Magnetic Fields and Properties

1998 Exam

• A power line consists of two cables, each with radius a, parallel to the z-axis.
• One cable's axis passes through (x, y, z) = (b, 0, 0) and carries a uniform current density J_1 = J \hat{z} A/m².
• The other cable's axis passes through (x, y, z) = (-b, 0, 0) and carries a uniform current density J_2 = -J \hat{z} A/m².
• The task is to find the magnetic flux density B at any point on the y-axis to determine a safe height for the power line installation.

Calculating Magnetic Flux Density B1

• J_1 = J \hat{z}
• Using Ampère's Law to find B1: B1 = \frac{\mu J a^2}{2 \pi r} \hat{\phi}
  where r = \sqrt{(b-y')^2 + h^2}, and y' is a point on the y-axis.
• B1 = \oint B1 \cdot dl = \iint \mu J \cdot dS

Calculating Magnetic Flux Density B2

• J_2 = -J \hat{z}
• Using Ampère's Law to find B2: B2 = -\frac{\mu J a^2}{2 \pi r} \hat{\phi}
  where r = \sqrt{(b+y'')^2 + h^2}, and y'' is a point on the y-axis.
• B2 = \oint B2 \cdot dl = \iint \mu J \cdot dS

Resultant Magnetic Fields

• B_1 = \frac{\mu J a^2}{2 \pi \sqrt{b^2 + h^2}} \hat{\phi}
• B_2 = -\frac{\mu J a^2}{2 \pi \sqrt{b^2 + h^2}} \hat{\phi}

Expressing B_1 in Cartesian Coordinates

• \hat{\phi} = -\hat{x} \sin(\alpha) + \hat{y} \cos(\alpha), where \alpha is the angle.
• \sin(\alpha) = \frac{h}{\sqrt{b^2 + h^2}}, \cos(\alpha) = \frac{b}{\sqrt{b^2 + h^2}}
• B_1 = \frac{\mu J a^2}{2 \pi (b^2 + h^2)} (-h \hat{x} + b \hat{y})

Expressing B_2 in Cartesian Coordinates

• \hat{\phi} = -\hat{x} \sin(\alpha) + \hat{y} \cos(\alpha), where \alpha is the angle.
• \sin(\alpha) = \frac{h}{\sqrt{b^2 + h^2}}, \cos(\alpha) = \frac{-b}{\sqrt{b^2 + h^2}}
• B_2 = \frac{\mu J a^2}{2 \pi (b^2 + h^2)} (h \hat{x} + b \hat{y})

Total Magnetic Flux Density

• B = B1 + B2 = \frac{\mu J a^2}{\pi (b^2 + h^2)} b \hat{y}

2017 Test 2 Q2

• An infinitely long parallel wire transmission line in free space carries a current I.
• Objective: Determine the magnetic flux density B at point P due to currents I and -I.

a) Magnetic Flux Density due to Current I

• Using Ampère's Law:
  B = \frac{\mu I}{2 \pi r} \hat{\phi}
  where r is the distance from the wire to point P.

b) Magnetic Flux Density due to Current -I

• B' = -\frac{\mu I}{2 \pi r'} \hat{\phi'}
• where r' = \sqrt{(b-x')^2 + h^2}

c) Total Magnetic Flux Density at P

• B = \frac{\mu I}{2 \pi} \frac{1}{b^2 + h^2} (b \hat{y} + h \hat{x})
• B(x, y) = \frac{\mu I}{ \pi} \frac{b}{(b^2 + h^2)} \hat{y}

Magnetic Vector Potential A

• Electrostatics: E = -\nabla V
• Magnetostatics: B = \nabla \times A (Wb/m²)
• \nabla^2 A = -\mu J
• A = \frac{\mu}{4 \pi} \int_V' \frac{J}{R'} dv' (Wb/m)

Magnetic Properties of Materials

• Magnetic behavior is governed by the interaction of atomic magnetic dipole moments with an external field.
• Classification: Diamagnetic, Paramagnetic, or Ferromagnetic.
• Magnetization vector: \mur \muo
• External field: B = \mu H
• Magnetic susceptibility: \chi_m

Magnetic Hysteresis

• Unmagnetized domains become magnetized domains when subjected to an external magnetic field
• Typical hysteresis curve for a ferromagnetic material.

Boundary Conditions

• Impressed source/Perfect conductors (Induced source) \neq 0
• n \times (H2 - H1) = J_s

2012 Exam Q5

• Given: \mu{r1} = 2, \mu{r2} = 4 and H_1 = 4\hat{x} + 2\hat{y} + 6\hat{z}
• Surface current density: J_s = 2\hat{y} + 2\hat{z}
• Normal vector: \hat{n} = -\hat{x}
• Boundary Conditions:
  \frac{B{x1}}{B{x2}} = \frac{\mu1}{\mu2}, \frac{H{x1}}{H{x2}} = \frac{\mu{r1}}{\mu{r2}} = \frac{1}{2}
  H_2 = 2\hat{x} + 4\hat{y} + 4\hat{z} \text{ A/m}

Magnetic Energy

• Magnetic Energy inside a solenoid:
  Wm = \intV \frac{1}{2} B \cdot H dV
  Wm = \intV \frac{1}{2} \mu H \cdot H dV = \int_V \frac{1}{2} \mu H^2 dV

2012 Exam Q6

• Infinitely long, solid nickel wire with radius r = a conducts current with volume current density J = r J_o \hat{z} A/m².
• Objective: Find the magnetic field intensity for region I (r < a) and region II (r > a).

Region I: r < a

• Using Ampère's Law:
  H = \frac{1}{r} \int0^r r' J(r') dr' \hat{\phi} H(r) = \frac{Jo r^2}{3} \hat{\phi}

Region II: r > a

• Using Ampère's Law:
  H = \frac{1}{r} \int0^a r' J(r') dr' \hat{\phi} H(r) = \frac{Jo a^3}{3r} \hat{\phi}

Magnetic Energy Stored Per Unit Length

• For r < a:
  B = \mu H = \mu \frac{Jo r^2}{3} \hat{\phi} \frac{W}{h} = \int0^a \int0^{2\pi} \frac{1}{2} \mu \frac{Jo^2 r^4}{9} r dr d\phi
  \frac{W}{h} = \frac{\pi \mu J_o^2 a^6}{27} \text{ J/m}