understand that the magnetic field due to the current in a solenoid is increased by a ferrous core
Magnetic Fields Due to Currents
Objective
To understand how electric currents produce magnetic fields, how the field pattern depends on the geometry of the conductor, why a ferrous (iron‑type) core increases the field inside a solenoid, and how these ideas link to the Lorentz force, force on a moving charge, and electromagnetic induction (Cambridge AS & A‑Level Physics 9702, topics 20.4 & 20.5).
1. What is a magnetic field?
A magnetic field B is a region of space in which a moving charge or a current‑carrying conductor experiences a force. The direction of the field at any point is defined by the direction of the force on a positive test charge moving perpendicular to the field.
Key definition (syllabus 20.4 1): “A magnetic field is a vector field that exerts a force on moving electric charges.”
2. Magnetic‑field patterns produced by currents
For the required sketching tasks, the idealised cases are:
- Straight long conductor – field lines are concentric circles centred on the wire. Direction given by the right‑hand rule (thumb = current, fingers = field).
- Single circular loop – field lines emerge from the centre of the loop, pass through the interior, and close outside the loop. Use the right‑hand rule for the current direction.
- Long solenoid (air core) – inside the coil the lines are almost parallel and uniform; outside they spread out and are very weak (similar to the fringe field of a bar magnet).
Diagram suggestion: three small sketches placed side‑by‑side, labelled “Straight wire”, “Current loop”, “Solenoid (air core)”.
3. Long solenoid – field without a core
Consider a solenoid of length l with N turns carrying a current I.
- Turns per unit length: \(n = \dfrac{N}{l}\) (turns m\(^{-1}\)).
- Magnetic field strength (magnetising field): \(H = nI\) (A m\(^{-1}\)).
- Magnetic flux density in air (or vacuum): \(B = \mu{0}H = \mu{0}nI\).
where \(\mu_{0}=4\pi\times10^{-7}\;\text{H m}^{-1}\) is the permeability of free space.
4. Effect of inserting a ferrous core
When the coil is filled with a magnetic material of relative permeability \(\mu_{r}\):
- Absolute permeability: \(\displaystyle \mu = \mu{0}\mu{r}\).
- Flux density: \(\displaystyle B{\text{core}} = \mu H = \mu{0}\mu_{r}nI.\)
- The field is amplified by the factor \(\mu_{r}\) (typical values for iron‑type materials are 100 – 10 000).
- As the flux density approaches the material’s saturation flux density, \(\mu_{r}\) falls sharply; the core then behaves almost like air.
5. Force on a moving charge
For a charge q moving with velocity \(\mathbf{v}\) in a magnetic field \(\mathbf{B}\), the magnetic part of the Lorentz force is
\[
\boxed{\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}}
\]
- Magnitude: \(F = qvB\sin\theta\), where \(\theta\) is the angle between \(\mathbf{v}\) and \(\mathbf{B}\).
- Direction given by the right‑hand rule for the cross‑product (or Fleming’s left‑hand rule for conventional current).
6. Force on a current‑carrying conductor
For a straight conductor of length L carrying current I in a magnetic field B:
\[
\boxed{F = BIL\sin\theta}
\]
- Direction obtained from Fleming’s left‑hand rule (thumb = force, fore‑finger = field, middle‑finger = current).
- When the conductor is perpendicular to the field (\(\theta=90^{\circ}\)), \(F = BIL\).
Worked example – A 0.10 m long wire is placed radially inside the solenoid of Section 7 (field parallel to the axis, \(\theta = 90^{\circ}\)).
- Calculate the field inside the solenoid (see Section 7).
- Use \(F = BIL\) to obtain the force.
7. Force between two parallel conductors
Two long, straight, parallel conductors separated by a distance d each carry currents \(I{1}\) and \(I{2}\). The magnetic force per unit length on each conductor is
\[
\boxed{\frac{F}{L}= \frac{\mu{0} I{1} I_{2}}{2\pi d}}
\]
This expression defines the ampere and is a frequent exam question (topic 20.4).
8. Magnetic flux and electromagnetic induction
Magnetic flux – \(\displaystyle \Phi = BA\cos\alpha\) (for a uniform field, \(\alpha\) is the angle between field and area normal).
