W9_parallel_resonance.eps

258 Week 8: Faraday’s Law and Induction

• From this we can compute theself-induced (loop) voltages for simple current-carrying loops,
  in particular solenoids. To compute the self-inductance of a solenoidwe begin with the result
  for the magnetic field inside an ideal solenoid from Ampere’s Law:

B=

μ 0 N I

L

(549)

(parallel to the solenoid axis). The currentIcreates a fluxper turnthat is equal to:

φt=BA=

μ 0 N AI

L

(550)

whereAis the cross-sectional area of the solenoid. The total flux is thus:

φ=N BA=μ^0 N

(^2) AI
L
=LsI (551)
whereLsis the self-inductance of the solenoid. Clearly:
Ls=
μ 0 N^2 A
L

(552)

which dependsonly on the geometry of the solenoidjust as the capacitance of an arrangement

of conductors depended only ontheirgeometry.

• The self-inductance of solenoids can be altered by wrapping them around suitablemagnetic
  materialsthat enhance (para) or reduce (dia) the magnetic fields inside. Solenoids so con-
  structed are ubiquitous in circuit design, where they are known asinductors; they are labelled
  with their inductanceLinHenries, the SI unit of inductance:

1 Henry =

1 Volt−Second

Ampere

= 1 Ohm−Second (553)

• In terms of inductance:
  VL=−L

dI

dt

(554)

is a statement of the voltage across an inductor using Faraday’s Law.

• Mutual inductance is the basis of a number of devices, in particular acenter-tap full-wave
  rectifier commonly used in e.g. DC power supplies or AM radios and intransformers, an
  essential component of the power distribution grid. If one imaginestwosolenoids, one with
  N 1 turns and cross sectional areaAand a second one withN 2 turns wrappedaroundthe first
  (so all of the flux (per turn) in the first passes through the loops of the second:

φt=

μ 0 N 1 AI 1

L (555)

for the first solenoid, so:

φ 2 =N 2

μ 0 N 1 AI 1

L

(556)

is the total flux through the second solenoid due to the current in the first. Thus:

M 21 =

φ 2

I 1

=

μ 0 N 1 N 2 A

L

=M 12 =M (557)