11. Electromagnetic inductionSavchenko Formulas, chapter 11 of 14, 26 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.
speed of the conductor across the field and itself
E
electric field in the conductor balancing the Lorentz force
The carriers in a moving conductor feel the Lorentz force $qvB$, which drives them to one end until a field $E = vB$ builds up, and the potential difference between the ends is $Blv$. The same follows from Faraday's law, since the loop's area changes at the rate $lv$. An aircraft wing, a spinning disc, a rod on rails and a current-carrying strip in a field all give the same emf, and the sign of the voltage across a strip with current tells the sign of the carriers.
$\displaystyle v(t) = v_{\text{уст}}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{mR}{B^2l^2}$$\displaystyle P = I^2R = \frac{B^2l^2v^2}{R}\ \text{(вся работа торможения идёт в тепло)}$
l
length of the bar on the rails
R
resistance of the loop
$v_{\text{уст}}$
terminal speed
$\tau$
time of approach to it
A moving bar drives a current $Blv/R$, and the Ampère force $IlB$ on that current opposes the motion in proportion to the speed, like viscous friction. Under gravity the bar accelerates to the speed at which $mg$ is balanced by this force, all of gravity's power going into heat. With a capacitor instead of a resistor in the loop the current is proportional to the acceleration and the bar moves with constant acceleration and an added mass $CB^2l^2$.
$\displaystyle F v = \mathcal{E}_{\text{инд}} I\ \text{(механическая мощность равна электрической)}$$\displaystyle \omega_{\text{хх}} = \frac{U}{k}\ \text{(холостой ход, }I \to 0)$
U
voltage applied to the motor
$\mathcal{E}_{\text{инд}}$
back emf, the induced emf in the moving winding
k
machine constant, $\mathcal{E} = k\omega$ and $M = kI$
$P_{\text{мех}}$
mechanical power at the shaft
The motor's winding moves in the field and a back emf proportional to the speed is induced in it, so the current is $(U - k\omega)/R$ and falls as the motor speeds up. The source's power splits into heat $I^2R$ and mechanical power $k\omega I = M\omega$, the same constant $k$ linking torque to current. Unloaded, $\omega \to U/k$, stalled, all the current becomes heat. A generator is the same machine driven from outside.
$\displaystyle \Phi = BS\cos\omega t, \qquad I = \frac{BS\omega}{R}\sin\omega t$$\displaystyle M = \frac{B^2S^2\omega}{R}\sin^2\omega t\ \text{(тормозящий момент)}$
S
area of the frame
$\omega$
angular velocity of rotation
r
length of a rod turning about its end
The flux through a frame spinning in a steady field varies as $\cos\omega t$, and its derivative gives a sinusoidal emf of amplitude $BS\omega$, which is the alternator. A rod turning about its end sweeps an area $\omega r^2/2$ per second, hence its emf, and the Faraday disc is the same. The current in the frame makes a braking torque, and the mechanical power spent against it equals the Joule heat.
The current is $-\frac{1}{R}\frac{d\Phi}{dt}$, and integrating over time gives a charge that depends only on the total change of flux, not on how fast it happened. The ballistic galvanometer rests on this, measuring a field by the charge when a coil is flipped or pulled out. The mean emf over a time $T$ is $\Delta\Phi/T$.
$\displaystyle F_x = I\frac{\partial\Phi}{\partial x}\ \text{(сила на контур с током)}$
I
current in the loop, held constant
$\Delta\Phi$
change of the external flux through the loop during the move
The work of Ampère forces in any displacement or turn of a loop with constant current equals the current times the change of the flux threading it. Hence the force and torque on a loop as derivatives of the flux, and the loop is pulled toward growing flux. The source keeping the current constant does twice this work, half of it changing the field energy.
$$\mathcal{E} = -\frac{d\Phi}{dt}, \qquad \oint\vec E\cdot d\vec l = -\frac{d\Phi}{dt}$$
$\displaystyle \mathcal{E} = -N\frac{d\Phi}{dt}\ \text{(катушка из }N\text{ витков)}$$\displaystyle \mathcal{E} = -S\frac{dB}{dt}\ \text{(неподвижный контур в меняющемся поле)}$
$\mathcal{E}$
induced emf in the loop
$\Phi$
magnetic flux through the loop
$\oint\vec E\cdot d\vec l$
circulation of the electric field round the loop
Any change of flux through a loop, from its motion, a changing field or a turn, induces in it an emf equal to the rate of change of flux, and the minus sign is Lenz's rule, the induced current opposes the change. In a loop at rest the emf comes from a vortex electric field whose circulation is not zero, and that field exists whether or not a conductor is placed there.
