Last lesson you saw why a solenoid's internal field is uniform. Now you can calculate it. Remarkably, the field strength does not depend on the coil's diameter or on where inside you measure, only on the current and how tightly the turns are packed. One short equation, and you can design a coil to hit any field you need.
You have a solenoid producing a certain field inside it. You then wind the same number of turns over a coil that is twice as long (so the turns are spread out more), keeping the current the same.
Before reading: does the internal field go up, down, or stay the same? Write your prediction and a one-sentence reason.
Field strength from turns per metre and current
The field inside a solenoid is set by how tightly the turns are packed and how much current flows, nothing else, not even the coil's width.
The uniform magnetic field along the axis inside a long solenoid is:
$$B = \mu_0 n I$$
Here $B$ is in tesla (T), $I$ in amperes (A), $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$, and $n$ is the number of turns per unit length, $n = N/L$ (units $\text{m}^{-1}$). Notice what is absent: there is no diameter and no position term, that is why a fat and a thin solenoid with the same $n$ and $I$ have the same internal field, the answer to the hook. The field is proportional to both the current and the turns per metre.
The uniform field inside a long solenoid is $B = \mu_0 n I$, where $n = N/L$ is the turns per unit length. It depends only on $n$ and $I$ (and $\mu_0$), not on the coil's diameter or where you measure inside.
Pause, copy the highlighted equation into your book before moving on.
Did you get this? True or false: in $B = \mu_0 n I$, the symbol $n$ is the total number of turns on the solenoid.
Turn N and L into n, then substitute
We just saw the equation. That raises a question: what if a problem gives the total turns and the length instead of $n$? This card answers it → first compute $n = N/L$, then substitute into $B = \mu_0 n I$.
The only setup step is converting the total turns over the coil length into turns per metre; after that it is a direct substitution.
Most problems quote the total number of turns $N$ and the coil length $L$. Find $n = N/L$ first (in turns per metre), then put it into the equation with the current.
A solenoid of $500$ turns is wound over a length of $0.25\ \text{m}$ and carries a current of $3.0\ \text{A}$. Find the magnetic field strength inside it.
To find the field inside a solenoid: compute $n = N/L$ (turns per metre), then $B = \mu_0 n I$. With $\mu_0 = 4\pi\times10^{-7}$, fields of a few millitesla are typical, much larger than a single straight wire's.
Add the highlighted method to your notes before the check below.
Quick check: A solenoid has $200$ turns over $0.40\ \text{m}$. What is its number of turns per unit length, $n$?
Pattern A, Calculate then reason
A solenoid has $800$ turns over $0.40\ \text{m}$ and carries $2.0\ \text{A}$. In your book: (1) find $n$ and then $B$ inside the coil; (2) recalculate $B$ if the current is doubled to $4.0\ \text{A}$; (3) write one sentence stating what happens to $B$ if the same $800$ turns are instead spread over $0.80\ \text{m}$, and explain why using $n = N/L$.
Two truths, one lie: three of these statements about $B = \mu_0 nI$ are correct. Which one is the lie?
Dot point: Calculate the strength of the uniform magnetic field inside a solenoid using B = μ₀nI
Rearranging the equation and reading the proportionalities
We just learned to compute $B$. That raises a question: what if you need a particular field and must choose the current or the winding? This card answers it → rearrange $B = \mu_0 n I$ and use the proportionalities.
Because $B$ is proportional to both $n$ and $I$, you can dial in any field you need by adjusting either the current or how tightly you wind the turns.
To find the current needed for a target field, rearrange: $I = \dfrac{B}{\mu_0 n}$. To find the turns per metre needed: $n = \dfrac{B}{\mu_0 I}$. The proportionalities make scaling quick: $B \propto I$, so doubling the current doubles the field; $B \propto n$, so doubling the turns per metre (winding twice as tightly, or the same turns over half the length) doubles the field. This control over $B$ is precisely why solenoids are used wherever a switchable, adjustable magnetic field is needed.
Rearrange $B = \mu_0 n I$ as $I = B/\mu_0 n$ or $n = B/\mu_0 I$ to design for a target field. Both $B \propto I$ and $B \propto n$, so doubling either the current or the turns-per-metre doubles the field.
Pause, copy the highlighted relationships into your book before the check below.
Last check: Decide if the statement is true or false.
Winding the same number of turns over half the length (doubling the turns per metre) doubles the magnetic field inside a solenoid, for the same current.
Fill the blanks: Complete the calculation setup.
