9. Constant magnetic fieldSavchenko Formulas, chapter 9 of 14, 18 formulas
Sections follow the book, and within a section the most used formulas come first. An italic problem number means the formula appears in its statement. Rest the cursor on a number to see the statement.
$$\vec F = q\,[\vec v\times\vec B], \qquad F = qvB\sin\alpha$$
$\displaystyle \vec F = q\vec E + q[\vec v\times\vec B]\ \text{(полная сила на заряд)}$$\displaystyle v = \frac{E}{B}\ \text{(скорость, при которой силы компенсируются)}$
q
charge of the particle
$\vec v$
its velocity
$\vec B$
magnetic induction
$\alpha$
angle between $\vec v$ and $\vec B$
The magnetic force is perpendicular to both the velocity and the field, so it does no work and changes only the direction of motion. Its direction follows the left-hand rule or the cross product, reversed for a negative charge. Together with the electric force it is the full force on a charge, and in crossed fields a particle flies straight when $v = E/B$.
$$d\vec F = I\,[d\vec l\times\vec B], \qquad F = IlB\sin\alpha$$
$\displaystyle F = \frac{\mu_0 I_1 I_2}{2\pi d}\,l\ \text{(два параллельных провода)}$$\displaystyle F = 2BIR\ \text{(полукольцо, как хорда)}$
I
current in the conductor
$d\vec l$
element of the conductor along the current
$\vec B$
magnetic induction
$\alpha$
angle between the conductor and the field
The Lorentz force on all carriers in an element of a conductor adds up to Ampère's force, perpendicular to the conductor and the field. The force on a bent conductor in a uniform field equals that on the chord joining its ends, and on a closed loop in a uniform field it is zero. Parallel currents attract with a force per unit length $\mu_0I_1I_2/2\pi d$, antiparallel ones repel.
$$\vec p_m = I\vec S, \qquad \vec N = [\vec p_m\times\vec B], \qquad N = ISB\sin\alpha$$
$\displaystyle U = -\vec p_m\cdot\vec B\ \text{(энергия контура в поле)}$$\displaystyle p_m = \frac{evr}{2}\ \text{(электрон на орбите)}$
$\vec p_m$
magnetic moment, current times area, along the loop's normal
$\vec N$
torque on the loop
$\alpha$
angle between the normal and the field
A uniform field does not pull a closed loop but turns it, trying to set its normal along the field, and the torque is $ISB\sin\alpha$ whatever the loop's shape. A coil of $n$ turns counts as one turn carrying $nI$. In equilibrium the magnetic torque balances that of gravity, and when tilted the loop oscillates at $\sqrt{p_mB/J}$. A nonuniform field pulls the loop toward the stronger field.
$\displaystyle T = BIR\ \text{(натяжение кольца с током в поле)}$$\displaystyle a = \frac{IlB - \mu mg}{m}\ \text{(стержень на рельсах)}$
I
current
l
length of the conductor in the field
T
tension of the suspension
$\alpha$
angle of deflection from the vertical
A suspended conductor swings out until the horizontal Ampère force balances a component of the tension, and the tangent of the angle is $IlB$ over the weight. For a current ring in a field the Ampère force on an arc is balanced by the tension $BIR$ at the arc's ends. A rod on rails in a field accelerates under $IlB$ less friction, and a frame on an axle turns under the torque $ISB$ against gravity's torque.
$\displaystyle B = \frac{\mu_0 I r}{2\pi R^2}\ \text{(внутри провода радиуса }R)$$\displaystyle B = \frac{\mu_0 I}{2\pi r}\tan\frac{\beta}{2}\ \text{(на оси конуса токов из вершины)}$
I
current in the wire
r
distance from the wire's axis
A straight current's field lines are circles around the wire by the right-hand rule, and the field falls as $1/r$, which follows from the circulation theorem for a circle of radius $r$. Inside a thick wire with uniform current only part of the current is enclosed and the field grows linearly. The fields of several wires add as vectors, for antiparallel wires they add between the wires and subtract outside.
$$d\vec B = \frac{\mu_0}{4\pi}\,\frac{I\,[d\vec l\times\vec r]}{r^3}, \qquad dB = \frac{\mu_0 I\,dl\sin\alpha}{4\pi r^2}$$
$\displaystyle B = \frac{\mu_0 I}{4\pi r}(\cos\alpha_1 - \cos\alpha_2)\ \text{(отрезок провода)}$$\displaystyle B = \frac{\mu_0 I}{4R}\ \text{(в центре полуокружности)}, \qquad B = \frac{\mu_0 I\varphi}{4\pi R}\ \text{(дуга)}$
I
current
$d\vec l$
element of the conductor along the current
$\vec r$
vector from the element to the observation point
$\alpha$
angle between $d\vec l$ and $\vec r$
Each current element contributes a field falling as $1/r^2$ and proportional to the sine of the angle between the element and the direction to the point, and the whole conductor's field is the integral of these contributions. Elements pointing at the point produce nothing there, so straight wires through the centre of an arc add nothing at the centre. A composite loop's field is built from the fields of arcs and segments.