$$E\cdot 2\pi r = -\frac{d\Phi}{dt}, \qquad E = \frac{r}{2}\frac{dB}{dt}\ (r < R), \qquad E = \frac{R^2}{2r}\frac{dB}{dt}\ (r > R)$$
$\displaystyle m\frac{dv}{dt} = eE = \frac{er}{2}\frac{dB}{dt}\ \text{(разгон электрона на орбите)}$$\displaystyle M = qEr = \frac{q}{2\pi}\frac{d\Phi}{dt}\ \text{(момент на заряженное кольцо)}$
r
radius of the circle round which the circulation is taken
R
radius of the region of changing field, of the solenoid
$dB/dt$
rate of change of the field
A changing magnetic field makes closed lines of electric field, by symmetry circles inside a solenoid, and Faraday's law for a circle of radius $r$ gives its size. Inside the region the field grows linearly with $r$, outside it falls as $1/r$, like a straight current's. This field accelerates charges round a circle, the betatron works on it, and it spins up a charged ring when the field is switched on.
$$m_{\text{эм}} = \frac{U}{c^2}, \qquad U = \frac{e^2}{8\pi\varepsilon_0 R}$$
U
field energy of the charged body or capacitor
c
speed of light
R
radius of the charged sphere
A moving charged capacitor carries a magnetic field along with its electric one, and the field momentum $\varepsilon_0\int[\vec E\times\vec B]\,dV$ comes out as $Uv/c^2$, so the field behaves like a mass $U/c^2$. This is a special case of $E = mc^2$, and for a sphere of charge $e$ it gives the classical electron radius if all the mass is ascribed to the field.
In a betatron the electron is accelerated by the vortex field $E = \frac{1}{2\pi R}\,d\Phi/dt$ and held on a circle by the Lorentz force, which needs $mv = eRB(R)$. The momentum grows as $e\Phi/2\pi R$, while holding the orbit needs $mv = eRB(R)$, and the orbit stays fixed only when the field on it equals half the mean field inside it. So the field $B_0(1 - r/r_0)$ holds the electron at radius $3r_0/4$.
$\displaystyle L\frac{dI}{dt} = U\ \text{(катушка на источнике, }I = \frac{Ut}{L})$$\displaystyle L\frac{dI}{dt} + IR = \mathcal{E}\ \text{(цепь с индуктивностью и сопротивлением)}$
L
inductance
I
current in the loop
$\mathcal{E}_L$
self-induction emf, opposing the change of current
The flux of a loop's own field is proportional to the current, $\Phi = LI$, and by Faraday's law a change of current induces an emf $-L\,dI/dt$. A coil across a source builds up current linearly, $I = \mathcal{E}t/L$, and on an alternating voltage its current lags by a quarter period. In a circuit equation the inductance enters as $L\,dI/dt$ beside $IR$ and $q/C$.
$$L = \mu\mu_0 n^2 S l = \frac{\mu\mu_0 N^2 S}{l}, \qquad \frac{L}{l} = \frac{\mu\mu_0}{2\pi}\ln\frac{r_2}{r_1}$$
$\displaystyle L' = \frac{\mu_0 h}{d}\ \text{(на единицу длины плоской линии, зазор }d\text{, ширина }h)$$\displaystyle L' = \frac{\mu_0}{\pi}\ln\frac{h}{r}\ \text{(двухпроводная линия)}$
n
turns per unit length
N
total number of turns
S, l
cross-section and length of the solenoid
r1, r2
radii of the core and sheath of the cable
$\mu$
permeability of the core or filling
Inductance is computed as flux per unit current, in a solenoid the field $\mu_0 nI$ threads $N$ turns, in a cable the flux is gathered in strips between core and sheath where $B = \mu_0 I/2\pi r$. Equivalently one computes the field energy $\int B^2/2\mu_0\,dV$ and sets it equal to $LI^2/2$, handy for lines whose flux is not obvious. A core of permeability $\mu$ multiplies the inductance by $\mu$, and $n$ times more turns on the same length by $n^2$.
