AcademySources of Magnetic Fields
Academy
Symmetry Applications of Ampere's Law
Level 1 - Physics topic page in Sources of Magnetic Fields.
Principle
Ampere's law becomes a field calculator when symmetry fixes the field direction and magnitude along a chosen loop.
Notation
\(r\)
radial distance from an axis
\(\mathrm{m}\)
\(R\)
radius of a current-carrying cylinder
\(\mathrm{m}\)
\(n\)
turns per unit length in a solenoid
\(\mathrm{m^{-1}}\)
\(N\)
number of turns
\(I\)
current
\(\mathrm{A}\)
\(\mu_0\)
permeability of free space
\(\mathrm{N\,A^{-2}}\)
Method
Derivation 1: Long straight wire
Choose a circle centered on the wire. The field is tangent and has constant magnitude on the circle.
Long wire
\[B=\frac{\mu_0 I}{2\pi r}\]
Derivation 2: Inside a uniform current cylinder
If current density is uniform through radius \(R\), only the current inside radius \(r\) is enclosed by a circular Amperian loop.
Enclosed current for \(r<R\)
\[I_{\mathrm{enc}}=I\frac{r^2}{R^2}\]
Inside field
\[B=\frac{\mu_0 I r}{2\pi R^2}\]
Outside field
\[B=\frac{\mu_0 I}{2\pi r}\]
Derivation 3: Solenoids and toroids
For a long solenoid, a rectangular Amperian loop gives a nearly uniform internal field and a negligible external field. For a toroid, a circular loop inside the core encloses \(NI\).
Long solenoid
\[B=\mu_0 nI\]
Toroid inside core
\[B=\frac{\mu_0NI}{2\pi r}\]
Rules
Long wire
\[B=\frac{\mu_0 I}{2\pi r}\]
Uniform current cylinder, inside
\[B=\frac{\mu_0 I r}{2\pi R^2}\quad(r<R)\]
Long solenoid
\[B=\mu_0 nI\]
Toroid
\[B=\frac{\mu_0NI}{2\pi r}\]
Examples
Question
A long solenoid has
\[n=800\,\mathrm{m^{-1}}\]
and current \[0.50\,\mathrm A\]
Find its internal field.Answer
\[B=\mu_0nI=(4\pi\times10^{-7})(800)(0.50)=5.0\times10^{-4}\,\mathrm T\]
Checks
- Pick the Amperian loop to match the symmetry.
- Include only the current enclosed by the loop.
- Ideal solenoid results assume the length is much larger than the radius.
- A toroid's field depends on \(r\) inside the core because the circular path length is \(2\\pi r\).