where $\mu_0 = 4\pi \times 10^{-7}$ T·m/A is permeability of free space.

Total field: $\vec{B} = \frac{\mu_0 I}{4\pi}\int \frac{d\vec{l} \times \hat{r}}{r^2}$

Infinite Straight Wire

Current $I$ in straight wire:

$B = \frac{\mu_0 I}{2\pi r}$

Field circles wire by right-hand rule.

Circular Loop

On axis at distance $x$ from center, loop radius $R$:

$B_x = \frac{\mu_0 IR^2}{2(x^2 + R^2)^{3/2}}$

At center ($x = 0$): $B = \frac{\mu_0 I}{2R}$

Far from loop ($x \gg R$): $B \approx \frac{\mu_0 I\pi R^2}{2\pi x^3} = \frac{\mu_0 \mu}{2\pi x^3}$

where $\mu = I\pi R^2$ is magnetic dipole moment.

Ampere's Law

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

Line integral around closed path equals $\mu_0$ times enclosed current.

Works best for:

• Infinite straight wire
• Solenoid
• Toroid
• Cylindrical symmetry

Long Solenoid

$n$ turns per unit length, current $I$:

Inside: $B = \mu_0 nI$ (uniform, parallel to axis)

Outside: $B \approx 0$

Toroid

$N$ total turns, inner radius $a$, outer radius $b$:

Inside toroid: $B = \frac{\mu_0 NI}{2\pi r}$

(varies with $r$)

Outside: $B = 0$

Magnetic Field of Moving Charge

Point charge $q$ moving with velocity $\vec{v}$:

$\vec{B} = \frac{\mu_0}{4\pi}\frac{q\vec{v} \times \hat{r}}{r^2}$

Two Parallel Wires

Wires separated by distance $d$, currents $I_1$ and $I_2$:

Force per unit length: $\frac{F}{L} = \frac{\mu_0 I_1I_2}{2\pi d}$

• Same direction: attractive
• Opposite direction: repulsive

This defines the ampere!

Magnetic Field Inside Conductor

For long straight conductor of radius $R$:

Outside ($r > R$): $B = \mu_0 I/(2\pi r)$

Inside ($r < R$), uniform current density: $B = \frac{\mu_0 Ir}{2\pi R^2}$

(Linear in $r$, like electric field in charged sphere)

Maxwell's Modification

With changing electric field:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{d\Phi_E}{dt}$

The $\mu_0\epsilon_0 d\Phi_E/dt$ term is displacement current.

This completes Maxwell's equations!

$B = \frac{(4\pi \times 10^{-7})(15)}{2\pi(0.02)}$

$B = \frac{(2 \times 10^{-7})(15)}{0.02} = \frac{3.0 \times 10^{-6}}{0.02}$

$B = \boxed{1.5 \times 10^{-4} \text{ T} = 0.15 \text{ mT}}$

(b) Field at r = 10 cm:

$B = \frac{(2 \times 10^{-7})(15)}{0.10}$

$B = \boxed{3.0 \times 10^{-5} \text{ T} = 30 \text{ μT}}$

(c) Distance for B = 1.0 × 10⁻⁵ T:

$1.0 \times 10^{-5} = \frac{(2 \times 10^{-7})(15)}{r}$

$r = \frac{(2 \times 10^{-7})(15)}{1.0 \times 10^{-5}} = \frac{3.0 \times 10^{-6}}{1.0 \times 10^{-5}}$

$r = \boxed{0.30 \text{ m} = 30 \text{ cm}}$

Note: B ∝ 1/r for long straight wire