#### Inductance
emd.Fl.i

Thinking of Faraday's experiments. Two loops of wire. From Biot-Savart, flux \(\Phi_2\) of \({\bf B}_1\) through loop 2 is (using fact that \({\bf B}_1\) is proportional to \(I_1\)) \[ \Phi_2 = \int {\bf B}_1 \cdot d{\bf a}_2 \Longrightarrow \Phi_2 = M_{21} I_1 \] where \(M_{21}\) is the mutual inductance of the two loops.

Useful formula: \[ \Phi_2 = \int {\bf B}_1 \cdot d{\bf a}_2 = \int ({\boldsymbol \nabla} \times {\bf A}_1) \cdot d{\bf a}_2 = \oint {\bf A}_1 \cdot d{\bf l}_2 \] But from A_CoulG, \[ {\bf A}_1 ({\bf r}) = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_1} \frac{d{\bf l}_1}{|{\bf r} - {\bf r}_1|} \] so \[ \Phi_2 = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_2} d{\bf l}_2 \cdot \left(\oint_{{\cal P}_1} \frac{d{\bf l}_1 }{|{\bf r}_2 - {\bf r}_1|}\right) \] and we can write the mutual inductance as the Neumann formula,

- Gr (7.22)

\[ M_{21} = \frac{\mu_0}{4\pi} \oint_{{\cal P}_1} \oint_{{\cal P}_2} \frac{d{\bf l}_1 \cdot d{\bf l}_2} {|{\bf r}_1 - {\bf r}_2|} \tag{Neumann_M}\label{Neumann_M} \] Two things: first, \(M_{21}\) is purely geometrical. Second,

- Gr (7.23)

\[ M_{12} = M_{21} \tag{Msym}\label{Msym} \]

Example: solenoid in solenoid

Consider a short solenoid (length \(l\), radius \(a\), \(n_1\) turns per unit length) which lies concentrically inside a very long solenoid (radius \(b\), \(n_2\) turns per unit length). Current \(I\) in short solenoid.

Task: compute the flux through the long solenoid.

Solution: it's complicated to calculate \({\bf B}_1\). Use mutual inductance, starting from the reverse situation: current \(I\) on outer solenoid, calculate flux through inner one. Field of outer solenoid: from Amp_int, \[ B = \mu_0 n_2 I \] so flux through a single loop of inner solenoid is \[ B \pi a^2 = \mu_0 n_2 I \pi a^2. \] For \(n_1 l\) turns in total, total flux through inner solenoid is \[ \Phi = \mu_0 \pi a^2 n_1 n_2 l I. \] Same as flux through outer solenoid if inner one has current \(I\). Mutual inductance is here \[ M = \mu_0 \pi a^2 n_1 n_2 l. \]

What if we vary current in loop 1? Flux in 2 will vary. Induces EMF in loop 2: \[ {\cal E} = -\frac{d\Phi_2}{dt} = -M \frac{dI_1}{dt}. \label{Gr(7.24)} \] Changing current also induces EMF in the source loop itself:

- Gr (7.25)

\[ \Phi = L I \tag{PLI}\label{PLI} \] where \(L\) is the self-inductance (or inductance) of the loop. Depends only on geometry. Changing current induces EMF of \[ {\cal E} = -L \frac{dI}{dt} \label{Gr(7.26)} \] Inductance: measured in henries (\(H\)). \(H = V s/A\).

Example: self-inductance of toroidal coil

Consider a toroidal coil with rectangular cross-section (inner radius \(a\), outer radius \(b\), height \(h\)) which carries total of \(N\) turns.

Task: find its self-inductance

Solution: magnetic field inside toroid is Btor \[ B = \frac{\mu_0 NI}{2\pi s} \] Flux through single turn: \[ \int {\bf B} \cdot d{\bf a} = \frac{\mu_0 N I}{2\pi} h \int_a^b \frac{ds}{s} = \frac{\mu_0 N I h}{2\pi} \ln \frac{b}{a}. \] Total flux: \(N\) times this, so self-inductance is \[ L = \frac{\mu_0 N^2 h}{2\pi} \ln \frac{b}{a} \label{Gr(7.27)} \]

Inductance (like capacitance) is intrinsically positive. Use Lenz law. Think of back EMF.

Example: circuit

Consider a circuit with inductance \(L\), resistor \(R\) and battery \({\cal E}_0\).

Task: find the current

Solution: Ohm's law: \[ {\cal E}_0 - L \frac{dI}{dt} = IR \Longrightarrow I(t) = \frac{{\cal E}_0}{R} + k e^{-(R/L)t}. \] If initial condition: \(I(0) = 0\), then \[ I(t) = \frac{{\cal E}_0}{R} \left[ 1 - e^{-(R/L)t} \right] \label{Gr(7.28)} \] where \(\tau \equiv L/R\) is the time constant of the circuit.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42