You already know a current wraps a wire in circular field lines. Now you can put a number on that field. One compact equation links the field strength to just two things you control, the current and how far away you are, and reveals that the field fades in proportion to distance, not to distance squared.
A wire carries a steady current and produces a magnetic field around it. You then double the current.
Before reading: what happens to the magnetic field strength at a fixed distance from the wire when the current doubles? Write your prediction and a one-sentence reason.
Field strength from current and distance
The strength of a wire's magnetic field grows with the current and weakens as you move away, and the equation captures both effects precisely.
The magnetic field strength a perpendicular distance $r$ from a long straight wire carrying current $I$ is:
$$B = \frac{\mu_0 I}{2\pi r}$$
Here $B$ is in tesla (T), $I$ in amperes (A), $r$ in metres (m), and $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$ is the permeability of free space. The equation says $B \propto I$ (double the current, double the field) and $B \propto 1/r$ (double the distance, halve the field). That gentle $1/r$ fall-off, rather than $1/r^2$, is the answer to the hook, and it comes from the field wrapping all the way around the wire.
The field a distance $r$ from a long straight wire is $B = \dfrac{\mu_0 I}{2\pi r}$, with $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$. So $B$ is proportional to the current and inversely proportional to the perpendicular distance ($B \propto 1/r$, not $1/r^2$).
Pause, copy the highlighted equation into your book before moving on.
Did you get this? True or false: doubling the perpendicular distance from a straight wire reduces its magnetic field to one quarter.
Substituting carefully into B = μ₀I/2πr
We just saw the equation and what it means. That raises a question: how do we get a number out of it? This card answers it → list $I$, $r$ and $\mu_0$, substitute, and watch the small powers of ten.
The only tricky part is keeping track of the constant $\mu_0$ and the small magnitudes, the substitution itself is straightforward.
The fields around ordinary wires are small (often microtesla), so expect answers with negative powers of ten. Keep $r$ in metres and carry $\mu_0 = 4\pi \times 10^{-7}$ through the calculation.
A long straight wire carries a current of $5.0\ \text{A}$. Find the magnetic field strength at a perpendicular distance of $0.10\ \text{m}$ from the wire.
To find the field of a wire, substitute into $B = \mu_0 I / 2\pi r$ with $r$ in metres. A useful shortcut is $\mu_0 / 2\pi = 2 \times 10^{-7}$, so $B = (2\times10^{-7}) I / r$. Expect small fields (microtesla range).
Add the highlighted method to your notes before the check below.
Quick check: A wire carries $10\ \text{A}$. The field at $0.20\ \text{m}$ is found using $B = (2\times10^{-7})I/r$. What is $B$?
Pattern A, Calculate then reason
A long straight wire carries $8.0\ \text{A}$. In your book: (1) calculate the field at $0.040\ \text{m}$ and at $0.080\ \text{m}$ using $B = \mu_0 I / 2\pi r$; (2) state the ratio of the two fields; (3) write one sentence explaining why the field at the larger distance is exactly half, linking it to $B \propto 1/r$.
Fill the blanks: Complete the relationship.
For a long straight wire, $B = \mu_0 I / 2\pi r$. The field is directly proportional to the and inversely proportional to the from the wire.
Dot point: Calculate the strength of the magnetic field produced by a current-carrying wire using B = μ₀I / 2πr
Predicting how B changes without recalculating
We just learned to compute $B$. That raises a question: can we predict changes without redoing the whole sum? This card answers it → yes, use the proportionalities $B \propto I$ and $B \propto 1/r$ directly.
Once you know how $B$ depends on $I$ and $r$, you can reason about changes by scaling, which is faster and reveals the physics.
Because $B \propto I$, multiplying the current by any factor multiplies the field by the same factor: triple the current, triple the field. Because $B \propto 1/r$, multiplying the distance by a factor divides the field by that factor: move three times farther, the field falls to a third. To rearrange for an unknown, treat the equation algebraically, for the current, $I = \dfrac{2\pi r B}{\mu_0}$; for the distance, $r = \dfrac{\mu_0 I}{2\pi B}$. These let you, for example, work out how much current a wire must carry to produce a required field at a given distance.
Use the proportionalities to predict changes: $B \propto I$ (scale the current, scale the field) and $B \propto 1/r$ (move farther, field falls in proportion). Rearrange for the current as $I = 2\pi r B / \mu_0$ or for distance as $r = \mu_0 I / 2\pi B$.
Pause, copy the highlighted relationships into your book before the check below.
