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}