Electromagnetism Concepts

Electric Fields

• Electric field at point P due to a single charge.
• Electric force on a test charge placed at P.

Magnetic Force on a Current Element

• dF_m = I dl \times B
• F_{in} = I \oint dl \times B

Magnetic Fields

• Two wires carrying current experience a force.
• Wires are intrinsically charge neutral, so the force cannot be due to the electric field.
• Magnetic field is related to the currents I1 and I2, and the forces F1 and F2.

Magnetic Torque on Current Loop

• Illustrations of a current loop in a magnetic field, showing the orientation of the loop and the magnetic field lines.
• AC Motor: Electrical to mechanical energy conversion.
• AC Generator: Mechanical to electrical energy conversion.

Electric vs Magnetic Comparison

Table 5-1: Attributes of electrostatics and magnetostatics.

Electric vs Magnetic Comparison

• Electric energy density: We = \frac{1}{2} \intV D \cdot E dV

• Magnetic energy density: Wm = \frac{1}{2} \intV B \cdot H dV

Magnetic Field due to Current Densities

• Illustrations of a volume dV and surface dS current density contributions to the magnetic field at a field point P(x, y, z).

Biot-Savart Law

• Assuming current flows in a thin wire with a constant cross-section S.
• Relationship between current density J and current I: J = \frac{I}{S} dl = J S dV

Biot-Savart Law

• Source coordinates: (x', y', z')
• Field point: P(x, y, z)
• Distance vector: R = R_p = R - R'
• Unit vector: \hat{R} = \frac{R}{|R|}

Example 5-2: Magnetic Field of Linear Conductor

• Differential length vector: dl = \hat{z} dz
• dl \times \hat{R} = dz (\hat{z} \times \hat{R}) = sin \theta dz
• Magnetic field intensity: H = \frac{I}{4\pi} \int{-l/2}^{l/2} \frac{dl \times \hat{R}}{R^2} = \frac{I}{4\pi} \int{-l/2}^{l/2} \frac{sin \theta}{R^2} dz
• Transformations:
  • R = r csc \theta
  • z = r cot \theta
  • dz = r csc^2 \theta d\theta

Example 5-3: Magnetic Field of a Loop

• dH is in the r-z plane, with components dHr and dHz
• z-components of the magnetic fields due to dl and dl' add because they are in the same direction, but their r-components cancel.
• Magnitude of field due to dl is discussed.

Applying Gauss’s Law

• Electric field due to an infinite line charge density \rho_l
• Gaussian surface: Cylinder
• Electric field: E = \frac{\rho_l}{2 \pi \epsilon r} \hat{r}

Ampère’s Law

• Integral form: \ointC H \cdot dl = \iintS J \cdot dS = I_{encl}

Magnetic Field of Long Conductor

• Magnetic field around an infinitely long wire: B = \frac{\mu_0 I}{2 \pi r} \hat{\phi}

Ampère’s Law

• Assuming I is in \hat{z} direction: H = H(r) \hat{\phi}
• \ointC H \cdot dl = \iintS J \cdot dS

Ampère’s Law

• With I in \hat{z}: H = H(r) \hat{\phi}
• \ointC H \cdot dl = H(r) \intC d l = H(r) 2 \pi r \hat{\phi}
• \iint_S J \cdot dS = I

Ampère’s Law

• I = \iint_S J \cdot dS
• dS = r dr d\phi \hat{z}

Ampère’s Law

• \ointC H \cdot dl = \iintS J \cdot dS

Internal Magnetic Field of Long Conductor: Example 5.4

• For r < a:
  • \oint{C1} H1 \cdot dl1 = I_1
  • H1 2 \pi r1 = I_1
  • I1 = I \frac{\pi r1^2}{\pi a^2}
  • H1 = \frac{I r1}{2 \pi a^2} \hat{\phi}

External Magnetic Field of Long Conductor: Example 5.4

• For r > a:
  • \oint{C2} H2 \cdot dl2 = I
  • H2 2 \pi r2 = I
  • H2 = \frac{I}{2 \pi r2} \hat{\phi}

Ampère’s Law

• Coaxial cable with current I.
• Current densities: J = I \hat{z}, -I

Ampère’s Law

• Current density: J = \frac{I}{\pi a^2} \hat{z}
• Magnetic field: H = H(r) \hat{\phi}
• Regions:
  • AP#1: r < a
  • AP#2: a < r < b
  • AP#3: r > b

Ampère’s Law (Region 1: r < a)

• \iintS J \cdot dS = \frac{I}{\pi a^2} \iintS dS = \frac{I}{\pi a^2} \int0^{2\pi} \int0^r r dr d\phi = \frac{I}{\pi a^2} \pi r^2 = I \frac{r^2}{a^2}
• \ointC H \cdot dl = H(r) \int0^{2\pi} r d\phi = H(r) 2 \pi r
• H(r) = \frac{I r}{2 \pi a^2} \hat{\phi}

Ampère’s Law (Region 2: a < r < b)

• \iintS J \cdot dS = \iintS J \cdot dS = I
• \ointC H \cdot dl = H(r) \int0^{2\pi} r d\phi = H(r) 2 \pi r
• H(r) = \frac{I}{2 \pi r} \hat{\phi}

Ampère’s Law (Region 3: r > b)

• \iint_S J \cdot dS = I - I = 0
• \ointC H \cdot dl = H(r) \int0^{2\pi} r d\phi = H(r) 2 \pi r
• H(r) = 0

Ampère’s Law

• H_1 = \frac{I r}{2 \pi a^2} \hat{\phi}
• H_2 = \frac{I}{2 \pi r} \hat{\phi}
• H_3 = 0

2018 Test 2 Q2

• Infinitely long co-axial transmission line in free space.
• Inner conductor: radius r = r1, uniform volume current density J1 \hat{z}
• Outer conductor: radius r = r2, uniform surface current density J2 \hat{z}
• Use Ampère's Law to determine the magnetic field intensity H in Cartesian coordinates.

2018 Test 2 Q2

• Magnetic field: H = H(r) \hat{\phi}
• \oint H \cdot dl = \iint Jv \cdot dS + \int Js \cdot dl

2018 Test 2 Q2

• \oint H \cdot dl = H(r) 2 \pi r
• \iint Jv \cdot dS = Jv \iint dS = J_v \pi r^2
• \int Js \cdot dl = Js \int dl = Js 2 \pi r2
• H(r) 2 \pi r = Jv \pi r^2 + Js 2 \pi r_2
• H(r) = \frac{Jv r}{2} + \frac{Js r_2}{r}

2018 Test 2 Q2

• Converting to Cartesian coordinates:
  • H = H(r) (-\hat{x} sin \phi + \hat{y} cos \phi)
  • sin \phi = \frac{y}{\sqrt{x^2 + y^2}}
  • cos \phi = \frac{x}{\sqrt{x^2 + y^2}}
• H = (-\hat{x} \frac{y}{\sqrt{x^2 + y^2}} + \hat{y} \frac{x}{\sqrt{x^2 + y^2}}) (\frac{Jv r}{2} + \frac{Js r_2}{r})
• H = (-\hat{x} \frac{y}{\sqrt{x^2 + y^2}} + \hat{y} \frac{x}{\sqrt{x^2 + y^2}}) (\frac{Jv \sqrt{x^2 + y^2}}{2} + \frac{Js r_2}{\sqrt{x^2 + y^2}})