻⁴ m²
Inductance and Induced EMF
Once the inductance is known, the induced EMF can be calculated using:
|ε| = L|ΔI/Δt|
Suppose:
L = 0.40 H
and the current changes from:
1.0 A to 5.0 A
in:
0.20 s
Then:
ΔI = 4.0 A
and:
ΔI/Δt = 20 A/s
Therefore:
|ε| = (0.40)(20)
|ε| = 8.0 V
The induced EMF acts to oppose the change in current.
Combining the Ideas
The two important equations are:
L = μN²A/l
and:
|ε| = L|ΔI/Δt|
Together they show how the physical construction of a solenoid affects its electrical behaviour.
For example:
more turns → greater L → greater self-induced EMF for the same rate of current change
Similarly:
larger area → greater L → greater self-induced EMF
Inductance and Energy Storage
A solenoid stores energy in its magnetic field.
The stored energy is:
E = ½LI²
where:
- E = energy in joules
- L = inductance in henries
- I = current in amperes
Increasing inductance allows more magnetic energy to be stored for the same current.
Worked Example: Energy Stored
A solenoid has:
L = 0.30 H
and carries:
I = 4.0 A
Use:
E = ½LI²
E = ½(0.30)(4.0²)
E = (0.15)(16)
E = 2.4 J
The energy is stored in the magnetic field of the solenoid.
What Happens When Current Increases?
When current begins increasing through a solenoid:
- its magnetic field strengthens
- magnetic flux increases
- a self-induced EMF appears
- the induced EMF opposes the increase
The larger the inductance, the greater this opposing EMF for a given rate of current change.
Therefore, a high-inductance solenoid tends to make rapid current changes more difficult.
What Happens When Current Decreases?
When current decreases:
- magnetic field weakens
- magnetic flux decreases
- self-induced EMF is produced
- the induced EMF acts to maintain the original current
The energy stored in the magnetic field can then be transferred back into the circuit.
This is why inductors can temporarily continue driving current when the original supply changes or is removed.
Inductors in Electrical Circuits
Real inductors are often constructed as coils or solenoids.
They are used because they can:
- oppose rapid changes in current
- store magnetic energy
- filter electrical signals
- smooth current variations
- transfer and control energy in power circuits
Inductors appear in:
- power supplies
- filters
- radios
- amplifiers
- motors
- transformers
- switching circuits
- electronic converters
Solenoids as Electromagnets and Inductors
The word solenoid is often used in two related ways.
In physics, a solenoid is a long coil of wire used to produce a magnetic field.
In practical devices, a solenoid may also contain a movable magnetic or iron component that converts electrical energy into mechanical motion.
Electrically, the coil still has inductance.
Therefore, when the current through a practical solenoid changes, self-induction occurs.
Practical Solenoid Devices
Solenoid coils are used in devices such as:
- electromagnetic valves
- relays
- door locks
- automotive systems
- actuators
- switching mechanisms
When these devices are switched on or off, their changing current produces self-induced EMF.
Circuit designers must account for this behaviour.
Ideal Solenoid vs Real Coil
The equation:
L = μN²A/l
is based on an ideal long solenoid.
Real coils may differ because:
- magnetic fields are not perfectly uniform
- some magnetic flux leaks outside the coil
- the wire has resistance
- core materials may behave nonlinearly
- core permeability may change
- coil geometry may not match the ideal model
Therefore, the equation provides an excellent model but may not exactly predict every real inductor.
Magnetic Saturation
In some magnetic-core inductors, increasing current eventually causes the core to approach magnetic saturation.
When this happens, the magnetic field no longer increases proportionally with current.
The effective inductance can decrease.
This means real magnetic-core inductors do not always have a perfectly constant inductance under all operating conditions.
Geometry Comparison Problem
Two solenoids use the same core material.
Solenoid B has:
- twice the number of turns
- twice the cross-sectional area
- twice the length
Compare their inductances.
Use:
L ∝ N²A/l
Therefore:
L_B/L_A = (2²)(2)/2
L_B/L_A = 4
So:
Solenoid B has four times the inductance of Solenoid A.
Multiple-Change Example
A solenoid initially has inductance:
L₁ = 0.25 H
Its design is changed so that:
- number of turns doubles
- area triples
- length doubles
- core remains unchanged
Use:
L ∝ N²A/l
Therefore:
L₂/L₁ = (2²)(3)/(2)
L₂/L₁ = 6
Therefore:
L₂ = 6(0.25)
L₂ = 1.50 H
Radius and Length Example
A solenoid is redesigned so that:
- its radius doubles
- its length triples
- number of turns remains constant
- core material remains unchanged
Because:
A ∝ r²
doubling radius makes:
A → 4A
Length becomes:
l → 3l
Therefore:
L₂/L₁ = 4/3
So:
L₂ = 4L₁/3
The inductance increases by a factor of approximately:
1.33
A Complete Solenoid Problem
An air-core solenoid has:
N = 1500 turns
radius:
r = 1.5 cm
length:
l = 0.40 m
First convert radius:
r = 0.015 m
Calculate area:
A = πr²
A = π(0.015)²
A ≈ 7.07 × 10⁻⁴ m²
Now use:
L = μ₀N²A/l
L = (4π × 10⁻⁷)(1500²)(7.07 × 10⁻⁴)/(0.40)
Therefore:
L ≈ 5.0 × 10⁻³ H
or:
L ≈ 5.0 mH
This example shows why converting centimetres to metres before calculating area is essential.
A Reliable Strategy for Solenoid Problems
First identify:
- N = number of turns
- A = cross-sectional area
- l = solenoid length
- μ = permeability
Use:
L = μN²A/l
If radius is given, calculate:
A = πr²
before calculating inductance.