Faraday’s law – \(\displaystyle \mathcal{E}= -\frac{d\Phi}{dt}\) (the negative sign shows the induced emf opposes the change in flux – Lenz’s law).
Factors that increase the induced emf (syllabus 20.5 5):
- Number of turns \(N\) in the coil ( \(\mathcal{E}\propto N\) ).
- Area of the coil (larger \(A\) gives larger \(\Phi\)).
- Rate of change of the magnetic field or of the current producing it (larger \(|dB/dt|\) or \(|dI/dt|\)).
- Orientation – maximum when the field is perpendicular to the coil ( \(\cos\alpha = 1\) ).
9. Real‑world applications
Transformers and electric generators exploit the same principle: a changing current in a primary coil creates a changing magnetic flux in a ferrous core, which induces an emf in a secondary coil. The high‑permeability core concentrates the flux, making the devices efficient.
10. Example calculation (solenoid with ferrous core)
Find the magnetic flux density inside a solenoid that is 0.50 m long, has 2000 turns, carries 2 A, and contains an iron core with \(\mu_{r}=500\).
- Turns per metre: \(n = \dfrac{2000}{0.50}=4000\;\text{turns m}^{-1}\).
- Magnetic field strength: \(H = nI = 4000\times2 = 8.0\times10^{3}\;\text{A m}^{-1}\).
- Flux density with core: \(B{\text{core}} = \mu{0}\mu_{r}H = (4\pi\times10^{-7})(500)(8.0\times10^{3})\approx5.0\times10^{-3}\;\text{T}\) (5 mT).
- For comparison, without core: \(B{\text{air}} = \mu{0}H \approx 1.0\times10^{-5}\;\text{T}\) (10 µT).
The iron core amplifies the field by roughly the factor \(\mu{r}=500\). If the current were increased until the core approached saturation, \(\mu{r}\) would fall and the field would no longer increase proportionally.
11. Summary of key formulae
| Quantity | Symbol | Expression | Units |
|---|---|---|---|
| Turns per unit length | n | \(\displaystyle n=\frac{N}{l}\) | turns m\(^{-1}\) |
| Magnetic field strength | H | \(\displaystyle H = nI\) | A m\(^{-1}\) |
| Flux density (air core) | B | \(\displaystyle B = \mu{0}nI = \mu{0}H\) | T |
| Flux density (ferrous core) | \(B_{\text{core}}\) | \(\displaystyle B{\text{core}} = \mu{0}\mu_{r}nI = \mu H\) | T |
| Absolute permeability | \(\mu\) | \(\displaystyle \mu = \mu{0}\mu{r}\) | H m\(^{-1}\) |
| Force on a moving charge | \(\mathbf{F}\) | \(\displaystyle \mathbf{F}=q\,\mathbf{v}\times\mathbf{B}\) | N |
| Force on a straight conductor | F | \(\displaystyle F = BIL\sin\theta\) | N |
| Force per unit length between parallel conductors | \(F/L\) | \(\displaystyle \frac{F}{L}= \frac{\mu{0}I{1}I_{2}}{2\pi d}\) | N m\(^{-1}\) |
| Magnetic flux | \(\Phi\) | \(\displaystyle \Phi = BA\cos\alpha\) | Wb |
| Induced emf (Faraday) | \(\mathcal{E}\) | \(\displaystyle \mathcal{E}= -\frac{d\Phi}{dt}\) | V |
12. Practice questions
A solenoid has N = 1500 turns, length l = 0.30 m and carries I = 3 A. Calculate the magnetic flux density:
- With an air core.
- With a steel core of \(\mu_{r}=200\).
- Explain qualitatively why inserting a ferrous core increases the magnetic field inside a solenoid.
- What happens to the magnetic field when the core material reaches magnetic saturation? State the effect on \(\mu_{r}\) and on the calculated \(B\).
- Two parallel wires are 5 mm apart and each carries 4 A in the same direction. Find the force per metre on each wire.
- A straight wire of length 0.12 m carrying 2 A is placed perpendicular to the field inside the solenoid of the example in Section 10. Determine the magnitude of the force on the wire.
- When the current in the solenoid of Section 10 is increased at a rate of 0.5 A s\(^{-1}\), a second coil of area 0.02 m\(^2\) placed around the solenoid experiences an induced emf. Calculate the emf assuming the second coil has a single turn.
13. Suggested diagram
