Inductance and Magnetic Fields

Two Wires Close Together

• Mutual Coupling: Magnetic flux fields of wire #1 couple with wire #2.
• Self Coupling: Magnetic flux fields of wire #1 couple with itself.
• The magnetic flux $\Phi$ is calculated by integrating the magnetic field B over the surface S:
  $\Phi = \int_{S} B \cdot dS$
• $\Phi<em>{12} = \int</em>{S2} B_1 \cdot dS$
• $\Phi<em>{11} = \int</em>{S1} B_1 \cdot dS$

Inductance

• Mutual Inductance: Deals with the flux linkage between two circuits.
• Self Inductance: Deals with the flux linkage within a single circuit.
• Flux Linkage is a key component in defining inductance.
• Mutual Inductance: $L<em>{12} = \frac{\int B</em>1 \cdot dS}{I_1}$
• Self Inductance: $L<em>{11} = \frac{\int B</em>1 \cdot dS}{I_1}$

Inductance Visuals

• Mutual Inductance: Demonstrated with a coil.
• Self Inductance: Shown with parallel-wire and coaxial transmission lines.
• Galvanometer: A predecessor to the ammeter, which deflects when magnetic flux changes through a loop.

Example 1

• Magnetic field intensity $\vec{H}$ at different regions:
  • Region 1 (AP#1): r < a, $\vec{H} = \frac{I}{2\pi r} \hat{\phi}$
  • Region 2 (AP#2): a < r < b, $\vec{H} = \frac{I}{2\pi r} \hat{\phi}$
  • Region 3 (AP#3): r > b, $\vec{H} = 0$
• Current density $J$ exists between $a$ and $b$.

Inductance of Coaxial Cable

• The magnetic field in the region $S$ between the two conductors is approximated.
• Total magnetic flux through $S$ is calculated.
• Inductance per unit length is determined.

Magnetic Energy

• The magnetic field in the insulating material of a coaxial cable: $B = \frac{\mu I}{2 \pi r}$
• Magnetic energy stored in the coaxial cable: $W<em>m = \int</em>V \frac{\mu H^2}{2} dV = \int_V \frac{\mu I^2}{8 \pi^2 r^2} dV$
• Calculating the integral: $W<em>m = \frac{\mu I^2}{8 \pi^2} \int</em>a^b \frac{2 \pi r l}{r^2} dr = \frac{\mu I^2 l}{4 \pi} \ln(\frac{b}{a})$
• Relating magnetic energy to inductance: $W_m = \frac{1}{2} L I^2$

2011 Exam Q6

• Problem: Determine the mutual inductance between two pairs of infinitely long transmission lines separated by distance $D$.
• Part a) & b) asks to determine the mutual inductance $L<em>{12}$ and $L</em>{21}$ per meter length.

2011 Exam Q6 - Solution

• Magnetic field $B$ due to wire #1: $B = \frac{\mu I}{2 \pi r} \hat{\phi}$
• Differential surface area element: $dS = drdz \hat{s} = drdz \hat{\phi}$
• Magnetic flux through the surface: $\Phi = \int \int B \cdot dS$
• $\Phi<em>{12} = \int \int B</em>1 \cdot dS = \int<em>0^1 \int</em>D^{D+d} \frac{\mu I}{2 \pi r} drdz = \frac{\mu I}{2 \pi} \ln(\frac{D+d}{D})$

2011 Exam Q6 - Continued

• Magnetic field due to wire #1 (with negative current): $B_1' = - \frac{\mu I}{2 \pi r'} \hat{\phi'}$
• Differential surface area element: $dS = dr'dz \hat{\phi'}$
• Magnetic flux calculation: $\Phi = \int \int B \cdot dS$
• $\Phi<em>{12} = \int \int B</em>1 \cdot dS = -\int<em>0^1 \int</em>D^{D+d} \frac{\mu I}{2 \pi r'} dr'dz = -\frac{\mu I}{2 \pi} \ln(\frac{D+d}{D})$

2011 Exam Q6 - Final Answer

• Calculating the total flux and mutual inductance.
• $\Phi_{12} = \frac{\mu I}{2 \pi} \ln(\frac{D+d}{D}) - \frac{\mu I}{2 \pi} \ln(\frac{D+d}{D})$
• $L = \frac{\Phi}{I} = \frac{\mu}{2 \pi} \ln(\frac{D+d}{D})$

2018 Test 2 Q2

• Problem: Co-axial transmission line with inner conductor (radius $r<em>1$) carrying uniform volume current density $J</em>{vz}$ and outer conductor (radius $r<em>2$) carrying uniform surface current density $J</em>{sz}$.
• Parts:
  • a) Use Ampère's Law to determine the magnetic field intensity $\vec{H}$ at point $P_1(x, y)$.
  • b) Determine the magnetic flux linkage $\Lambda_{12}$ (per unit length) between the two transmission lines.
  • c) Given total current $I_n$ on the inner conductor and $\vec{H}(x, y) = 0$ at point $P(x, y)$, determine the surface current density on the outer conductor.

