Motional EMF and Maxwell's Equations

Motional EMF

• Magnetic force on charge $q$ moving with velocity $u$ in a magnetic field $B$: This magnetic force is equivalent to the electrical force that would be exerted on the particle by the electric field $E_m$ given by
• This, in turn, induces a voltage difference between ends 1 and 2, with end 2 being at the higher potential. The induced voltage is called a motional emf.

General Formula

• If any segment of a closed circuit with contour $C$ moves with a velocity $u$ across a static magnetic field $B$, then the induced motional emf is given by
  $V<em>{emf} = \oint</em>C (u \times B) \cdot dl$
• Only those segments of the circuit that cross magnetic field lines contribute to $V_{emf}$.

Example 6-3: Sliding Bar

• The length of the loop is related to $u$ by $x_0 = ut$.
• Note that $B$ increases with $x$

Example 6-5: Moving Rod Next to a Wire

• The wire carries a current $I = 10 A$. A 30-cm-long metal rod moves with a constant velocity $u = 25 m/s$.
• Find $V_{12}$.
• $B = \frac{\mu_0 I}{2 \pi r} \hat{z}$
• $V<em>{12} = \int</em>{10 cm}^{40 cm} (u \times B) \cdot dl = \int<em>{10 cm}^{40 cm} (25 \hat{x} \times \frac{\mu</em>0 I}{2 \pi r} \hat{z}) \cdot \hat{x} dr$
• $V<em>{12} = \int</em>{10 cm}^{40 cm} (25 \frac{\mu<em>0 I}{2 \pi r}) dr = \frac{25 \mu</em>0 I}{2 \pi} \int_{0.1}^{0.4} \frac{1}{r} dr$
• $V_{12} = \frac{25 \times 4 \pi \times 10^{-7} \times 10}{2 \pi} \ln(\frac{0.4}{0.1}) = 5 \times 4 \pi \times 10^{-7} \times 10 \times ln(4) = 13.9 \mu V$

EM Motor/Generator Reciprocity

• Motor: Electrical to mechanical energy conversion
• Generator: Mechanical to electrical energy conversion

EM Generator EMF

• As the loop rotates with an angular velocity $\omega$ about its own axis, segment 1–2 moves with velocity $u$.
• Segment 3-4 moves with velocity $-u$.

Displacement Current

• This term is conduction current $I_C$.
• This term must represent a current.
• Application of Stokes’s theorem gives:

Define Displacement Current

• The displacement current does not involve real charges; it is an equivalent current that depends on the rate of change of the electric field.

Capacitor Circuit

• Given: Wires are perfect conductors and capacitor insulator material is perfect dielectric.
• For Surface S1: $I<em>1 = I</em>{1c} + I_{1d}$. ($D = 0$ in perfect conductor)

Capacitor Circuit

• For Surface S2: $I<em>2 = I</em>{2c} + I_{2d}$
• $I_{2c} = 0$ (perfect dielectric)
• Conclusion: $I<em>1 = I</em>2$

Example 6-7: Displacement Current Density

• Given: Conductivity $"sigma = 2 \times 10^7 S/m$ and relative permittivity $\epsilon_r = 1$.
• Conduction current: $I_c = 2 \sin(\omega t) (mA)$, where $\omega = 10^9 rad/s$.
• Find the displacement current.

Solution

• $I_c = JA = \sigma EA$, where A is the cross-section of the wire.

• $E = \frac{I_c}{\sigma A} = \frac{2 \times 10^{-3} \sin(\omega t)}{2 \times 10^7 A} = \frac{10^{-10} \sin(\omega t)}{A} (V/m)$.

• $D = \epsilon E$, so $I_d = \frac{\partial D}{\partial t} A = \epsilon A \frac{\partial E}{\partial t}$.

• $I_d = \epsilon A \frac{\partial}{\partial t} (\frac{10^{-10}}{A} \sin(\omega t)) = \epsilon \omega \times 10^{-10} \cos(\omega t) = 8.85 \times 10^{-12} \times 10^9 \times 10^{-10} \cos(\omega t) = 0.885 \times 10^{-12} \cos(\omega t) (A)$.

• $I<em>c$ and $I</em>d$ are in phase quadrature (90° phase shift between them).

