Year 11 Physics Module 4 ⏱ ~40 min 5 MC · 3 Short Answer Lesson 14 of 17

Magnetic Fields in Solenoids

Philips Ingenia 3 T MRI solenoid (2022): 150,000 turns wound over a 3 m length, carrying 480 A persistent superconducting current. n = N/L = 150,000/3 = 50,000 turns/m. B = μ₀nI = 4π × 10⁻⁷ × 50,000 × 480 ≈ 3.02 T, matching the rated field. Stores 400 MJ of magnetic energy. The solenoid formula explains why MRI machines need so many turns at high current.

Today's hook: The Philips Ingenia 3 T MRI solenoid (2022) has 150,000 turns wound over a 3 m length and carries 480 A of persistent superconducting current. Using B = μ₀nI: n = 150,000 ÷ 3 = 50,000 turns/m, B = 4π × 10⁻⁷ × 50,000 × 480 ≈ 3.02 T. A single straight wire at 10 A produces only 200 μT at 1 cm, the solenoid's 50,000 turns/m stack 50,000 field contributions in the same direction. That is why MRI uses a solenoid, not a straight wire.
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Warm up first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read, predict

A solenoid has 500 turns wound over a length of 25 cm and carries a current of 2.0 A.

Predict 1: How many turns per metre does this solenoid have?

Predict 2: Would doubling the number of turns (keeping the same length and current) double or quadruple the field inside?

Inside a long solenoid, the magnetic field is:

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Learning Intentions

Learning Intentions

Know

  • Field inside a solenoid: $B = \mu_0 n I$
  • $n = N/L$ is the number of turns per metre (m⁻¹)
  • Right-hand grip rule for solenoid: fingers curl in current direction, thumb points to north pole
  • Inside: uniform; outside: like a bar magnet

Understand

  • Why $B \propto n$ and $B \propto I$ in a solenoid
  • How fields from individual loops superpose to create a uniform interior field
  • Why the solenoid is an effective electromagnet

Can Do

  • Calculate $B$ inside a solenoid given $N$, $L$, $I$
  • Determine the north/south poles of a solenoid using the right-hand rule
  • Compare field strength for different solenoid configurations
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Key Terms

Key Terms
SolenoidA coil of wire wound in a helix. When current flows, the field contributions from each turn add constructively to produce a strong, nearly uniform field inside.
Turn density ($n$)The number of turns per unit length: $n = N/L$ (turns m⁻¹). More turns per metre → stronger field.
ElectromagnetA solenoid with a soft iron core. The iron enhances the field strength significantly and can be switched on/off with the current.
Right-hand grip rule (solenoid)Curl the fingers of the right hand in the direction of conventional current around the coil. The thumb points toward the north pole of the solenoid.
Cross-lesson links: L13 established B = μ₀I/(2πr) for a straight wire. L14 extends this: winding that wire into a solenoid with n turns/m multiplies the field n-fold. The Philips Ingenia 2022 MRI solenoid (150,000 turns, 3 m, 480 A, B = 3.02 T) is the capstone example for the entire magnetism section of M4.
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Solenoid Field Formula

Core Content

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Solenoid Field Formula
+5 XP

Take a straight piece of wire and carry 10 A through it: a compass 1 cm away deflects slightly, measuring 200 μT. Now wind that same wire into 10 tight loops, the compass deflects roughly 10 times as much. Wind it into 100 loops, 100 times stronger. Each loop adds its field in the same direction along the axis, and the contributions stack. Stretch those loops along a 1-metre tube so they are evenly spaced, you have a solenoid. The field inside is strong, uniform, and parallel to the axis. That is the principle behind every electromagnet, loudspeaker, and MRI machine.

Magnetic field inside a solenoid

$B = \mu_0 n I = \mu_0 \dfrac{N}{L} I$

$n = N/L$: turns per metre (m⁻¹); $N$: total turns; $L$: length of solenoid (m); $I$: current (A)

Key proportionalities

$B \propto n$, doubling turns per metre doubles field (more turns per length = more field contributions per metre).

$B \propto I$, doubling current doubles field.

$B$ is independent of the solenoid's diameter (for ideal long solenoid).

Worked example, Philips 3 T MRI solenoid

$N = 150{,}000$ turns, $L = 3.0\ \text{m}$, $I = 480\ \text{A}$. Calculate $B$.

  1. $n = N/L = 150{,}000/3.0 = 5.0 \times 10^4\ \text{m}^{-1}$
  2. $B = \mu_0 n I = 4\pi \times 10^{-7} \times 5.0 \times 10^4 \times 480$
  3. $= 4\pi \times 10^{-7} \times 2.4 \times 10^7 = 4\pi \times 2.4 = 9.6\pi \approx 3.02\ \text{T}$ ✓

Note: real MRI magnets use superconductors cooled to 4 K so resistance is zero; the persistent current needs no ongoing power supply. Actual systems include shielding and correction coils.

Right-hand grip rule for solenoids

North pole direction

Curl the fingers of the right hand in the direction conventional current flows around the coil windings. The thumb points toward the north pole of the solenoid.

