0195136047.pdf

536 ELECTROMECHANICS

Solution

(a) The force in the air gap can be expressed by

f=−

1

2

φ^2

dR

dx

(See Example 12.4.2.) The gap reluctance, which in this case is due to the nonmagnetic

bushing, is given by

R= 2 Rg=

2 lg

μ 0 ax

where the areaaxis variable, depending on the position of the plunger, andlgis the

thickness of the bushing on either side of the plunger. The air-gap magnetic flux is given

by

φ=

F

R

=

Ni

R

=

μ 0 Niax

2 lg

Now,

dR

dx

=

d

dx

(

2 lg

μ 0 ax

)

=

− 2 lg

μ 0 ax^2

The force is then given by

f=−

1

2

φ^2

dR

dx

=−

1

2

(

μ 0 Niax

2 lg

) 2 (

− 2 lg

μ 0 ax^2

)

or

f=

1

4

μ 0 a

lg

(N i)^2

which is independent of the positionx, and is a constant for a given exciting mmf and

geometry.

(b) Forx=a=1cm= 0 .01 m,

f=kx= 1 × 0. 01 =

1

4

4 π× 10 −^7 × 0. 01

0. 001

( 100 i)^2

or

i^2 =

4 × 0. 001 × 0. 01

4 π× 10 −^7 × 0. 01

1

( 100 )^2

= 0. 3182

or

i= 0 .564 A

The student should recognize the practical importance of determining the approximate

mmf or current requirements for electromechanical transducers.