Chapter 5.7 and 5.8

Inductance

Magnetic FIeld in a Solenoid

Total field B at P is obtained by integrating the contributions form the entire length of the solenoid

Cross section, where $L = \mu \frac{N^2}{l}S$

$$ B = \hat z \mu n I = \frac{\hat z \mu N I}{l} \newline \text{(long solenoid with l/a >> 1)} $$

Self-Inductance of a Solenoid

• Magnetic flux $\Phi$ linking a surface $S$:

  $$ \Phi = \int _s B \cdot ds \text{ (Wb)} $$

• Self-inductance of any conducting structure is the ratio of the magnetic flux linkage $\Lambda$ to the current $I$ flowing through the structure:

  $$ L = \frac{\Lambda}{I} \text{ (H)} $$

• For a solenoid:

  $$ L = \mu \frac{N^2}{l}S $$

Self Inductance of Other Conductors

• For the parallel-wire transmission line, ac currents flow on the outer surfaces of the wires
• For the coaxial line, the current flows on the outer surface of the inner conductor and on the inner surface of the outer one
  • Current-carrying surfaces are those adjacent to the electric and magnetic fields present in the region between the conductors

For a two-conductor configuration similar to this:

$$ L = \frac{\Lambda}{I} = \frac{\Phi}{I} = \frac{1}{I} \int _S B \cdot ds $$

Inductance of Coaxial Transmission Line

$$ L' = \frac{L}{l} = \frac{\Phi}{lI} = \frac{\mu}{2\pi} \ln (\frac{b}{a}) $$

Mutual Inductance

Magnetic coupling between two different conducting structures

$$ L_{12} = \frac{\Lambda_{12}}{I_1} = \frac{N_2}{I_1} \int _{S_2}B_1 \cdot ds \text{ (H)} $$

Magnetic Energy

Magnetic Energy Density

$$ w_m = \frac{W_m}{\nu} = \frac{1}{2} \mu H^2 \text{ (J/}m^3) $$