• $I<em>d$ is about nine orders of magnitude smaller than $I</em>c$, so the displacement current usually is ignored in good conductors.

Boundary Conditions

• Table summarizing boundary conditions for electric and magnetic fields between different media (Dielectric, Conductor).

Tangential E

• General Form: $\hat{n}<em>2 \times (E</em>1 - E_2) = 0$
  • Dielectric-Dielectric: $E<em>{1t} = E</em>{2t}$
  • Dielectric-Conductor: $E<em>{1t} = E</em>{2t} = 0$

Normal D

• General Form: $\hat{n}<em>2 \cdot (D</em>1 - D<em>2) = \rho</em>s$
  • Dielectric-Dielectric: $D<em>{1n} - D</em>{2n} = \rho_s$
  • Dielectric-Conductor: $D<em>{1n} = \rho</em>s, D_{2n} = 0$

Tangential H

• General Form: $\hat{n}<em>2 \times (H</em>1 - H<em>2) = J</em>s$
  • Dielectric-Dielectric: $H<em>{1t} = H</em>{2t}$
  • Dielectric-Conductor: $H<em>{1t} = J</em>s, H_{2t} = 0$

Normal B

• General Form: $\hat{n}<em>2 \cdot (B</em>1 - B_2) = 0$

  • Dielectric-Dielectric: $B<em>{1n} = B</em>{2n}$
  • Dielectric-Conductor: $B<em>{1n} = B</em>{2n} = 0$

• Notes:

  • $\rho_s$ is the surface charge density at the boundary.
  • $J_s$ is the surface current density at the boundary.
  • Normal components are along $\hat{n}_2$, the outward unit vector of medium 2.
  • $E<em>{1t} = E</em>{2t}$ implies equal magnitude and parallel direction.
  • Direction of $J<em>s$ is orthogonal to $(H</em>1 - H_2)$.

Charge Current Continuity Equation

• Current $I$ out of a volume is equal to the rate of decrease of charge $Q$ contained in that volume:
• $\sum I = 0$

EM Potentials

Static Condition

• $V(R) = \frac{1}{4 \pi \epsilon} \int<em>{V'} \frac{\rho</em>v(R')}{R'} dv'$

Dynamic Condition

• $V(R, t) = \frac{1}{4 \pi \epsilon} \int<em>{V'} \frac{\rho</em>v(R', t)}{R'} dv'$

Dynamic condition with propagation delay

• $V(R,t) = \frac{1}{4 \pi \epsilon} \int<em>{V'} \frac{\rho</em>v(R<em>i, t - R'/u</em>p)}{R'} dv'$
• $A(R,t) = \frac{\mu}{4 \pi} \int<em>{V'} \frac{J(R</em>i, t - R'/u_p)}{R'} dv'$

Time Harmonic Potentials

• If charges and currents vary sinusoidally with time:
• we can use phasor notation: e.g., $\frac{\partial}{\partial t} \Rightarrow j \omega$
• Expressions for potentials become:

Maxwell’s Equations Time Harmonic

• Time harmonic implies $\frac{\partial}{\partial t} \Rightarrow j \omega$

Maxwell's Equations Time Harmonic

• $\nabla \times \tilde{E} = -j \omega \mu \tilde{H}$
• $\tilde{H} = -\frac{1}{j \omega \mu} \nabla \times \tilde{E}$
• $\nabla \times \tilde{H} = j \omega \epsilon \tilde{E}$
• $\tilde{E} = \frac{1}{j \omega \epsilon} \nabla \times \tilde{H}$

Example 6-8: Relating E to H

• Given a nonconducting medium with $\epsilon = 16 \epsilon<em>0$ and $\mu = \mu</em>0$, the electric field intensity of an electromagnetic wave is
• $E(z, t) = \hat{x} 10 \sin(10^{10} t - kz) (V/m)$
• Determine the associated magnetic field intensity H and find the value of k.