Two solenoids compare clockwise and anticlockwise conventional current as viewed from the left end. Clockwise current makes the left end south and the internal field point right toward the north pole. Reversing the current makes the left end north and reverses every field arrow.
Trace it: start at the left-end current arrow, identify the north pole with the right-hand grip rule, then follow the internal field from south to north. Reversing current reverses both poles and every field arrow.

The magnetic field inside a solenoid is uniform and parallel to the axis: $B = \mu_0 n I = \mu_0 NI/L$ (T), where $n = N/L$ is the turn density (turns m⁻¹). $B$ is proportional to both $n$ and $I$. Right-hand grip rule: curl fingers in current direction, thumb points to north pole.

Add the highlighted formula and rule to your notes before the check below.

Doubling the number of turns per metre in a solenoid (while keeping current constant) doubles the field inside.

The field inside a solenoid is strongest at the ends and weakest in the middle.

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Ferromagnetic Materials & Magnetic Domains

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Ferromagnetic Materials & Magnetic Domains
+5 XP

Push an unmagnetised nail near a bar magnet and it snaps into place as if it had been magnetic all along, yet a moment earlier it produced no field of its own. Inside that nail, billions of microscopic regions called magnetic domains, each a cluster of atoms whose electron spins are already aligned with each other, are normally jumbled in every possible direction, so their individual fields cancel out overall. The solenoid's own field does not create these domains; it rotates and grows them into alignment, turning a disordered patchwork of tiny magnets into one large, unified magnet. This is exactly what happens inside the soft iron core of an electromagnet.

How a ferromagnetic material becomes magnetised
  1. Unmagnetised iron: domains point in random directions; their fields cancel and the material shows no net magnetism.
  2. External field applied (e.g. from a solenoid): domains aligned with the field grow at the expense of neighbouring domains, and misaligned domains rotate to point more closely along the field direction.
  3. Fully magnetised: nearly all domains point the same way, and the material now produces a strong external field of its own, on top of the solenoid's field.
  4. Field removed: some alignment persists (retained magnetism). How much depends on whether the material is magnetically soft or hard.

Soft vs. hard ferromagnetic materials

Temporary vs. permanent magnetisation

Soft ferromagnetic materials (e.g. soft iron): domains align easily under an external field, but randomise again almost as soon as the field is removed. This gives temporary magnetisation, ideal for electromagnet and transformer cores that must switch on and off with the current.

Hard ferromagnetic materials (e.g. steel, alnico): domains are harder to align in the first place, but once aligned they resist randomising and stay locked in place after the field is removed. This gives permanent magnetisation, used for compass needles and fridge magnets.

Unmagnetised iron already contains magnetic domains, regions where atomic magnetic moments point the same way, oriented randomly so their fields cancel overall. An external magnetic field aligns these domains, producing a net field. Soft ferromagnetic materials (soft iron) lose this alignment quickly once the field is removed (temporary magnetisation); hard ferromagnetic materials (steel) retain it (permanent magnetisation).

Pause, add the domain-alignment mechanism and the soft/hard comparison to your notes.

Before an external magnetic field is applied, the magnetic domains inside a piece of unmagnetised iron are:

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Activity 1, Solenoid Field Calculations

Activity 1, Solenoid Field Calculations
ApplyBand 3

A solenoid has 800 turns wound over a length of 40 cm and carries a current of 3.0 A.

  1. Calculate the turn density $n$ (turns per metre).
  2. Calculate the magnetic field strength inside the solenoid.
  3. If the current is reduced to 1.5 A, what is the new field strength?

A solenoid has $n = 2000\ \text{turns m}^{-1}$ and carries 5.0 A. The field inside is approximately:

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Activity 2, Solenoid vs Straight Wire

Activity 2, Solenoid vs Straight Wire
AnalyseBand 4

Compare two configurations both carrying 10 A:

  • Configuration A: A straight wire; calculate $B$ at $r = 1.0\ \text{cm}$.
  • Configuration B: A solenoid with $n = 1000\ \text{turns m}^{-1}$; calculate $B$ inside.
  1. Calculate $B$ for each configuration.
  2. Calculate the ratio $B_{\text{solenoid}}/B_{\text{wire}}$ and explain physically why a solenoid produces a stronger, more useful field.

Three of these are correct statements about solenoids. Pick the odd one out.

A solenoid has 400 turns, length 0.20 m, and current 5.0 A. The field inside is approximately _____ mT.

In a solenoid, using the right-hand grip rule: if conventional current flows clockwise when viewed from the left end, that end is the:

✓

Show what you have learned

Quick recall, solenoids
+5 XP
Short Answer, 10 marks
+5 XP

ApplyBand 3(3 marks) 3. A solenoid has 600 turns wound over a 30 cm length, carrying 4.0 A. Calculate (a) the turn density, (b) the magnetic field strength inside, and (c) the field strength if the current is increased to 8.0 A.

