Ampère's Law

Go once round any closed loop adding up the component of B along it, and you get μ₀ times the current that passes through the loop. For steady currents it is always true, and like Gauss's law only sometimes useful: where symmetry makes B the same size all along a loop and parallel to it, the sum is B times the loop's length and B follows in one line. Choose a current, size the loop, and read off the current enclosed and the field.

Ampère's law, with Maxwell's displacement current	$\oint\vec B\cdot d\vec l=\mu_0\Bigl(I_{\mathrm{enc}}+\varepsilon_0\frac{d\Phi_E}{dt}\Bigr)$
Long solenoid and toroid	$B=\mu_0nI,\qquad B=\frac{\mu_0NI}{2\pi r}$

Worked example — Ampère's Law

Current through the loop	1.25 A
Length of the loop	0.06283 m
μ₀ times the current	1.571×10⁻⁶ T·m
B on the loop	25 μT

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Ampère's Law

Ampère's law: going round any closed loop, the sum of the components of B along the loop equals μ₀ times the current passing through it, ∮B·dl = μ₀I_enc. With Maxwell's correction the current includes the displacement current ε₀ dΦ_E/dt.

What each symbol means

I_enc: the net current through any surface the loop bounds, counted by the right-hand rule
n: turns per metre of a solenoid
K: current per metre of width of a sheet

Where the formula comes from

Choose a loop where symmetry fixes B. Round a straight wire, B is the same size all round a circle about it and points along the circle, so the sum is B·2πr.
$\oint\vec B\cdot d\vec l=B\cdot2\pi r=\mu_0I_{\mathrm{enc}}$
Count the current through it. Inside a wire with its current spread evenly, the loop encloses the fraction r²/R². Outside, all of it.
$I_{\mathrm{enc}}=I\,\frac{r^{2}}{R^{2}}\ \Rightarrow\ B=\frac{\mu_0Ir}{2\pi R^{2}},\qquad r\ge R:\ B=\frac{\mu_0I}{2\pi r}$
A long solenoid. Take a rectangle with one side inside, one outside. Outside B is negligible and the short sides cross B at right angles, so only the inside side counts, and it encloses nℓ turns.
$B\ell=\mu_0n\ell I\ \Rightarrow\ B=\mu_0nI$
A toroid. A circle inside the core encloses all N turns once, so B·2πr = μ₀NI, and B varies as 1/r.
$B\cdot2\pi r=\mu_0NI\ \Rightarrow\ B=\frac{\mu_0NI}{2\pi r}$
When the current is not steady. Between the plates of a charging capacitor no current flows, but the electric flux changes, and the displacement current ε₀ dΦ_E/dt takes its place.
$I_{d}=\varepsilon_0\frac{d\Phi_E}{dt},\qquad\oint\vec B\cdot d\vec l=\mu_0\left(I_{\mathrm{enc}}+I_{d}\right)$

Assumes: Steady currents for steps 1 to 4, and long, symmetric arrangements: a real solenoid's field is μ₀nI only well inside it, away from its ends, and outside it is small rather than zero. Air or vacuum; iron inside a coil multiplies B.

Problems solved in full

Inside and outside a thick wire 3 steps
A wire of radius 2.0 cm carries 5.0 A spread evenly over its cross-section. Find B at 1.0 cm and at 4.0 cm from its axis.

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1. Inside, the loop encloses the fraction r²/R² of the current:
  
  $I_{\mathrm{enc}}=I\,\frac{r^{2}}{R^{2}}=(5{.}0)\,\frac{(0{.}01)^{2}}{(0{.}02)^{2}}=1{.}25\ \mathrm{A}$
2. Ampère's law round the circle:
  
  $B=\frac{\mu_0I_{\mathrm{enc}}}{2\pi r}=\frac{(4\pi\times10^{-7})(1{.}25)}{2\pi\,(0{.}01)}=2{.}5\times10^{-5}\ \mathrm{T}$
3. Outside, all of I is enclosed:
  
  $B=\frac{\mu_0I}{2\pi r}=\frac{(4\pi\times10^{-7})(5{.}0)}{2\pi\,(0{.}04)}=2{.}5\times10^{-5}\ \mathrm{T}$
Answer

25 μT at both radii. Inside, B grows in proportion to r up to 50 μT at the surface; outside it falls as 1/r.