$$\vec M = n\vec p_m, \qquad i = M\ \text{(поверхностный ток на единицу длины)}, \qquad \vec B = \mu\mu_0\vec H, \quad \mu = 1 + \chi$$
$\displaystyle \vec B = \mu_0(\vec H + \vec M), \qquad \oint\vec H\cdot d\vec l = I_{\text{пров}}$$\displaystyle B = \mu_0 M\ \text{(внутри длинного намагниченного стержня)}$
$\vec M$
magnetization, magnetic moment per unit volume
i
equivalent surface current per unit length
$\chi$
magnetic susceptibility
$\mu$
permeability
A uniformly magnetized body is equivalent to a surface current $i = M$ running round its side, so a long rod with magnetization $M$ acts as a solenoid with $nI = M$ and a ball as a set of loops carrying $M\sin\theta$. In weak fields the magnetization is proportional to the field, $M = \chi H$, and the field in matter is $\mu$ times the outer one. The circulation of $\vec H$ is set by the conduction currents alone.
$$\vec B = \varepsilon_0\mu_0[\vec v\times\vec E] = \frac{\mu_0}{4\pi}\,\frac{q[\vec v\times\vec r]}{r^3}$$
$\displaystyle B = \frac{\mu_0}{2\pi}\,\frac{\rho v}{r}\ \text{(заряженная нить, движущаяся вдоль себя)}$$\displaystyle B = \frac{\mu_0\sigma v}{2}\ \text{(движущаяся заряженная плоскость)}$
q
the charge
$\vec v$
its velocity, much less than $c$
$\vec E$
the electric field of the same charge at that point
$\vec r$
vector from the charge to the observation point
A slowly moving charge produces a magnetic field obtained from its electric field by a cross product with $\vec v/c^2$. Hence the field of a moving charged body of any shape can be taken from electrostatics, a line gives the field of a straight current, a plane the field of a sheet current. The magnetic force is of order $v^2/c^2$ of the electric one, which is why magnetism is a relativistic correction.
$\displaystyle B \approx \frac{\mu_0 I R^2}{2h^3} = \frac{\mu_0 p_m}{2\pi h^3}\ \text{(далеко на оси, }h \gg R)$$\displaystyle B = \frac{\mu_0 I}{4R}\ \text{(центр полуокружности)}$
I
current in the loop
R
radius of the loop
h
distance from the centre along the axis
On the axis of a loop the transverse contributions cancel and the longitudinal ones add with the factor $R/\sqrt{R^2+h^2}$, giving the formula. At the centre the field is $\mu_0I/2R$, an arc gives the same times its fraction of the circle. Far along the axis the field falls as $1/h^3$ and is set by the magnetic moment $I\pi R^2$, like a dipole's field.
$\displaystyle F = p_m\frac{\partial B}{\partial x}\ \text{(сила на диполь в неоднородном поле)}$$\displaystyle F = \frac{3\mu_0 p_1p_2}{2\pi r^4}\ \text{(два диполя на одной оси)}$
$\vec p_m$
magnetic moment, $IS$ for a loop
$\vec r$
vector from the dipole to the point
$B_\parallel, B_\perp$
field on the dipole's axis and in its equatorial plane
Far from any closed current the field is that of an electric dipole with $p/\varepsilon_0$ replaced by $\mu_0 p_m$, twice as large on the axis as on the perpendicular at the same distance, and falling everywhere as $1/r^3$. A magnetized ball or a short magnet has the same field. The force between dipoles follows from the field's derivative and falls as $1/r^4$.
$$B = \frac{\mu_0 i}{2}, \qquad B = \mu_0 j x\ \text{(внутри слоя)}, \qquad B = \frac{\mu_0 j d}{2}\ \text{(снаружи слоя)}$$
$\displaystyle i = \sigma v\ \text{(движущаяся заряженная плоскость)}$$\displaystyle B_{1\tau} - B_{2\tau} = \mu_0 i\ \text{(скачок поля на слое тока)}$
i
surface current density, current per unit width
j
volume current density in the slab
x
distance from the slab's mid-plane
d
thickness of the slab
An infinite current sheet gives on both sides a uniform field $\mu_0 i/2$, parallel to the sheet and perpendicular to the current, which follows from the circulation round a rectangle across the sheet. Inside a slab with volume current the field grows linearly from the middle. Across a current sheet the tangential $B$ jumps by $\mu_0 i$ while the normal component is continuous, hence the refraction of field lines. Two sheets with opposite currents give a field only between them, like a parallel-plate capacitor.
$$\oint \vec B\cdot d\vec l = \mu_0 I_{\text{охв}}$$
$\oint\vec B\cdot d\vec l$
circulation of the induction round a closed loop
$I_{\text{охв}}$
total current through the loop, signed by the direction of traversal
The circulation of the magnetic field round any closed loop equals $\mu_0$ times the current threading the loop, whatever currents flow outside. For symmetric currents one picks a loop along which $B$ is constant, a circle round a wire, a rectangle across a sheet or a solenoid, and the field comes out in one line. Like Gauss's law for $E$, it is the shortest route to the field of a wire, a solenoid, a torus and a sheet.