Last check: Decide if the statement is true or false.
If the current in a straight wire is tripled while the distance is unchanged, the magnetic field at that point also triples.
Two truths, one lie: three of these statements about $B = \mu_0 I / 2\pi r$ are correct. Which one is the lie?
Dot point: Calculate the strength of the magnetic field produced by a current-carrying wire using B = μ₀I / 2πr
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 long straight wire carries a current of $4.0\ \text{A}$. Calculate the magnetic field strength at a perpendicular distance of $0.050\ \text{m}$, then state and justify what happens to the field if the distance is doubled to $0.10\ \text{m}$.
1 mark: correct equation $B = \mu_0 I / 2\pi r$ with values substituted · 1 mark: $B = 1.6 \times 10^{-5}\ \text{T}$ · 1 mark: doubling $r$ halves the field to $8.0 \times 10^{-6}\ \text{T}$ · 1 mark: justified by $B \propto 1/r$
ApplyBand 5(3 marks) 2. A long straight wire carries a current of $12\ \text{A}$. Calculate the perpendicular distance from the wire at which the magnetic field strength is $3.0 \times 10^{-5}\ \text{T}$. Take $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$.
1 mark: rearranges $B = \mu_0 I/2\pi r$ to $r = \mu_0 I/2\pi B$ · 1 mark: uses $\mu_0/2\pi = 2 \times 10^{-7}$ · 1 mark: $r = 0.080\ \text{m}$
Dot point: Calculate the strength of the magnetic field produced by a current-carrying wire using B = μ₀I / 2πr
AnalyseBand 5(3 marks) 3. Contrast the way the magnetic field of a straight current-carrying wire depends on distance with the way the electric field of a point charge depends on distance, and state the consequence for how quickly each field falls away.
1 mark: $B \propto 1/r$ for the wire · 1 mark: $E \propto 1/r^2$ for a point charge · 1 mark: the point-charge field falls off much more rapidly with distance
Dot point: Calculate the strength of the magnetic field produced by a current-carrying wire using B = μ₀I / 2πr
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): Using $B = \dfrac{\mu_0 I}{2\pi r}$ with the shortcut $\mu_0 / 2\pi = 2 \times 10^{-7}$: $B = \dfrac{(2\times10^{-7})(4.0)}{0.050} = 1.6 \times 10^{-5}\ \text{T}$. If the distance is doubled to $0.10\ \text{m}$, the field becomes $B = \dfrac{(2\times10^{-7})(4.0)}{0.10} = 8.0 \times 10^{-6}\ \text{T}$, which is exactly half the original value. This is because the magnetic field of a long straight wire is inversely proportional to the perpendicular distance ($B \propto 1/r$), so doubling $r$ halves $B$.
Q2 (3 marks): Rearranging the equation for the distance gives $r = \mu_0 I/(2\pi B)$. The grouping $\mu_0/2\pi$ simplifies to $2 \times 10^{-7}\ \text{T m A}^{-1}$, because the $\pi$ cancels and $4/2 = 2$. Substituting, $r = (2 \times 10^{-7})(12)/(3.0 \times 10^{-5}) = (2.4 \times 10^{-6})/(3.0 \times 10^{-5}) = 0.080\ \text{m}$, that is $8.0\ \text{cm}$ from the wire. Note that because $B$ depends on the first power of $r$, no square root is needed here, unlike the corresponding calculation for the electric field of a point charge.
Q3 (3 marks): For a long straight wire, $B = \mu_0 I/(2\pi r)$, so the magnetic field is inversely proportional to the first power of the distance. Doubling the distance from the wire halves the field. For a point charge, $E = kq/r^2$, so the electric field is inversely proportional to the square of the distance, and doubling the distance reduces the field to one quarter. The consequence is that the point-charge field falls away much more rapidly than the field of the straight wire. Tripling the distance divides the wire's field by three but divides the point charge's field by nine. The geometrical reason is that the wire's field lines spread over the curved surface of a cylinder, whose area grows in proportion to $r$, whereas the point charge's lines spread over a sphere, whose area grows in proportion to $r^2$.
You can now calculate the field around a straight wire with $B = \mu_0 I / 2\pi r$ and predict how it scales. Next lesson bends wires into a solenoid and looks at its structure and field.
↩ Back to focus-area hubAt the start you predicted the effect of doubling the current. The answer: the field doubles, because $B \propto I$. And on the hook, the wire's field falls as $1/r$ (not $1/r^2$) because the field wraps around the whole wire, so it fades more gently than a point charge's field. If either 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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