Inductance and Magnetic properties of materials

Inductance and Magnetic properties of materials

Inductance

Mutual Inductance

Suppose \(S_1\) generate a magnetic field \(B_1\), then the flux linkage in \(S_2\) due to \(S_1\) is

\[ \begin{gathered} \Psi_{12} \propto N_2A_2B_1 \propto N_2\Phi_{12} \\\\ \Rightarrow \Psi_{12} = M_{12} i_1 \end{gathered} \]

From Faraday's Law, we have

\[ \begin{gathered} M_{12} = \frac{\Psi_{12}}{i_1} = \frac{N_2\Phi_{12}}{i_1} \\\\ \epsilon_2 = - \frac{d\Psi_{12}}{dt} = -M_{12} \frac{di_1}{dt} \end{gathered} \]

\(M_{12}, M_{21}\) are called mutual inductance constant. Generally, \(M_{12} = M_{21} = M\).

Self Inductance

The magnetic field produced by the current in the loop is proportional to that current. The flux, therefore, is also proportional to the current: \(\(\Phi_B = \iint B \bullet dA \propto i\)\)

Therefore, we can define this constant of proportionality between flux and current to be the self-inductance as \(\(L = \frac{\Phi_B}{i}\)\)

Combining with Faraday’s Law gives the emf induced by a changing current: \(\(\epsilon = - \frac{d\Psi}{dt} = -L \frac{di}{dt}\)\)

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For a solenoid, \(\(L = \frac{\Psi}{i} = \frac{N \cdot \mu_0ni \cdot A}{i} = \mu_0n^2lA = \mu_0n^2V\)\)

The self-inductance per unit volume is \(\mu_0n^2\). The self-inductance per unit length is \(\mu_0n^2A\).

For a toroid of rectangular,

\[ \begin{gathered} \oint B\bullet dl = \mu_0Ni \\\\ \Phi_B = \iint B\bullet dA = \int_a^b \frac{\mu_0Ni}{2\pi r} hdr = \frac{\mu_0Nhi}{2\pi} \ln \frac{b}{a} \\\\ L = \frac{N\Phi_B}{i} = \frac{\mu_0N^2h}{2\pi} \ln \frac{b}{a} \end{gathered} \]

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When no flux leakage, \(M = \sqrt{L_1L_2}\)

When direct in series \(L = L_1 + L_2 + 2M = L_1 + L_2 + 2\sqrt{L_1L_2}\)

When opposite in series \(L = L_1 + L_2 - 2M = L_1 + L_2 - 2\sqrt{L_1L_2}\)

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For inductors with magnetic material \(\(L = \kappa_m L_0\)\) where \(\kappa_m\) is permeability constant.

Magnetic properties of materials

Nuclear Magnetism

\[\mu = iA = \frac{ev}{2\pi r} \cdot \pi r^2 = \frac{1}{2} erv = \frac{1}{2} er^2 \omega\]

Suppose the angular momentum is \(l = mvr\), the orbital magnetic dipole moment can be \(\(\vec{\mu}_L = - \frac{e}{2m} \vec{L}\)\)

By Quantum mechanism, the smallest \(L\) is \(h/2\pi\), so the Bore Magneton is \(\(\mu_B = \frac{eh}{4\pi m} = 9.274 \times 10^{-24} Am^2\)\)

Suppose the intrinsic angular momentum is \(S\), the intrinsic magnetic dipole moment can be \(\(\vec{\mu}_S = - \frac{e}{m} \vec{S}\)\)

Then we can define

\[ \begin{gathered} \vec{J} = \vec{L} + 2\vec{S} \\\\ \vec{\mu}_J = - \frac{e}{m} \vec{J} \end{gathered} \]

Magnetization of material

After magnetization, induced current \(i'\) will be generated on the surface.

Define the magnetization vector \(\(\vec{M} = \frac{\sum \mu_m}{\Delta V}\)\)

For uniform magnetization,

\[ \begin{gathered} j' = \frac{i'}{\Delta z} \\\\ \Delta m = i' \Delta A = j' \Delta x \Delta y \Delta z \\\\ M = \frac{\Delta m}{\Delta V} = j' \\\\ M\Delta z = i' \end{gathered} \]

For non-uniform magnetization, we can proof \(\(\oint \vec{M}\bullet d\vec{l} = \sum_{inl} i'\)\)

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By Ampere’s Loop Law, we have

\[ \begin{gathered} \oint \vec{B} \bullet d\vec{l} = \mu_0 \sum_{inl} (i_0 + i') \\\\ \Rightarrow \oint ( \frac{\vec{B}}{\mu_0} - \vec{M} ) \bullet d\vec{l} = \sum_{inl} i_0 \end{gathered} \]