$B(1\ \mathrm{cm})=B(4\ \mathrm{cm})=25\ \mu\mathrm{T}$
A coaxial cable 2 steps
A coaxial cable carries 5.0 A along an inner wire of radius 0.50 cm and back along a thin outer conductor of radius 3.0 cm. Find B at 2.0 cm and at 4.0 cm from the axis.

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Show solution
1. Between the conductors the loop encloses the inner current:
  
  $B\cdot2\pi r=\mu_0I\ \Rightarrow\ B=\frac{(4\pi\times10^{-7})(5{.}0)}{2\pi\,(0{.}02)}=5{.}0\times10^{-5}\ \mathrm{T}$
2. Outside the cable it encloses both, which cancel:
  
  $B\cdot2\pi r=\mu_0\,(I-I)=0$
Answer

50 μT at 2.0 cm, and zero at 4.0 cm.

$B(2\ \mathrm{cm})=50\ \mu\mathrm{T},\qquad B(4\ \mathrm{cm})=0$
A long solenoid 2 steps
A long solenoid has 2,000 turns per metre and carries 3.0 A. Find B inside it.

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Show solution
1. A rectangle with an inside side of length ℓ encloses nℓ turns, each carrying I; only the inside side contributes:
  
  $B\ell=\mu_0\,(n\ell)\,I\ \Rightarrow\ B=\mu_0nI$
2. In numbers:
  
  $B=(4\pi\times10^{-7})(2000)(3{.}0)\approx7{.}540\times10^{-3}\ \mathrm{T}$
Answer

7.54 mT, whatever the solenoid's radius and wherever inside it, away from the ends.

$B=\mu_0nI\approx7{.}54\ \mathrm{mT}$
The end of a finite solenoid 3 steps
The solenoid of problem 3 is 50 cm long with a radius of 2.0 cm. Find B on its axis at the middle and at an open end.

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Show solution
1. Adding the field of its loops (the Biot–Savart rung) gives, on the axis,
  
  $B=\frac{\mu_0nI}{2}\left(\cos\alpha_1-\cos\alpha_2\right)$
2. At the middle, cos α₁ = −cos α₂ = (L/2)/√(L²/4 + R²):
  
  $B(0)=\mu_0nI\,\frac{L/2}{\sqrt{L^{2}/4+R^{2}}}=(7{.}540\times10^{-3})\,\frac{0{.}25}{\sqrt{0{.}0625+0{.}0004}}\approx7{.}516\times10^{-3}\ \mathrm{T}$
3. At the end, cos α₁ = L/√(L² + R²) and cos α₂ = 0:
  
  $B(\tfrac{L}{2})=\frac{\mu_0nI}{2}\,\frac{L}{\sqrt{L^{2}+R^{2}}}=(3{.}770\times10^{-3})\,\frac{0{.}50}{\sqrt{0{.}25+0{.}0004}}\approx3{.}767\times10^{-3}\ \mathrm{T}$
Answer

7.516 mT at the middle, 3.767 mT at the end: half of μ₀nI to within 0.1%.

$B(0)\approx7{.}516\ \mathrm{mT},\qquad B(\tfrac{L}{2})\approx3{.}767\ \mathrm{mT}$
A toroid 2 steps
A toroid has 500 turns carrying 2.0 A on a core from 5.0 cm to 10 cm. Find B at the inner and outer walls of the core, and outside it.

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1. A circle of radius r inside the core encloses all N turns:
  
  $B\cdot2\pi r=\mu_0NI\ \Rightarrow\ B=\frac{\mu_0NI}{2\pi r}$
2. At the two walls:
  
  $B(0{.}05)=\frac{(2\times10^{-7})(500)(2{.}0)}{0{.}05}=4{.}0\times10^{-3}\ \mathrm{T},\qquad B(0{.}10)=2{.}0\times10^{-3}\ \mathrm{T}$
Answer

4.0 mT at 5.0 cm, 2.0 mT at 10 cm, and zero outside the core.

