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 $B<em>1$: $B</em>1 = \frac{\mu J a^2}{2 \pi r} \hat{\phi}$
  where $r = \sqrt{(b-y')^2 + h^2}$, and $y'$ is a point on the y-axis.
• $B<em>1 = \oint B</em>1 \cdot dl = \iint \mu J \cdot dS$

Calculating Magnetic Flux Density B2

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

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

• $B = B<em>1 + B</em>2 = \frac{\mu J a^2}{\pi (b^2 + h^2)} b \hat{y}$

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: $\mu<em>r \mu</em>o$
• 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) $\neq 0$
• $n \times (H<em>2 - H</em>1) = J_s$

2012 Exam Q5

• Given: $\mu<em>{r1} = 2$, $\mu</em>{r2} = 4$ 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:
  $\frac{B<em>{x1}}{B</em>{x2}} = \frac{\mu<em>1}{\mu</em>2}$, $\frac{H<em>{x1}}{H</em>{x2}} = \frac{\mu<em>{r1}}{\mu</em>{r2}} = \frac{1}{2}$
  $H_2 = 2\hat{x} + 4\hat{y} + 4\hat{z} \text{ A/m}$

Magnetic Energy

• Magnetic Energy inside a solenoid:
  $W<em>m = \int</em>V \frac{1}{2} B \cdot H dV$
  $W<em>m = \int</em>V \frac{1}{2} \mu H \cdot H dV = \int_V \frac{1}{2} \mu H^2 dV$

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} \int<em>0^r r' J(r') dr' \hat{\phi}$$H(r) = \frac{J</em>o r^2}{3} \hat{\phi}$

Region II: r > a

• Using Ampère's Law:
  $H = \frac{1}{r} \int<em>0^a r' J(r') dr' \hat{\phi}$$H(r) = \frac{J</em>o a^3}{3r} \hat{\phi}$

Magnetic Energy Stored Per Unit Length

• For r < a:
  $B = \mu H = \mu \frac{J<em>o r^2}{3} \hat{\phi}$$\frac{W}{h} = \int</em>0^a \int<em>0^{2\pi} \frac{1}{2} \mu \frac{J</em>o^2 r^4}{9} r dr d\phi$
  $\frac{W}{h} = \frac{\pi \mu J_o^2 a^6}{27} \text{ J/m}$