Solution:

• Rewrite $E(z, t)$ as
• $E(z, t) = \hat{x} 10 \cos(10^{10} t - kz - \pi/2) = Re[\tilde{E}(z) e^{j \omega t}]$, with $\omega = 10^{10} rad/s$
• $\tilde{E}(z) = \hat{x} 10 e^{-jkz} e^{-j\pi/2} = -\hat{y} j 10 e^{-jkz}$

Example 6-8: Relating E to H (Continued)

• To find both $\tilde{H}(z)$ and $k$, use Faraday's law to find $\tilde{H}(z)$; then Ampère's law to find $\tilde{E}(z)$, which we will then compare with the original expression for $\tilde{E}(z)$ to find $k$.
• Applying Faraday's Law:
• $\tilde{H}(z) = \frac{1}{j \omega \mu} \nabla \times \tilde{E} = \frac{1}{j \omega \mu} \begin{vmatrix} \hat{x} &amp; \hat{y} &amp; \hat{z} \ \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \ 0 &amp; -j 10 e^{-jkz} &amp; 0 \end{vmatrix}$
• $\tilde{H}(z) = \frac{1}{j \omega \mu} \hat{x} \frac{\partial}{\partial z} (-j 10 e^{-jkz}) = \hat{x} \frac{10k}{\omega \mu} e^{-jkz}$

Example 6-8 (Continued)

• Use $\tilde{H}(z)$ in Ampère's law to find $\tilde{E}(z)$:
• $\tilde{E}(z) = \frac{1}{j \omega \epsilon} \nabla \times \tilde{H} = \frac{1}{j \omega \epsilon} \begin{vmatrix} \hat{x} &amp; \hat{y} &amp; \hat{z} \ \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \ \frac{10k}{\omega \mu} e^{-jkz} &amp; 0 &amp; 0 \end{vmatrix}$
• $\tilde{E}(z) = \frac{1}{j \omega \epsilon} (-\hat{y}) \frac{\partial}{\partial z} (\frac{10k}{\omega \mu} e^{-jkz}) = \frac{1}{j \omega \epsilon} (-\hat{y}) (\frac{10k}{\omega \mu} (-jk) e^{-jkz}) = -\hat{y} j \frac{10 k^2}{\omega \mu \epsilon} e^{-jkz}$
• Equating the two expressions for $\tilde{E}(z)$ leads to:
• $k^2 = \omega^2 \mu \epsilon$
• $k = \omega \sqrt{\mu \epsilon} = 4 \omega \sqrt{\mu<em>0 \epsilon</em>0} = \frac{4 \omega}{c} = \frac{4 \times 10^{10}}{3 \times 10^8} = 133 (rad/m)$

Example 6-8 (Final)

• $H(z, t) = Re[\tilde{H}(z) e^{j \omega t}] = Re[\hat{x} \frac{10k}{\omega \mu} e^{-jkz} e^{j \omega t}] = \hat{x} \frac{10k}{\omega \mu} \cos(\omega t - kz)$
• $H(z, t) = \hat{x} \frac{10 \times 133}{10^{10} \times 4 \pi \times 10^{-7}} \cos(10^{10} t - 133z) = \hat{x} 0.11 \cos(10^{10} t - 133z) (A/m)$

Ulaby 6.26

• Given the electric field radiated by a short dipole antenna in spherical coordinates:
• $E(R, \theta, t) = \hat{\theta} \frac{6 \times 10^8 \pi \sin(\theta)}{R} \cos(2 \pi \times 10^8 t - 2 \pi R)$
• Find $H(R, \theta, t)$.

Ulaby 6.26 Solution

• Radiated electric field in phasor form: $\tilde{E}(R, \theta) = \hat{\theta} \frac{6 \times 10^8 \pi \sin(\theta)}{R} e^{-j2 \pi R}$
• $\nabla \times \tilde{E} = -j \omega \mu \tilde{H}$
• $\tilde{H} = \frac{1}{-j \omega \mu} \nabla \times \tilde{E}$
• $\nabla \times \tilde{E} = \frac{1}{R} [\frac{\partial}{\partial R} (R \tilde{E_\theta})] \hat{\phi}$
• $\tilde{H} = \hat{\phi} \frac{1}{-j \omega \mu} \frac{1}{R} \frac{\partial}{\partial R} (R \tilde{E_\theta})$
• $\tilde{H} = \hat{\phi} \frac{1}{-j \omega \mu} \frac{1}{R} \frac{\partial}{\partial R} (6 \times 10^8 \pi \sin(\theta) e^{-j2 \pi R})$
• $\tilde{H} = \hat{\phi} \frac{6 \times 10^8 \pi \sin(\theta) (-j2 \pi)}{-j \omega \mu R} e^{-j2 \pi R}$
• $\tilde{H} = \hat{\phi} \frac{12 \pi^2 \times 10^8 sin(\theta)}{\omega \mu R} e^{-j2 \pi R}$
• $\tilde{H} = \hat{\phi} \frac{12 \pi^2 \times 10^8 \sin(\theta)}{2 \pi \times 10^8 \times 4 \pi \times 10^{-7} R} e^{-j2\pi R} =\frac{1.53 \sin(\theta)}{R} e^{- j2 \pi R}$

