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{12} = \int{S2} B_1 \cdot dS
\Phi{11} = \int{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{12} = \frac{\int B1 \cdot dS}{I_1}
Self Inductance: L{11} = \frac{\int B1 \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: Wm = \intV \frac{\mu H^2}{2} dV = \int_V \frac{\mu I^2}{8 \pi^2 r^2} dV
Calculating the integral: Wm = \frac{\mu I^2}{8 \pi^2} \inta^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{12} and L{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{12} = \int \int B1 \cdot dS = \int0^1 \intD^{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{12} = \int \int B1 \cdot dS = -\int0^1 \intD^{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 r1) carrying uniform volume current density J{vz} and outer conductor (radius r2) carrying uniform surface current density J{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}
Js = - \frac{I0}{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 P1(x, 0) and P2(x+d, 0).
Parts:
- a) Use Ampère's Law to determine the total magnetic field intensity \vec{H}(P1) at point P1(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 P1(x, 0) due to current I and surface current density Js.

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 = \intS \vec{J} \cdot d\vec{a} = \int0^a \int0^{2 \pi} r J0 r dr d\phi = \frac{2 \pi J_0 a^3}{3}
L{21} = \frac{\Phi{21}}{I} = \frac{B0 h w \ln(b/a)}{(2 \pi J0 a^3)/3}