AnalyseBand 4(3 marks) 4. Two solenoids both carry 2.0 A. Solenoid P has 500 turns over 25 cm; Solenoid Q has 1000 turns over 50 cm. (a) Calculate the field inside each solenoid. (b) Explain why they produce the same field, even though solenoid Q has twice as many turns. (c) How would you modify solenoid P to double its field strength without changing current?

EvaluateBand 6(4 marks) 5. An MRI machine requires a highly uniform field of 1.5 T inside a 0.60 m long bore. Using $B = \mu_0 nI$, evaluate the design choices available to an engineer to achieve this field. In your response, include a worked calculation for one valid configuration, and discuss a practical constraint that limits each design parameter.

Show all answers

Activity 1, Model Answers

  1. $n = N/L = 800/0.40 = 2000\ \text{turns m}^{-1}$
  2. $B = \mu_0 n I = 4\pi \times 10^{-7} \times 2000 \times 3.0 = 4\pi \times 6.0 \times 10^{-3} \approx 7.54 \times 10^{-3}\ \text{T} \approx 7.5\ \text{mT}$
  3. $B$ is proportional to $I$; halving $I$ halves $B$: new $B = 3.77\ \text{mT} \approx 3.8\ \text{mT}$

Activity 2, Model Answers

  1. $B_{\text{wire}} = \mu_0 I/(2\pi r) = (4\pi \times 10^{-7} \times 10)/(2\pi \times 0.01) = 2.0 \times 10^{-4}\ \text{T} = 0.20\ \text{mT}$. $B_{\text{solenoid}} = \mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 10 = 1.26 \times 10^{-2}\ \text{T} = 12.6\ \text{mT}$.
  2. Ratio = 12.6/0.20 = 63. A solenoid is 63 times stronger than a single wire at 1 cm because each of the 1000 turns per metre adds its field contribution in the same direction along the axis. All contributions superpose constructively. Additionally, the solenoid field is uniform inside (all parallel), whereas the wire field diverges in circles, the solenoid is far more useful for applications requiring a strong, directed, uniform field (electromagnets, MRI, motors).

Short Answer, Model Answers

Q3 (3 marks): (a) $n = 600/0.30 = 2000\ \text{turns m}^{-1}$. (b) $B = 4\pi \times 10^{-7} \times 2000 \times 4.0 = 1.005 \times 10^{-2}\ \text{T} \approx 10.1\ \text{mT}$. (c) Doubling $I$ doubles $B$: new $B \approx 20.1\ \text{mT}$.

Q4 (3 marks): (a) $n_P = 500/0.25 = 2000\ \text{m}^{-1}$; $B_P = 4\pi \times 10^{-7} \times 2000 \times 2.0 = 5.03\ \text{mT}$. $n_Q = 1000/0.50 = 2000\ \text{m}^{-1}$; $B_Q = 5.03\ \text{mT}$. (b) $B = \mu_0 nI$ depends on $n = N/L$, not $N$ alone. Both solenoids have the same $n$ (2000 turns/m), so they produce the same field. Solenoid Q is longer and has more turns, but also proportionally more length. (c) To double $B_P$ without changing $I$: double the turn density $n$. This could be achieved by adding 500 more turns over the same 25 cm length (wind it twice) to give $n = 4000\ \text{m}^{-1}$.

Q5 (4 marks): To achieve $B = 1.5\ \text{T}$: $1.5 = \mu_0 n I = 4\pi \times 10^{-7} \times n \times I$. The engineer can trade off $n$ vs $I$. Example configuration: $I = 500\ \text{A}$. Then $n = 1.5/(4\pi \times 10^{-7} \times 500) = 1.5/(6.28 \times 10^{-4}) \approx 2387\ \text{turns m}^{-1}$. For $L = 0.60\ \text{m}$: $N = n \times L = 2387 \times 0.60 \approx 1432\ \text{turns}$. Constraints: (1) Increasing $I$ increases $P = I^2 R$, resistive heating causes catastrophic damage unless superconductors are used (NbTi cooled to 4 K). (2) Increasing $n$ requires more wire in the same space, limited by conductor cross-section, insulation, and winding precision. (3) Uniform field requires a long solenoid ($L \gg$ diameter), but patient bores must be wide enough for human anatomy, limiting length. In practice, MRI magnets use superconductors to eliminate resistive heating, enabling the very high $nI$ products required for 1.5–3 T clinical fields.

✓

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How did your thinking change?

The Philips Ingenia 3 T MRI solenoid (2022) has 150,000 turns over 3 m = 50,000 turns/m, carrying 480 A of persistent superconducting current. B = μ₀nI = 4π × 10⁻⁷ × 50,000 × 480 ≈ 3.02 T, matching the rated field. The superconducting wire has zero resistance, so the 480 A circulates without any power supply once established. Every MRI image in that hospital is produced by this single solenoid equation.

Now check your Think First answers: n = 500 ÷ 0.25 = 2000 turns/m. Doubling the number of turns while keeping the same length doubles n and therefore doubles B, not quadruples. B ∝ n ∝ N when length and current are constant.

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