$$B = \mu_0 n I, \qquad B = \frac{\mu_0 N I}{2\pi r}\ \text{(тороид)}$$
$\displaystyle B = \frac{\mu_0 n I}{2}(\cos\alpha_1 - \cos\alpha_2)\ \text{(на оси конечного соленоида)}$$\displaystyle B = \frac{\mu_0 nI}{2}\ \text{(на торце длинного соленоида)}$
n
turns per unit length
I
current in the winding
N
total number of turns of the torus
r
distance from the torus's axis
A long solenoid is a rolled-up current sheet with $i = nI$, the field inside is uniform, $\mu_0 nI$, and zero outside, as the circulation round a rectangle with one side inside shows. At the end face the field is half as large, since half the solenoid is missing, and on the axis of a finite solenoid it is expressed through the angles subtended by its ends. A torus keeps its field inside, falling as $1/r$. A solenoid's winding is pushed outward by the pressure $B^2/2\mu_0$.
$$p = w = \frac{B^2}{2\mu_0}, \qquad w = \frac{B^2}{2\mu\mu_0}$$
$\displaystyle F = \frac{B^2}{2\mu_0}S\ \text{(сила на стенку соленоида, на торец сердечника)}$$\displaystyle w = \frac{\varepsilon_0E^2}{2} + \frac{B^2}{2\mu_0}\ \text{(полная плотность энергии поля)}$
w
energy density of the magnetic field
p
pressure of the field on a current-carrying conductor, outward
$\mu$
permeability of the medium
As for the electric field, the energy sits in the field with density $B^2/2\mu_0$, and where there is field on one side of a current surface and none on the other this same quantity is the pressure on the surface. So a solenoid stretches itself, plates with opposite currents repel, and currents in one direction attract, the field between them being weakened. The force on a core equals the difference of pressures on its ends.
solid angle the current surface subtends at the point
$B_\parallel$
field component perpendicular to the current lines within the surface
The field of a surface current comes from the field of a charged surface multiplied by $v/c^2$, and the normal field of a charged surface is $\sigma\Omega/4\pi\varepsilon_0$, so the field of the current $i = \sigma v$ is expressed through the same solid angle. For an infinite sheet $\Omega = 2\pi$ gives $\mu_0 i/2$, for the end of a long solenoid also $2\pi$, for a closed surface round the point $4\pi$ and $B = \mu_0 i$. Part of a surface is handled by subtraction, like the field of a cavity.
$$\Phi = \int \vec B\cdot d\vec S = BS\cos\alpha$$
$\displaystyle \Phi = \frac{\mu_0 I l}{2\pi}\ln\frac{r_2}{r_1}\ \text{(рамка у прямого провода)}$$\displaystyle \Phi = BS\cos\omega t\ \text{(вращающаяся рамка)}$
$\Phi$
magnetic flux through a surface
S
area of the surface spanning the loop
$\alpha$
angle between $\vec B$ and the surface's normal
Flux is the number of field lines through a surface, in a uniform field the induction times the projected area. In a nonuniform field it is gathered by an integral, near a straight wire in strips $B(r)\,l\,dr$. For a coil of $N$ turns the turns' fluxes add. Through a closed surface the flux is always zero, so every surface spanning one loop gives the same flux.
$$L = \frac{\Phi}{I}, \qquad L = \mu_0 n^2 S l\ \text{(соленоид)}$$
L
inductance of the loop
$\Phi$
flux of the loop's own field through it, for a coil through all turns
n
turns per unit length
The flux of a loop's own field is proportional to its current, with a coefficient set by geometry and the medium alone. In a solenoid the field $\mu_0 nI$ threads $nl$ turns of area $S$, hence $\mu_0 n^2 Sl$. The mutual inductance of two loops is defined the same way through the flux of one loop through the other and is the same both ways, for a small loop on the axis of a large one it is computed from the large loop's field at the small one's centre.
$$\oint \vec B\cdot d\vec S = 0, \qquad \operatorname{div}\vec B = 0$$
$\displaystyle B_r = -\frac{r}{2}\frac{\partial B_z}{\partial z}\ \text{(радиальное поле у оси)}$$\displaystyle B_1S_1 = B_2S_2\ \text{(трубка линий поля)}$
$\oint\vec B\cdot d\vec S$
flux through a closed surface
Br
radial field component near the symmetry axis
There are no magnetic charges, field lines close on themselves, and the flux through any closed surface is zero. Hence the flux along a tube of lines is constant, the field is inversely proportional to the tube's cross-section, and the normal component of $B$ is continuous across any boundary. For an axially symmetric field the flux through a cylinder round the axis gives the radial component $-\frac{r}{2}\,\partial B_z/\partial z$ wherever the longitudinal field changes, exactly as for the electric field near a lens axis.