A long wire carries 5 A. What is B at 2 m from the wire?

B = μ₀I/(2πr) = (4π×10⁻⁷)(5)/(2π×2) = 5×10⁻⁷ T = 0.5 μT.

For a solenoid with n=100 turns/m and I=5 A, what is B inside?

B = μ₀nI = (4π×10⁻⁷)(100)(5) = 6.28×10⁻⁴ T ≈ 0.628 mT.

A coaxial cable has +10 A inner and -10 A outer. What is B outside the cable?

B = 0 — the net enclosed current is zero, so ∮B·dl = μ₀(0) = 0.

Pro 🔒~30 min

Ampère's Law

Magnetic field circulation around current-carrying conductors

How it works

Ampère's Law is one of Maxwell's equations relating the circulation of the magnetic field around a closed loop to the current enclosed by that loop. For a long straight wire carrying current I, the magnetic field at distance r forms concentric circles with magnitude B = μ₀I/(2πr), where μ₀ = 4π×10⁻⁷ T·m/A. Using the right-hand rule: thumb along current direction, fingers curl in the direction of B. For a coaxial cable (inner current +I, outer current -I), the field outside is zero because the net enclosed current is zero. Inside a solenoid with n turns per unit length, B = μ₀nI (uniform and parallel to the axis), while outside it is approximately zero. Ampère's Law is most useful when the magnetic field has high symmetry (cylindrical for wires, rectangular for solenoids), allowing B to be pulled out of the integral.

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Step-by-step

Select a current configuration and set the current magnitude.
Adjust the Amperian loop radius to see how the enclosed current changes.
Observe the 3D magnetic field vectors (color-coded by magnitude, following the right-hand rule).
The circulation panel shows ∮B·dl and verifies Ampère's Law.
Rotate the 3D view with mouse drag.

Key formulas

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ Ampère's Law: the line integral of B around a closed loop equals μ₀ times the enclosed current
$B = \frac{\mu_0 I}{2\pi r}$ Magnetic field around an infinite straight wire at distance r
$B = \mu_0 n I$ Magnetic field inside an ideal solenoid: n = turns per unit length, I = current

Frequently asked questions

A long wire carries 5 A. What is B at 2 m from the wire?: B = μ₀I/(2πr) = (4π×10⁻⁷)(5)/(2π×2) = 5×10⁻⁷ T = 0.5 μT.
For a solenoid with n=100 turns/m and I=5 A, what is B inside?: B = μ₀nI = (4π×10⁻⁷)(100)(5) = 6.28×10⁻⁴ T ≈ 0.628 mT.
A coaxial cable has +10 A inner and -10 A outer. What is B outside the cable?: B = 0 — the net enclosed current is zero, so ∮B·dl = μ₀(0) = 0.