$B_{a}=4\ \mathrm{mT},\qquad B_{b}=2\ \mathrm{mT}$
A current sheet 1 steps
A large flat sheet carries 100 A per metre of its width. Find B near it.

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1. A rectangle crossing the sheet, with sides of length ℓ parallel to it, encloses Kℓ; B runs along both sides, opposite ways, and the crossing sides contribute nothing:
  
  $2B\ell=\mu_0K\ell\ \Rightarrow\ B=\frac{\mu_0K}{2}=\frac{(4\pi\times10^{-7})(100)}{2}\approx6{.}283\times10^{-5}\ \mathrm{T}$
Answer

62.83 μT on each side, parallel to the sheet, and the same at any distance for as long as the sheet looks infinite from there.

$B=\frac{\mu_0K}{2}\approx62{.}83\ \mu\mathrm{T}$
Between the plates of a charging capacitor 2 steps
Circular plates of radius 10 cm are charged by a current of 2.0 A. Find B between them 5.0 cm from the axis.

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Show solution
1. Between the plates E = σ/ε₀ grows as the charge does; through a disk of radius r the flux is (Q/ε₀)(r²/R²), so the displacement current is
  
  $I_{d}=\varepsilon_0\frac{d\Phi_E}{dt}=\varepsilon_0\,\frac{d}{dt}\Bigl(\frac{Q}{\varepsilon_0}\,\frac{r^{2}}{R^{2}}\Bigr)=I\,\frac{r^{2}}{R^{2}}=0{.}5\ \mathrm{A}$
2. Ampère's law with Maxwell's term:
  
  $B=\frac{\mu_0I_{d}}{2\pi r}=\frac{(4\pi\times10^{-7})(0{.}5)}{2\pi\,(0{.}05)}=2{.}0\times10^{-6}\ \mathrm{T}$
Answer

2.0 μT. Outside the plates, at r ≥ R, the field is μ₀I/(2πr), the same as round the wire.

$B=\frac{\mu_0Ir}{2\pi R^{2}}=2\ \mu\mathrm{T}$
A wire whose current is not spread evenly 3 steps
A wire of radius 2.0 cm carries 5.0 A with current density J = J₀r/R. Find J₀ and B at 1.0 cm from the axis.
Show solution
1. The current inside radius r, adding rings of area 2πr′ dr′:
  
  $I(r)=\int_{0}^{r}J_0\frac{r'}{R}\,2\pi r'\,dr'=\frac{2\pi J_0r^{3}}{3R}$
2. Setting r = R gives J₀, and Ampère's law gives B:
  
  $J_0=\frac{3I}{2\pi R^{2}}=\frac{3\,(5{.}0)}{2\pi\,(0{.}02)^{2}}\approx5968\ \mathrm{A/m^{2}},\qquad B=\frac{\mu_0I(r)}{2\pi r}=\frac{\mu_0I\,r^{2}}{2\pi R^{3}}$
3. At r = 1.0 cm:
  
  $B=\frac{(4\pi\times10^{-7})(5{.}0)(0{.}01)^{2}}{2\pi\,(0{.}02)^{3}}=1{.}25\times10^{-5}\ \mathrm{T}$
Answer

J₀ = 5,968 A/m²; B = 12.5 μT, half the 25 μT of an even spread, because more of the current is near the surface.

$J_0\approx5968\ \mathrm{A/m^{2}},\qquad B=12{.}5\ \mu\mathrm{T}$

Supplementary problems, answers only

A long straight wire of radius 1.0 mm carries 10 A spread evenly. Find B at its surface.

Show answer

2.0 mT.

$B=\frac{\mu_0I}{2\pi R}=2{.}0\ \mathrm{mT}$
A solenoid 30 cm long has 600 turns and carries 2.0 A. Find B inside it, away from the ends.

Show answer

5.027 mT.

$B=\mu_0\,\frac{N}{L}\,I\approx5{.}027\ \mathrm{mT}$
A toroid of 1,000 turns carries 1.5 A. Find B on the centre line of its core, at radius 8.0 cm.

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3.75 mT.

$B=\frac{\mu_0NI}{2\pi r}=3{.}75\ \mathrm{mT}$
Two parallel current sheets each carry 50 A per metre of width, in opposite directions. Find B between them and outside.