If self-induced EMF is required, use:
|ε| = L|ΔI/Δt|
If stored energy is required, use:
E = ½LI²
Always convert measurements into SI units before substituting.
Typical conversions include:
1 cm = 10⁻² m
1 cm² = 10⁻⁴ m²
1 mm = 10⁻³ m
1 mH = 10⁻³ H
1 μH = 10⁻⁶ H
Checking Your Answer
Ask:
- Does increasing N increase L strongly?
- Does increasing A increase L?
- Does increasing l reduce L if N is fixed?
- Does increasing μ increase L?
- Did I square N?
- Did I convert area correctly?
- If radius changed, did I remember that A ∝ r²?
- Are all quantities in SI units?
- Does a larger L produce a larger induced EMF for the same ΔI/Δt?
These checks can catch many common errors.
Common Mistakes
Forgetting to square the number of turns
The equation is:
L = μN²A/l
not μNA/l.
Confusing inductance with induced EMF
Inductance is a property of the coil. Induced EMF occurs when the current changes.
Thinking inductance depends directly on current in the ideal model
For an ideal linear solenoid, L is determined by geometry and permeability, not by the instantaneous current.
Using diameter instead of radius in A = πr²
Always divide diameter by two first.
Failing to square the radius
Area depends on r².
Using centimetres directly in the equation
Convert lengths to metres and areas to square metres.
Thinking a longer coil always has greater inductance
For fixed N, A, and μ:
L ∝ 1/l
Ignoring what is held constant
If both N and l change, use the full relationship rather than considering only one variable.
Assuming real cores always have constant permeability
Real magnetic materials can show saturation and other nonlinear behaviour.
Key Terms
Inductance — property of a circuit describing its ability to produce self-induced EMF when current changes.
Henry (H) — SI unit of inductance.
Solenoid — long coil containing many closely spaced turns of wire.
Inductor — component designed to provide inductance in an electrical circuit.
Self-induction — production of an induced EMF by a circuit's own changing current.
Self-induced EMF — EMF generated by changing magnetic flux caused by changing current in the same circuit.
Magnetic flux — measure of magnetic field passing through a surface.
Flux linkage — magnetic flux linked with all turns of a coil.
Magnetic permeability — measure of how a material responds to and supports a magnetic field.
Relative permeability — permeability of a material compared with the permeability of free space.
Cross-sectional area — area enclosed by each turn perpendicular to the solenoid's axis.
Air-core solenoid — solenoid without a strongly magnetic core material.
Magnetic core — material placed inside a coil to modify and often strengthen its magnetic field.
Magnetic saturation — condition in which increasing magnetizing influence produces progressively less increase in a magnetic material's response.
Magnetic energy — energy stored in the magnetic field associated with current in an inductor.
Key Takeaways
- Inductance measures how strongly a circuit produces self-induced EMF when its current changes.
- Inductance is represented by L and measured in henries.
- Solenoids behave as inductors because current produces magnetic flux through many turns.
- Changing current produces changing magnetic flux and therefore self-induced EMF.
- The induced EMF opposes the change in current according to Lenz's law.
- For an ideal long solenoid, L = μN²A/l.
- Inductance increases with magnetic permeability.
- Inductance is proportional to the square of the number of turns.
- Doubling the number of turns produces four times the inductance if other factors remain constant.
- Inductance is directly proportional to cross-sectional area.
- Because A = πr², increasing radius can strongly increase inductance.
- For fixed N, A, and μ, increasing solenoid length decreases inductance.
- Magnetic cores can greatly increase inductance by increasing magnetic flux.
- Air-core and magnetic-core inductors are useful for different applications.
- A larger inductance produces a larger self-induced EMF for the same rate of current change.
- The magnitude of self-induced EMF is |ε| = L|ΔI/Δt|.
- Inductors store energy in magnetic fields.
- Stored energy is E = ½LI².
- Coil geometry has a direct effect on electrical behaviour.
- The ideal solenoid equation is a model; real coils can experience flux leakage, resistance, core losses, and magnetic saturation.
Check Your Understanding
- Define inductance and state its SI unit.
- Explain why a solenoid behaves as an inductor.
- Write the equation for the inductance of an ideal long solenoid and identify each quantity.
- A solenoid has its number of turns doubled while all other factors remain unchanged. By what factor does its inductance change?
- A solenoid has its number of turns tripled. What happens to its inductance if all other factors remain constant?
- Explain why increasing the cross-sectional area of a solenoid increases its inductance.
- A solenoid's radius is doubled while all other properties remain unchanged. By what factor does its inductance change?
- Explain what happens to inductance if the length of a solenoid is doubled while its number of turns, area, and core material remain constant.
- An air-core solenoid has 800 turns, area 3.0 × 10⁻⁴ m², and length 0.30 m. Using μ₀ = 4π × 10⁻⁷ H/m, calculate its inductance.
- A solenoid has L = 0.20 H and experiences a current change of 6.0 A in 0.30 s. Calculate the magnitude of its self-induced EMF.
- A 0.50 H solenoid carries a current of 3.0 A. Calculate the energy stored in its magnetic field.
- Two solenoids have identical area, length, and core material. One has 300 turns and the other has 900 turns. Compare their inductances.
- A solenoid is redesigned so that its number of turns doubles, its cross-sectional area doubles, and its length becomes four times greater. Determine the factor by which its inductance changes.
- Explain why inserting a suitable magnetic core into a solenoid can increase its inductance.
- An engineer wants to design a solenoid with a greater inductance without changing its core material. Describe at least three geometrical changes that could increase the inductance and explain each using L = μN²A/l.