$\displaystyle F_x = \frac{I^2}{2}\frac{dL}{dx}\ \text{(сила при постоянном токе)}$$\displaystyle Q = \frac{LI^2}{2}\ \text{(тепло после размыкания цепи)}$
L
inductance
I
current
$\Phi$
flux through the coil
B
field inside, $\mu_0 nI$ for a solenoid
To bring the current up to $I$ the source works against the self-induction emf, and this work $LI^2/2$ is stored in the magnetic field with density $B^2/2\mu_0$, the two forms agreeing for a solenoid. When the circuit is broken it all goes into heat or a spark, in a loop with a capacitor it is traded for $CU^2/2$. The force on a core or a movable part at constant current is $\frac{I^2}{2}\,dL/dx$ and pulls toward larger inductance.
$$\Phi_2 = M I_1, \qquad \mathcal{E}_2 = -M\frac{dI_1}{dt}, \qquad M = \mu\mu_0 n_1 n_2 S l$$
$\displaystyle L = L_1 + L_2 \pm 2M\ \text{(две катушки последовательно, согласно или встречно)}$$\displaystyle M = \frac{\mu_0\pi r^2 R^2}{2 d^3}\ \text{(малый виток на оси большого, далеко)}$
M
mutual inductance, the same both ways
I1
current in the first loop
$\Phi_2$
flux of the first loop through the second
n1, n2
turns per unit length of two windings on one core
The flux that one loop's current makes through another is proportional to that current, and the coefficient is symmetric, so it is computed in whichever direction the field is simpler, a long coil's field through a short one or a large loop's field at a small loop's centre. A changing current in the first loop induces in the second an emf $-M\,dI_1/dt$. For two coils in series $2M$ is added to or subtracted from the sum of inductances.
Both windings on a common core are threaded by one flux, so the emf per turn is the same and the voltages are as the numbers of turns. The net magnetizing current of the core is small, whence $N_1I_1 + N_2I_2 \approx 0$, the currents are inversely proportional to the turns and power passes without loss. A load $R$ on the secondary looks from the primary like $R(N_1/N_2)^2$.
The capacitor's charge and the coil's current vary as in a spring pendulum, $L$ playing the mass and $1/C$ the stiffness, and the frequency is $1/\sqrt{LC}$. Current and charge are a quarter period apart, and the current amplitude is $\omega_0 q_0$. A circuit connected to a constant source oscillates about the new equilibrium $q = C\mathcal{E}$ at the same frequency, with a peak charge twice the equilibrium one. Two parallel coils are replaced by one with $L_1L_2/(L_1+L_2)$.
With a sinusoidal current the voltage across a coil leads the current by a quarter period with amplitude $I_0\omega L$, across a capacitor it lags with amplitude $I_0/\omega C$, across a resistor it is in phase. They add as vectors, hence the impedance and phase shift, and with complex amplitudes the whole circuit follows the rules of direct current with $Z_L = i\omega L$ and $Z_C = 1/i\omega C$. At $\omega L = 1/\omega C$ the reactive voltages cancel, which is resonance.
$\displaystyle Q = \frac{L_1I_1^2}{2} - \frac{(L_1 + L_2)I'^2}{2}\ \text{(тепло в искре при переключении)}$
L1, L2
inductances of the coils
I1, I2
their currents before switching
$I'$
common current right after joining
In an instantaneous switch the self-induction emfs are enormous and no charge has time to pass through resistances and capacitances, so the total flux linkage $\sum L_kI_k$ is conserved, like momentum in an inelastic collision. The currents after switching follow from this, and the lost energy goes into a spark. For two parallel coils the difference $L_1I_1 - L_2I_2$ never changes, since their voltages are equal.
The steady current in a circuit with an applied emf is a forced oscillation at the emf's frequency with an amplitude that grows as the natural frequency is approached, which is resonance, limited only by the resistance. The general solution is the forced plus the free oscillation, and with initial conditions at $\omega \approx \omega_0$ one gets beats in which the amplitude grows linearly. A chain of LC sections passes oscillations like a line with phase velocity $l/\sqrt{LC}$.
$$L\frac{di}{dt} + iR = U, \qquad i = \frac{U}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}$$
$\displaystyle i = i_0 e^{-Rt/L}\ \text{(ток после отключения источника через сопротивление)}$$\displaystyle u_L = U e^{-t/\tau}\ \text{(напряжение на катушке при включении)}$
L
inductance
R
resistance of the circuit
U
voltage of the source
$\tau$
time constant $L/R$
Inductance keeps the current from jumping, on switch-on the current approaches $U/R$ exponentially with time constant $L/R$, and when a coil is shorted through a resistor it decays by the same exponential, giving its energy $LI^2/2$ to heat. In a time $\tau$ the quantities change by a factor $e$. At the first instant the whole source voltage sits across the coil.
