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.
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:
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
Core Content
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.
$B = \mu_0 n I = \mu_0 \dfrac{N}{L} I$
$n = N/L$: turns per metre (m⁻¹); $N$: total turns; $L$: length of solenoid (m); $I$: current (A)
$B \propto n$ — 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).
$N = 150{,}000$ turns, $L = 3.0\ \text{m}$, $I = 480\ \text{A}$. Calculate $B$.
- $n = N/L = 150{,}000/3.0 = 5.0 \times 10^4\ \text{m}^{-1}$
- $B = \mu_0 n I = 4\pi \times 10^{-7} \times 5.0 \times 10^4 \times 480$
- $= 4\pi \times 10^{-7} \times 2.4 \times 10^7 = 4\pi \times 2.4 = 9.6\pi \approx 3.02\ \text{T}$ ✓
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
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.
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.
A solenoid has 800 turns wound over a length of 40 cm and carries a current of 3.0 A.
- Calculate the turn density $n$ (turns per metre).
- Calculate the magnetic field strength inside the solenoid.
- 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:
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.
- Calculate $B$ for each configuration.
- 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:
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
- $n = N/L = 800/0.40 = 2000\ \text{turns m}^{-1}$
- $B = \mu_0 n I = 4\pi \times 10^{-7} \times 2000 \times 3.0 = 4\pi \times 6.0 \times 10^{-3} \approx 7.54 \times 10^{-3}\ \text{T} \approx 7.5\ \text{mT}$
- $B$ is proportional to $I$; halving $I$ halves $B$: new $B = 3.77\ \text{mT} \approx 3.8\ \text{mT}$
Activity 2 — Model Answers
- $B_{\text{wire}} = \mu_0 I/(2\pi r) = (4\pi \times 10^{-7} \times 10)/(2\pi \times 0.01) = 2.0 \times 10^{-4}\ \text{T} = 0.20\ \text{mT}$. $B_{\text{solenoid}} = \mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 10 = 1.26 \times 10^{-2}\ \text{T} = 12.6\ \text{mT}$.
- 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.
Final five questions — solenoids and Module 4 wrap-up.
⚔ Enter the arenaThe 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.