12.4 FORCES AND TORQUES IN MAGNETIC-FIELD SYSTEMS 535|f|=φ^2
2dR(x)
dx=1
2(
Ni
R) 2
dR
dx=i^2
2N^2
R^2dR
dx=1. 52
2100020. 6285 × 10 −^8
x^2
Forx= 1 .5mm= 1. 5 × 10 −^3 m,|f|=1. 52
2100020. 6285 × 10 −^8
( 1. 5 × 10 −^3 )^2= 3142 .5NThe student should recognize the practical importance of force-generating capabilities
of electromechanical transducers.EXAMPLE 12.4.3
Solenoids find application in a variety of electrically controlled valves. The magnetic structure
shown in Figure E12.4.3 is a simplified representation of a solenoid in which the flux in the air
gap activates the motion of the iron plunger.
(a) Develop a general expression for the force exerted on the iron plunger and comment on
its dependence on positionx.
(b) Determine the current through the coil ofN=100 turns to pull the plunger tox=a,
given thata=1 cm,lg=1 mm, and the spring constant isk=1 N/m. Assume the
permeability of the magnetic structure to be infinite and neglect fringing.Spring 1)Nonmagnetic
Movable iron bushing material
plungerN turnsi2 aaa ax x = 0lgaFixed magnetic
core structure
(μ ∞)(μrFigure E12.4.3Simplified representation of a
solenoid.