Resistance enters the circuit equation as friction enters a pendulum, the amplitude decays exponentially with exponent $R/2L$ and the frequency is slightly below the natural one. Successive amplitudes are in the ratio $e^{\gamma T}$, and the quality factor tells over how many periods the oscillation dies by $e^\pi$. For $R \ge 2\sqrt{L/C}$ there is no oscillation and the charge decays monotonically.
effective value, giving the same power as a direct current
The instantaneous power $iu$ oscillates, and its average over a period is half the product of the amplitudes times the cosine of the phase shift, since the mean of $\sin^2$ is one half. Heat appears only in the resistance, $\langle P\rangle = I_0^2R/2$, a coil and a capacitor consume no power on average. Effective values $I_0/\sqrt2$ and $U_0/\sqrt2$ let the formulas be written as for direct current.
$\displaystyle U_{\max} = I_{\max}\sqrt{\frac{L}{C}}\ \text{(волновое сопротивление контура)}$$\displaystyle Q = \frac{q_0^2}{2C} - \frac{LI^2}{2}\ \text{(тепло, если в контуре есть сопротивление)}$
q0
peak charge of the capacitor
$I_{\max}$
peak current
U0
capacitor voltage when the current is zero
In a circuit without resistance the energy sloshes between capacitor and coil and their sum is constant. When the current peaks the capacitor is empty, so the amplitudes obey $LI_{\max}^2 = CU_0^2$. With resistance the difference of energies of two states is the heat released, and if the circuit is rearranged by switching part of the energy is lost in a spark, found through conservation of flux or charge.
In a loop without resistance the induced emf cannot be balanced by a voltage drop, so $d\Phi/dt = 0$ and the flux through it never changes, any external change being cancelled by a current. Hence the current when a field is switched on, when the ring is squeezed or a core inserted, and the growth of the field inside a squeezed ring as $r_0^2/r^2$. A ring in a nonuniform field oscillates about the position where the flux equals its initial value, with an extra stiffness $B^2S^2/L$.
$\displaystyle F = \frac{\mu_0 I^2 l}{4\pi h}\ \text{(провод над сверхпроводящей плоскостью, изображение тока)}$$\displaystyle \frac{B^2}{2\mu_0} = \rho g h\ \text{(сверхпроводящая жидкость поднята полем)}$
B
field at the surface, tangential to it
p
pressure of the field on the superconductor, away from the field
S, l
cross-section and length of the region the field is expelled from
A superconductor keeps the field out, at its surface the normal component of $B$ is zero and the tangential one is carried by a surface current $i = B/\mu_0$. The field presses on the surface with $B^2/2\mu_0$ per unit area, hence the levitation of a magnet or a wire above a superconductor, where the field follows from the method of images with a reversed current. Pushing a superconducting body into a field costs the energy of the expelled field, $B^2/2\mu_0$ per volume.
$\displaystyle B = \frac{\mu_0\varepsilon_0 r}{2}\frac{dE}{dt}\ \text{(внутри заряжаемого плоского конденсатора)}$$\displaystyle B\cdot2\pi r = \mu_0 I\frac{r^2}{r_0^2}\ \text{(ток смещения между обкладками равен току в проводе)}$
$\Phi_E$
flux of the electric field through the loop
$I_{\text{см}}$
displacement current, $\varepsilon_0\,d\Phi_E/dt$, in a medium $d\Phi_D/dt$
I
conduction current through the loop
Between the plates of a charging capacitor there is no conduction current, but the current in the wire continues without a break as the displacement current $\varepsilon_0\,d\Phi_E/dt$, which Maxwell added to the circulation theorem. So a changing electric field makes a magnetic field just as a current does, inside the capacitor $B = \frac{\mu_0\varepsilon_0 r}{2}\,dE/dt$, and in a dielectric the displacement current is $\varepsilon$ times larger. A moving charged capacitor carries the field $\mu_0\varepsilon_0 vE$, as follows from the field of moving charges.