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62.83 μT between them, zero outside.

$B=2\cdot\frac{\mu_0K}{2}=\mu_0K\approx62{.}83\ \mu\mathrm{T}$
A coaxial cable carries 4.0 A in a hollow inner conductor of radius 1.0 mm and back in an outer conductor of radius 5.0 mm. Find B at 0.5 mm, 3.0 mm and 6.0 mm from the axis.

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0, 266.7 μT and 0.

$0,\quad\frac{\mu_0I}{2\pi r}\approx266{.}7\ \mu\mathrm{T},\quad0$
A capacitor with circular plates of radius 5.0 cm is charged so that dE/dt = 1.0 × 10¹² V/(m·s) between them. Find the displacement current and B at the edge of the plates.

Show answer

0.06954 A and 0.2782 μT.

$I_{d}=\varepsilon_0\pi R^{2}\frac{dE}{dt}\approx0{.}06954\ \mathrm{A},\qquad B=\frac{\mu_0I_{d}}{2\pi R}\approx0{.}2782\ \mu\mathrm{T}$

At the mouth of a long solenoid the field is half its inside value

Press The mouth of a solenoid. 50 cm long, 2,000 turns per metre, 3 A: in the middle the field is 7.516 mT, near enough μ₀nI = 7.54 mT, and at the open end it is 3.767 mT. The half comes from symmetry. Put a second, identical solenoid against the open end: now the joint is deep inside one coil twice as long, where the field is very nearly μ₀nI, and the two halves contribute equally to it. Take one away and half is left; for this coil 49.96% of μ₀nI, and exactly half in the limit of an endless one. Nor does it fade gradually all the way from the middle: drag P along the axis and the field stays within 1% of μ₀nI until P is about five radii, 10 cm, from the end, and loses the rest of its half over that last stretch.

A toroid's field is not the same across its core

Press A toroid. A circle of radius r inside the core encloses all N turns, so B = μ₀NI/(2πr), and r changes across the core: 4 mT at the inner wall at 5 cm, 2 mT at the outer at 10 cm. Bending a solenoid into a ring crowds its turns on the inside. Outside the core the loop either encloses no turns at all or encloses every turn twice, once going each way, and the field is zero: a toroid keeps all its field inside, which is why so many transformers and chokes are wound on rings.

A charging capacitor breaks Ampère's law as he wrote it

Press A charging capacitor. No charge crosses the gap, yet a loop of radius 5 cm between the plates has a field of 2 μT along it. Stretch a flat disk over that loop and no current passes through it; stretch a bag over the same loop so that it passes round one plate and the wire's 2 A passes through it. Ampère's law cannot give two answers for one loop. Maxwell's correction adds the rate of change of electric flux, ε₀ dΦ_E/dt, as a displacement current: through the flat disk it is (r/R)² of 2 A, 0.5 A, and it gives the 2 μT. That term lets a changing electric field make a magnetic one, and with it Maxwell's equations predicted waves travelling at the speed of light; Hertz began confirming them in 1887.

Learning path

Where fields come from

References (4)

Ampère's law over an arbitrary closed path, and the thick wire inside (μ₀I₀r/(2πa²)) and outside (Example 12.7): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 12.5, “Ampère's Law”.
The long solenoid μ₀nI (zero outside), the toroid μ₀NI/(2πr) (zero outside), and the finite solenoid on its axis in the form (μ₀NI/2L)(sin θ₂ − sin θ₁): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 12.6, “Solenoids and Toroids”.
Insight block 3: the charging capacitor with two surfaces on one loop, the displacement current I_d = ε₀ dΦ_E/dt, and Maxwell's prediction of electromagnetic waves travelling at the speed of light, confirmed by Hertz from 1887: S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 16.1, “Maxwell's Equations and Electromagnetic Waves”.
Insight block 3: the paper in which Maxwell's equations, with the displacement current, give electromagnetic waves: J. C. Maxwell, “A Dynamical Theory of the Electromagnetic Field”, Philosophical Transactions of the Royal Society of London 155 (1865), pp. 459-512. DOI 10.1098/rstl.1865.0008

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.