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{emf} = \ointC (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{12} = \int{10 cm}^{40 cm} (u \times B) \cdot dl = \int{10 cm}^{40 cm} (25 \hat{x} \times \frac{\mu0 I}{2 \pi r} \hat{z}) \cdot \hat{x} dr
• V{12} = \int{10 cm}^{40 cm} (25 \frac{\mu0 I}{2 \pi r}) dr = \frac{25 \mu0 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: I1 = I{1c} + I_{1d}. (D = 0 in perfect conductor)

Capacitor Circuit

• For Surface S2: I2 = I{2c} + I_{2d}
• I_{2c} = 0 (perfect dielectric)
• Conclusion: I1 = I2

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).

• Ic and Id are in phase quadrature (90° phase shift between them).

• Id is about nine orders of magnitude smaller than Ic, 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}2 \times (E1 - E_2) = 0
  • Dielectric-Dielectric: E{1t} = E{2t}
  • Dielectric-Conductor: E{1t} = E{2t} = 0

Normal D

• General Form: \hat{n}2 \cdot (D1 - D2) = \rhos
  • Dielectric-Dielectric: D{1n} - D{2n} = \rho_s
  • Dielectric-Conductor: D{1n} = \rhos, D_{2n} = 0

Tangential H

• General Form: \hat{n}2 \times (H1 - H2) = Js
  • Dielectric-Dielectric: H{1t} = H{2t}
  • Dielectric-Conductor: H{1t} = Js, H_{2t} = 0

Normal B

• General Form: \hat{n}2 \cdot (B1 - B_2) = 0

  • Dielectric-Dielectric: B{1n} = B{2n}
  • Dielectric-Conductor: B{1n} = B{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{1t} = E{2t} implies equal magnitude and parallel direction.
  • Direction of Js is orthogonal to (H1 - 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{V'} \frac{\rhov(R')}{R'} dv'

Dynamic Condition

• V(R, t) = \frac{1}{4 \pi \epsilon} \int{V'} \frac{\rhov(R', t)}{R'} dv'

Dynamic condition with propagation delay

• V(R,t) = \frac{1}{4 \pi \epsilon} \int{V'} \frac{\rhov(Ri, t - R'/up)}{R'} dv'
• A(R,t) = \frac{\mu}{4 \pi} \int{V'} \frac{J(Ri, 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 \epsilon0 and \mu = \mu0, 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} & \hat{y} & \hat{z} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 0 & -j 10 e^{-jkz} & 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} & \hat{y} & \hat{z} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ \frac{10k}{\omega \mu} e^{-jkz} & 0 & 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{\mu0 \epsilon0} = \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} & \hat{y} & \hat{z} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 40 e^{-j10 \pi z} & -j40 e^{-j10 \pi z} & 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