E and MMagnetic Fields

Fields of Loops and Arcs

At the center of a loop B = μ₀I/(2R), and on the axis the field falls off with distance.

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With Biot–Savart we can find the field of a current loop, the building block of solenoids, electromagnets and, on a tiny scale, magnetic atoms. As with the charged ring, symmetry does most of the work.

Why it works

At the center: every piece $d\ell$ is the same distance $R$ away and at right angles to the line to the center, and every piece's field points the same way (along the axis). So $B = \dfrac{\mu_0 I}{4\pi R^2}\oint d\ell = \dfrac{\mu_0 I}{4\pi R^2}(2\pi R) = \dfrac{\mu_0 I}{2R}$. An arc covering angle $\phi$ gives the same fraction of that: $\dfrac{\mu_0 I\phi}{4\pi R}$.
On the axis: each piece is $\sqrt{z^2 + R^2}$ away; the sideways parts cancel around the loop, and the parts along the axis carry a factor $R/\sqrt{z^2 + R^2}$.

Key idea

Center of a loop: $B = \dfrac{\mu_0 I}{2R}$ (times $N$ for $N$ turns). Center of an arc of angle $\phi$ (radians): $B = \dfrac{\mu_0 I\phi}{4\pi R}$. On the axis: $$B = \frac{\mu_0 IR^2}{2(z^2 + R^2)^{3/2}},$$ which far away falls as $1/z^3$, the field of a magnetic dipole.
The field of a current loop: lines through the middle, curling back around the wire, like the field of a short bar magnet.

Worked examples

Worked example: A flat coil

A flat coil of 20 turns with radius $5.0\ \text{cm}$ carries $2.0\ \text{A}$. Find $B$ at its center.
  1. $B = \dfrac{\mu_0 NI}{2R} = \dfrac{(4\pi \times 10^{-7})(20)(2.0)}{2(0.050)} = 5.03 \times 10^{-4}\ \text{T}$, about $503\ \mu\text{T}$, ten times Earth's field.

Answer: $B \approx 503\ \mu\text{T}$

Worked example: A quarter circle

A quarter-circle arc of radius $4.0\ \text{cm}$ carries $5.0\ \text{A}$. Find its field at the center.
  1. $\phi = \pi/2$: $B = \dfrac{\mu_0 I\phi}{4\pi R} = \dfrac{(4\pi \times 10^{-7})(5.0)(\pi/2)}{4\pi(0.040)} = 1.96 \times 10^{-5}\ \text{T}$.
  2. That is one quarter of a full loop's $\mu_0 I/(2R)$, as it should be.

Answer: $B \approx 19.6\ \mu\text{T}$

Worked example: Moving out along the axis

How strong is a loop's field on its axis at $z = 2R$, compared with the center?
  1. $\dfrac{B(z)}{B(0)} = \dfrac{R^3}{(z^2 + R^2)^{3/2}} = \dfrac{1}{(4 + 1)^{3/2}} = 0.0894$.
  2. Less than a tenth, two radii out. Far away, $B \approx \dfrac{\mu_0 IR^2}{2z^3}$.

Answer: about $0.0894$ of the center value

Far from a loop, its field has the same shape as a bar magnet's, and both are called magnetic dipoles. Earth's field far out in space looks much like this. Close to a pair of coils the field can be shaped instead: two identical coils, one radius apart and carrying the same current, make a field that is very nearly uniform in the space between them, a common way to make a known field in a lab.

Common mistake

Using the long-straight-wire formula at the center of a loop. $\mu_0 I/(2\pi R)$ is the field a distance $R$ from a straight wire, a different shape. At the center of a loop, all the wire is the same distance away and bent around the point, so its contributions add more effectively, giving $\mu_0 I/(2R)$, larger by a factor of $\pi$. Check: which shape is the wire? A loop's formula has no $\pi$ in it.

Practice

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Builds on

Leads to

This is one of the top rungs of Electricity and Magnetism.

Sources and further reading

  • OpenStax University Physics Volume 2, 12.4 Magnetic Field of a Current Loop
  • MIT OpenCourseWare, 8.02 Physics II: Electricity and Magnetism
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