Ulaby 6.26 Solution(continued)

• $H(R,\theta,t) = Re[ \tilde{H}(R,\theta)e^{j \omega t}]$
• $H(R,\theta,t) = \hat{\phi} \frac{1.53 \sin(\theta)}{R} Re[ e^{- j2\pi R} e^{j 2\pi \times 10^8 t} ]$
• $H(R,\theta,t) = \hat{\phi} \frac{1.53 \sin(\theta)}{R} cos( 2\pi \times 10^8 t - 2\pi R )$

2015 Test 2 Question 3

• $\nabla \times \tilde{E} = - j \omega \mu \tilde{H}$
• $\nabla \times \tilde{H} = \tilde{J} + j \omega \tilde{D}$

2015 Test 2 Question 3(Continued)

• $\tilde{E}(z) = [40 \hat{x} - j 40 \hat{y}] e^{-j10 \pi z}$
• $\tilde{H} = \frac{1}{- j \omega \mu } \nabla \times \tilde{E}$
• $\nabla \times \tilde{E} = \begin{bmatrix} \hat{x} &amp; \hat{y} &amp; \hat{z} \ \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \ 40 e^{-j10 \pi z} &amp; -j40 e^{-j10 \pi z} &amp; 0 \end{bmatrix}$
• $\nabla \times \tilde{E} = \hat{x} [ \frac{\partial}{\partial z} ( - j 40 e^{- j10 \pi z}) ] - \hat{y} [ \frac{\partial}{\partial z} ( 40 e^{- j10 \pi z}) ]$
• $\nabla \times \tilde{E} = 400 \pi e^{- j10 \pi z} \hat{x} + j 400 \pi e^{-j10 \pi z} \hat{y}$

2015 Test 2 Question 3(Continued)

• $\tilde{H} = \frac{1}{- j \omega \mu} \nabla \times \tilde{E} = \frac{1}{- j \omega \mu } [ 400 \pi e^{- j10 \pi z} \hat{x} + j 400 \pi e^{- j10 \pi z} \hat{y}]$
• $\tilde{H} = \frac{400 \pi }{ \omega \mu } [ j e^{- j10 \pi z} \hat{x} + e^{- j10 \pi z} \hat{y} ]$
• $\tilde{H} = \hat{x} \frac{400 \pi }{ 8 \pi \times 10^7 } j e^{- j10 \pi z} + \hat{y} \frac{400 \pi }{ 8 \pi \times 10^7 } e^{- j10 \pi z} = \hat{x} Re[ 5 \times 10^{-6} j e^{-j10 \pi z + j \omega t} ] + \hat{y}Re[ 5 \times 10^{-6} e^{-j10 \pi z + j \omega t} ]$
• $H = \hat{x} 5 \times 10^{-6} sin( \omega t -10 \pi z) + \hat{y} 5 \times 10^{-6}cos( \omega t - 10 \pi z)$
• Given $\mu = 4 \pi \times 10^{-7}$ therefore $\omega \mu = 8 \pi \times 10^7$

2015 Test 2 Question 3(Continued)

• $\tilde{E}(z) = 40 \hat{x} e^{-j10 \pi z}$
• Assuming displacement current
• $\tilde{J} = j \omega \epsilon \tilde{E} = j 8 \pi \times 10^7 ( 10^{-10} / (36 \pi)) (40 e^{-j10 \pi z} \hat{x} )$
• $\tilde{J} = j 0.089 \hat{x} e^{-j10 \pi z}$
• $\omega \epsilon = 8 \pi \times 10^7 ( 10^{-9} / 36 \pi ) = j 0.698$
• $u_p = {\omega \over k } = 1$