A solenoid has $N$ turns over a length $L$. The turns per unit length is $n = N/$ , and the internal field is $B = \mu_0 n$ .
Dot point: Calculate the strength of the uniform magnetic field inside a solenoid using B = μ₀nI
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 4(4 marks) 1. A solenoid has $600$ turns wound over a length of $0.30\ \text{m}$ and carries a current of $2.5\ \text{A}$. Calculate the turns per unit length and the magnetic field inside the solenoid, then state what happens to the field if the current is halved.
1 mark: $n = N/L = 2000\ \text{m}^{-1}$ · 1 mark: correct equation $B = \mu_0 n I$ with substitution · 1 mark: $B \approx 6.3 \times 10^{-3}\ \text{T}$ · 1 mark: halving the current halves the field (to $\approx 3.1 \times 10^{-3}\ \text{T}$) because $B \propto I$
ApplyBand 5(4 marks) 2. A solenoid is to be designed to produce a uniform internal field of $5.0 \times 10^{-3}\ \text{T}$ using a current of $2.0\ \text{A}$. Calculate the number of turns per unit length required, and hence the total number of turns needed if the coil is to be $0.25\ \text{m}$ long. Take $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$.
1 mark: rearranges $B = \mu_0 nI$ to $n = B/(\mu_0 I)$ · 1 mark: substitutes correctly · 1 mark: $n \approx 2.0 \times 10^{3}\ \text{m}^{-1}$ · 1 mark: $N = nL \approx 500$ turns
Dot point: Calculate the strength of the uniform magnetic field inside a solenoid using B = μ₀nI
AnalyseBand 5(3 marks) 3. A designer needs to double the magnetic field inside a solenoid without changing the current. Describe two distinct ways of achieving this using $B = \mu_0 nI$, and explain why simply adding more turns is not by itself sufficient.
1 mark: double the number of turns while keeping the length the same · 1 mark: halve the length while keeping the same number of turns · 1 mark: explains that $B$ depends on $n = N/L$, so adding turns that also lengthen the coil leaves $n$, and therefore $B$, unchanged
Dot point: Calculate the strength of the uniform magnetic field inside a solenoid using B = μ₀nI
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): The number of turns per unit length is $n = N/L = 600 / 0.30 = 2000\ \text{m}^{-1}$. The field inside the solenoid is $B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(2.5) \approx 6.3 \times 10^{-3}\ \text{T}$. If the current is halved to $1.25\ \text{A}$, then because $B$ is directly proportional to the current ($B \propto I$), the field also halves, to about $3.1 \times 10^{-3}\ \text{T}$. The turns per unit length and $\mu_0$ are unchanged, so only the current's effect on $B$ matters.
Q2 (4 marks): Rearranging the solenoid equation for the winding density gives $n = B/(\mu_0 I)$. Substituting, the denominator is $(4\pi \times 10^{-7})(2.0) = 2.513 \times 10^{-6}$, so $n = (5.0 \times 10^{-3})/(2.513 \times 10^{-6}) = 1.99 \times 10^{3}$, which is about $2.0 \times 10^{3}$ turns per metre. The total number of turns follows from $n = N/L$, rearranged to $N = nL = (1.99 \times 10^{3})(0.25) = 497$, so about $500$ turns are needed. Note that the radius of the coil never enters the design, because the internal field of a long solenoid does not depend on it.
Q3 (3 marks): The internal field is $B = \mu_0 nI$, and with the current fixed the only variable is $n$, the number of turns per unit length. One way to double the field is to double the number of turns while keeping the overall length of the solenoid the same, for example by winding a second layer over the first, which doubles $N$ with $L$ unchanged and so doubles $n$. A second way is to halve the length of the coil while keeping the same total number of turns, packing the windings twice as densely, which also doubles $N/L$. Simply adding more turns is not by itself sufficient, because if the extra turns are wound on the end they increase $L$ in the same proportion as $N$. The ratio $n = N/L$ is then unchanged, so the field stays exactly as it was. It is the density of the winding that matters, not the total count.
You can now calculate the uniform field inside a solenoid with $B = \mu_0 n I$ and design a coil for a target field. The final lesson uses an iron core to multiply this field and build a practical electromagnet.
↩ Back to focus-area hubAt the start you predicted the effect of spreading the same turns over twice the length. The answer: the field halves, because spreading the turns halves $n = N/L$, and $B \propto n$. The total number of turns alone does not set the field, the turns per metre does. If that surprised you, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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