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