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Topic 5 of 9

Magnetostatics & the Biot-Savart Law

Steady currents create static magnetic fields. The Biot-Savart law gives the field from any current distribution; for symmetric geometries, Ampere's law (the magnetic analog of Gauss's law) is far more efficient. The divergence of B is always zero — there are no magnetic monopoles.

Key Concepts

Biot-Savart Law

d\vec B=\frac{\mu_0}{4\pi}\frac{I\,d\vec l\times\hat{\mathscr r}}{\mathscr r^2}

. The magnetic field from a current element

Id\vec l

. Direction: right-hand rule (fingers along current, curl toward field point).

Ampere's Law

\oint\vec B\cdot d\vec l=\mu_0 I_{\rm enc}

(for steady currents). Like Gauss's law, use when symmetry makes

|\vec B|

constant on an Amperian loop. Choose a circular loop for solenoids and infinite wires.

Vector Potential

\vec B=\nabla\times\vec A

. Ensures

\nabla\cdot\vec B=0

automatically. In the Coulomb gauge (

\nabla\cdot\vec A=0

\nabla^2\vec A=-\mu_0\vec J

, the magnetic analog of Poisson's equation.

Magnetic Dipole

A small current loop with magnetic moment

\vec m=I\vec a

(area vector). Far-field: same form as electric dipole but with

\vec E\to\vec B

1/(4\pi\varepsilon_0)\to\mu_0/(4\pi)

\vec p\to\vec m

Key Equations

Biot-Savart

d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}

Magnetic field from a current element; μ₀ = 4π×10⁻⁷ T·m/A.

Infinite wire

B = \frac{\mu_0 I}{2\pi s}

Field at distance s from a long straight wire carrying current I.

Solenoid

B = \mu_0 nI

Interior field of a solenoid with n turns/meter, current I.

Worked Example

Field of a Long Straight Wire

Problem

A long straight wire carries $I=10$ A. Find $B$ at $r=0.05$ m from the wire.

Solution

By symmetry, use Ampere's law with a circular loop of radius $r$ :

\oint\vec B\cdot d\vec l = B\cdot2\pi r = \mu_0 I

B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi\times10^{-7}\times10}{2\pi\times0.05} = \frac{4\times10^{-6}}{0.10} = 4.0\times10^{-5}\text{ T} = 40\,\mu\text{T}

Answer

B=40

μT.

Practice

Exercises

7 problems

1 of 7

Drag the Amperian loop radius and watch $B = \mu_0 I/(2\pi r)$ trace the 1/r curve. For $I = 5.0$ A, find B at $r = 0.10$ m (in μT).

Magnetic field — long wire

Loop radius r 0.15 m → 6.67 μT

Drag the Amperian loop inward — B grows as 1/r. The curve on the right shows this falloff. At r = 0.10 m and I = 5.0 A, find B.

Magnetic field

B = μT

Formula: B = μ₀I/(2πr) | μ₀ = 4π×10⁻⁷ T·m/A, I = 5.0 A, r = 0.10 m

2 of 7

Adjust the current and see the uniform field inside the solenoid grow — while outside it stays near zero. For $n = 2000$ turns/m, find B inside at $I = 3.0$ A (in mT).

Solenoid — uniform B field

Current I 3.0 A → 7.540 mT

Inside a solenoid the field is perfectly uniform — proportional to n and I. Outside it's nearly zero. At n = 2000/m, I = 3.0 A, find B inside.

Magnetic field

B = mT

Formula: B = μ₀nI | μ₀ = 4π×10⁻⁷ T·m/A, n = 2000/m, I = 3.0 A

3 of 7

A circular loop (radius $R=0.20$ m) carries $I=4.0$ A. Find $B$ at the center (in μT). $B=\mu_0I/(2R)$ .

Unlock Exercise 3