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}.
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}