2018 Test 2 Q2a - Solution

• Using Ampère's Law to find the magnetic field intensity $\vec{H}$.

2018 Test 2 Q2b - Solution

• Calculating magnetic flux linkage $\Lambda_{12}$ by integrating the magnetic flux density over the area between the transmission lines.
• Considering magnetic field $H$ at point $P(x,y)$.

2018 Test 2 Q2b - Detailed Solution

• $\Phi = \int \int B \cdot dS = \int \int \mu H \cdot dS$
• Breaking down the integral and solving for the flux linkage $\Lambda_{12}$.
• $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan(\frac{x}{a})$

2018 Test 2 Q2c - Solution

• Using Ampère's Law to determine the surface current density $J_s$ on the outer conductor when the magnetic field $\vec{H}(x, y) = 0$.
• $I = \int \vec{J} \cdot d\vec{S}$
• $J<em>s = - \frac{I</em>0}{2 \pi r_2}$

2022 Test 2 Q2

• Problem: Infinitely long co-axial structure with offset inner current distribution (radius $a$, current $I$ in the $\hat{z}$ direction) and outer current distribution (radius $b$, uniform surface current density $-J_s$).
• Second infinitely long parallel wire transmission line (separation $d$) coincides with $P<em>1(x, 0)$ and $P</em>2(x+d, 0)$.
• Parts:
  • a) Use Ampère's Law to determine the total magnetic field intensity $\vec{H}(P<em>1)$ at point $P</em>1(x, 0)$.
  • b) Determine the magnetic flux linkage $\Lambda_{12}$ (per unit length) between the two transmission lines.

2022 Test 2 Q2 - Setup

• Visual representation of the co-axial structure and the placement of the transmission lines.
• Applying Ampère's Law to calculate magnetic field intensity.
• $\oint H \cdot dl = I_{encl}$
• $H = \frac{I}{2 \pi r} \hat{\phi}$

2022 Test 2 Q2 - Continued

• Calculating magnetic field intensity $\vec{H}$ at point $P<em>1(x, 0)$ due to current $I$ and surface current density $J</em>s$.

2022 Test 2 Q2 - Solution

• Finding magnetic flux linkage $\Lambda_{12}$ between the two transmission lines.
• Considering magnetic field $\vec{B}$ and integrating over the surface area.
• $\Lambda_{12} = \int \int B \cdot dS = \int \int \mu H \cdot dS$

2015 Test 2 Q2

• Problem: Two pairs of infinitely long transmission lines separated by distance $D$.
• Task: Determine the distance $D$ to achieve a mutual inductance of $L_{12} = 2 \times 10^{-8}$ H per meter length.

2015 Test 2 Q2 - Setup

• Visual representation of the two pairs of transmission lines.
• Using the formula for magnetic field due to a long wire: $B = \frac{\mu I}{2 \pi r} \hat{\phi}$
• Differential surface area element: $dS = dydz \hat{x}$

2015 Test 2 Q2 - Calculations

• Calculating magnetic flux $\Phi$ by integrating the magnetic field over the surface.
• Considering the geometry and using appropriate trigonometric relations.
• $\Phi = \int \int B \cdot dS$

2015 Test 2 Q2 - Continued

• Solving for the distance $D$ using the calculated flux and the given mutual inductance.

2012 Exam Q6

• Problem: Infinitely long, solid nickel wire with radius $r = a$, centered on the z-axis, conducts current with a volume current density $J = r J_0 \hat{z}$ A/m².
• Given: $J = r J_0 \hat{z}$ for r < a
• Magnetic field $\vec{H} = 0$ for r > a

2012 Exam Q6b

• A rectangular loop (width $w$, height $h$) is placed in the xz-plane close to the nickel wire.
• Given magnetic flux density for r > a: $B = \frac{B_0}{r} \hat{\phi}$ T.
• Task: Determine the mutual inductance between the nickel wire and the rectangular loop.

2012 Exam Q6b - Solution

• Calculate the magnetic flux $\Phi_{21}$ through the rectangular loop due to the current in the nickel wire, using the given magnetic field $B$.
• $\Phi_{21} = \int \int B \cdot dS$

2012 Exam Q6b - Continued

• $I = \int<em>S \vec{J} \cdot d\vec{a} = \int</em>0^a \int<em>0^{2 \pi} r J</em>0 r dr d\phi = \frac{2 \pi J_0 a^3}{3}$
• $L<em>{21} = \frac{\Phi</em>{21}}{I} = \frac{B<em>0 h w \ln(b/a)}{(2 \pi J</em>0 a^3)/3}$