Materials Science and Engineering/Equations/Magnetism

Force of Charged Particle

When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:

${\vec {F}}=q{\vec {v}}\times {\vec {B}}$ Force on Current-Carrying Wire

The formula for the total force is as follows:

$\mathbf {F} =I\mathbf {L} \times \mathbf {B} \,$ where

F = Force, measured in newtons
I = current in wire, measured in amperes
B = magnetic field vector, measured in teslas
$\times$ = vector cross product
L = a vector, whose magnitude is the length of wire (measured in metres), and whose direction is along the wire, aligned with the direction of conventional current flow.

The magnetic field generated by a steady current (a continual flow of electric charge, for example through a wire, which is constant in time and in which charge is neither building up nor depleting at any point), is described by the Biot-Savart law:

$d\mathbf {B} ={\frac {\mu _{0}}{4\pi }}{\frac {Id\mathbf {l} \times \mathbf {\hat {r}} }{r^{2}}}$ (in SI units), where

$I$ is the current,
$d\mathbf {l}$ is a vector, whose magnitude is the length of the differential element of the wire, and whose direction is the direction of conventional current,
$d\mathbf {B}$ is the differential contribution to the magnetic field resulting from this differential element of wire,
$\mu _{0}$ is the magnetic constant,
$\mathbf {\hat {r}}$ is the unit displacement vector from the wire element to the point at which the field is being computed, and
$r$ is the distance from the wire element to the point at which the field is being computed.

Magnetic Field Inside Coil - Empty Inductor

$B=\mu _{0}nI\;$ Energy per Unit Volume of Empty Inductor

${\frac {B^{2}}{2\mu _{0}}}={\frac {\mu _{0}n^{2}I^{2}}{2}}$ Total Stored Energy in an Empty Inductor

${\frac {\mu _{0}n^{2}AlI^{2}}{2}}={\frac {LI^{2}}{2}}$ Magnetic Field

$B=\mu _{0}nI+\mu _{0}M\;$ $B=\mu _{0}(H+M)\;$ $B=\mu _{0}\mu _{r}H\;$ Relative Permeability of a Material

$\mu _{r}={\frac {\mathbf {B} }{\mu _{0}\mathbf {H} }}$ $\mu _{r}=1+{\frac {\mathbf {M} }{\mathbf {H} }}$ $\mu _{r}=1+\mathrm {X} _{m}\;$ Anisotropy Energy

$E_{an}=K\sin ^{2}\phi \;$ 