Therefore, we can define magnetic field strength \(\(\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M}\)\)

The new Ampere’s Loop Law is \(\(\oint \vec{H}\bullet d\vec{l} = \sum_{inl} i_0\)\)

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Define susceptibility magnetization coefficient: \(\(\vec{M} = \chi_m \vec{H}\)\)

Define permeability constant: \(\(\vec{B} = \kappa_m \mu_0\vec{H}\)\)

For magnetic materials

\[ \begin{gathered} B = \mu_0(H + M) = \mu_0(1 + \chi_m)H = \kappa_m \mu_0H \\\\ \Rightarrow \kappa_m = 1 + \chi_m \end{gathered} \]

Paramagnetic materials: \(\chi_m > 0, \kappa_m > 1, \kappa_m \approx 1\) Diamagnetic materials: \(\chi_m < 0, \kappa_m < 1, \kappa_m \approx 1\) Ferromagnetic materials: \(\chi_m(H), \kappa_m(H)\)

RL circuits

For RC circuits, when K is closed,

\[ \begin{gathered} iR + \frac{q}{C} = \epsilon \\\\ \Rightarrow \frac{dq}{dt} + \frac{q}{RC} = \frac{\epsilon}{R} \\\\ \Rightarrow q = C\epsilon(1 - e^{- \frac{1}{RC}t}) \end{gathered} \]

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For RL circuits, when K is connected to seat,

\[ \begin{gathered} iR + L\frac{di}{dt} = \epsilon \\\\ \Rightarrow i = \frac{\epsilon}{R} (1 - e^{- \frac{R}{L} t}) \\\\ \Rightarrow V_L = -L \frac{di}{dt} = \epsilon e^{- \frac{R}{L} t} = \epsilon e^{- t/\tau_L} \end{gathered} \]

where \(\tau_L = \frac{L}{R}\) is induct time constant.

When K is disconnected from seat,

\[ \begin{gathered} iR + L\frac{di}{dt} = 0 \\\\ \Rightarrow i = \frac{\epsilon}{R} e^{- \frac{R}{L} t} \\\\ \Rightarrow V_L = -\epsilon e^{- \frac{R}{L} t} \end{gathered} \]

Energy storage in a magnetic field

The magnetic energy in a self-inductance is

\[ \begin{gathered} dW = -\epsilon_L dq = L \frac{di}{dt} \cdot idt = Lidi \\\\ \Rightarrow W = \int_0^I Lidi = \frac{1}{2} LI^2 \end{gathered} \]

The magnetic energy stored in two solenoid is

\[ \begin{aligned} W &= W_1 + W_2 \\ &= - \int_0^{\infty} \epsilon_{21} i_1 dt - \int_0^{\infty} \epsilon_{12} i_2 dt \\ &= \int_0^{I_1I_2} (M_{21}i_1di_2 + M_{12}i_2di_1) \\ &= M \int_0^{I_1I_2} d(i_2i_1) \\ &= MI_1I_2 \end{aligned} \]

So the total magnetic energy in \(k\) solenoids is \(\(U_m = \frac{1}{2} \sum_{i=1}^{k} L_iI_i^2 + \frac{1}{2} \sum_{i,j=1}^{k} M_{ij}I_iI_j\)\)

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The energy density in a magnetic field is \(\(u_B = \frac{U}{V} = \frac{\frac{1}{2}LI^2}{V} = \frac{\frac{1}{2} \cdot \mu_0 n^2 V \cdot I^2}{V} = \frac{1}{2} \mu_0n^2I^2 = \frac{B^2}{2\mu_0}\)\)

Therefore,

\[ \begin{gathered} u_B = \frac{B^2}{2\mu_0} = \frac{1}{2} \vec{B} \bullet \vec{H} \\\\ u_E = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \vec{D} \bullet \vec{E} \end{gathered} \]

Electromagnetic Oscillation

For capacitor and inductor, there are

\[ \begin{gathered} U_E = \frac{1}{2} \frac{q^2}{C} \\\\ U_B = \frac{1}{2} Li^2 \end{gathered} \]

Because the total energy is constant, we have

\[ \begin{gathered} U = U_B + U_E = \frac{1}{2} \frac{q^2}{C} + \frac{1}{2} Li^2 \\\\ \Rightarrow \frac{dU}{dt} = Li \frac{di}{dt} + \frac{q}{C} \frac{dq}{dt} = Li \frac{d^2q}{dt^2} + \frac{q}{C} i = 0 \\\\ \Rightarrow \frac{d^2q}{dt^2} + \frac{1}{LC} q = 0 \end{gathered} \]

So \(\(\omega = 2\pi f = \sqrt{\frac{1}{LC}}\)\)