0195136047.pdf

12.4 FORCES AND TORQUES IN MAGNETIC-FIELD SYSTEMS 537

EXAMPLE 12.4.4

A relay is essentially an electromechanical switch that opens and closes electrical contacts. A
simplified relay is represented in Figure E12.4.4. It is required to keep the fenomagnetic plate at a
distance of 0.25 cm from the electromagnet excited by a coil ofN=5000 turns, when the torque
is 10 N·m at a radiusr=10 cm. Estimate the current required, assuming infinitely permeable
magnetic material and negligible fringing as well as leakage.

(a) Express the stored magnetic energyWmas a function of the flux linkageλand the position

x.

(b) Express the coenergyWm′ as a function of the currentiand the positionx.

x = lg

i

Radius r

Cross–sectional area

Ag = 1 cm^2

Electromagnet

Ferromagnetic plate

N turns

Figure E12.4.4Simplified representation of a relay.

Solution

For the stated torque at a radius of 10 cm, the force on the plate is

f=

10

0. 1

=100 N

The magnitude of the force developed by the electromagnet must balance this force. The reluctance
due to the two air gaps is given by

Q=

2 x

μ 0 Ag

=

2 x

4 π× 10 −^7 × 1 × 10 −^4

=

x

2 π× 10 −^11

The inductance is then given by

L=

N^2

R

=

50002 × 2 π× 10 −^11

x

=

5 π× 10 −^4

x

=

1. 57 × 10 −^3
   x
   In a magnetically linear circuit, the stored magnetic energyWmis equal to the coenergyWm′,

Wm=Wm′=

1

2

Li^2 =

1

2

λ^2

L

(a)Wm(λ, x)=

1

2

λ^2

x

5 π× 10 −^4

=

λ^2 x

π× 10 −^3

Sincef=−∂Wm(λ, x)/∂x, the magnitude of the developed force is given by

|f|=

λ^2

π× 10 −^3

=

(Li)^2

π× 10 −^3

=

( 5 π× 10 −^4 )^2 i^2

π× 10 −^3 ×x^2

= 2. 5 π× 10 −^4

i^2

x^2