Chapter 12 - Sources of Magnetic Fields

Chapter 12.1 - The Bio-Savart Law

At point $P$, the magnetic field $\vec{dB}$ due to a small segment $d\vec{l}$ of a current carrying wire $I$ is:

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

If you're only concerned with the magnitude: $$ B = \frac{\mu_0}{4\pi} \int_{wire} \frac{I dl \sin(\theta)}{r^2} $$

$\mu_0$ is the permeability of free space
$I$ is the current
$d\vec{l}$ is the differential length
$r$ is the distance from the wire to the point $P$.
$\hat{r}$ is the unit vector pointing from the wire to the point $P$
$\theta$ is the angle between $d\vec{l}$ and $\hat{r}$

The magnetic field at the center of a circular arc of wire, radius $R$ and angle $\theta$ is:

$$ B = \frac{\mu_0 I \theta}{4\pi r} $$

Chapter 12.2 - Magnetic Field of a Straight Wire

The magnetic field $B$ at a distance $R$ away from a thin straight wire carrying current $I$:

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

Chapter 12.3 - Magnetic Force between Parallel Currents

The magnetic field $\vec{B}_1$ created by the first wire, experienced by the other wire, separated by distance $r$:

$$ \vec{B}_1 = \frac{\mu_0 I_1}{2\pi r} $$

The force per unit length between two conducting wires separated by distance $r$:

$$ \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r} $$

$$ \vec{F}_1 = -\vec{F}_2 $$

If the current is in the same direction, the force is attractive.
- [↑ + ↑] = [→ ←]
If the current is in opposite directions, the force is repulsive.
- [↑ + ↓] = [← →]

Chapter 12.4 - Magnetic Field of a Current Loop

Helmholtz Coils

A flat, circular coil with N turns of wire, radius $R$ and $N$ turns, is known as a "Helmholtz Coil"

The magnetic field at the center of a flat, circular coil:

$$ \vec{B} = \frac{\mu_0 N I}{2 R} \hat{n} $$

$N$ is the number of turns in the loop.
$\hat{n}$ is the unit vector perpendicular to the plane of the loop.

Magnetic field at an axial displacement $x$ from the center of a flat, circular coil:

$$ B = \frac{\mu_0 N I R^2}{2 (R^2 + x^2)^{3/2}} $$

Magnetic field at an axial displacement $z$ from the center of a flat, square coil of side-length $a$:

$$ B = \frac{\mu_0 N I a^2}{2 \pi \sqrt{z^2 + \frac{a^2}{2}}(z^2 + \frac{a^2}{4})} $$

Chapter 12.5 - Ampère's Law

Over an arbitrary closed path, the line integral of the magnetic field $\vec{B}$ is:

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

$I_{\text{enc}}$ is the current enclosed by the path.
$d\vec{l}$ is the differential length of a small segment of the path.

The magnetic field due to a thick conductive wire:

Inside the wire ($r \le R$)

$$ B = \frac{\mu_0 I}{2\pi R^2}r \ \ (r \le R) $$

Outside the wire ($r \ge a $):
$$ B = \frac{\mu_0 I}{2\pi r} \ \ (r \ge R) $$
$r$ the radial distance of the Amperian loop used as our surface.
$R$ is the radius of the thick wire
$I$ is the total current flowing through the entire conducting wire.

Note that for situations where the wire is hollow, or contains another wire within it, this formula will not work. You will need to determine the right formula using:

$$ I_{enc} = J * A $$

$J$ is the current density, (Units: $\frac{A}{m^2}$)
$A$ is the cross-sectional area of wire enclosed by the surface (Units: $m^2$)

Once you have these values, you can use the following revised formula:

$$ B = \frac{\mu_0 I_{enc}}{2\pi r} \ \ (r \le R) $$

Chapter 12.6 - Solenoids and Toroids

The magnetic field along the central axis of an infinite solenoid:

$$ B = \mu_{0} \frac{N}{l} I = \mu_{0} n I $$

$N$ is the number of turns
$l$ is the length of the solenoid
$n$ ($\frac{N}{l}$) is the number of turns per unit length
$\hat{n}$ is the unit vector

The magnetic field of a toroid, in particular, inside the torus, at a distance $r$ from the center of the ring of empty space that is being enclosed by the torus:

$$ B = \frac{u_{0} N I}{2\pi r} $$

Chapter 12.7 - Gauss's Law of Magnetism

The net magnetic flux through any closed surface is zero:

$$ \oint \vec{B} \cdot d\vec{A} = 0 $$