0195136047.pdf

534 ELECTROMECHANICS

Solution

(a) A change in the energy stored in the electromagnetic fielddWmis equal to the sum of

the incremental work done by the electric circuit and the incremental work done by the

mechanical system. Thus,

dWm=ei dt−fedx=

dλ

dt

idt−fedx=idλ−fedx

or

fedx=idλ−dWm

whereeis the electromotive force across the coil and the negative sign is due to the

sign convention shown in Figure E.12.4.2. Noting that the flux in the magnetic structure

depends on two independent variables, namely, the currentithrough the coil and the

displacementxof the bar, one can rewrite the equation,

fedx=i

(

∂λ

∂i

di+

∂λ

∂x

dx

)

−

(

∂Wm

∂i

di+

∂Wm

∂x

dx

)

whereWmis a function ofiandx. Sinceiandxare independent variables, one gets

fe=i

∂λ

∂x

−

∂Wm

∂x

and 0 =i

∂λ

∂i

−

∂Wm

∂i

We can then see that

fe=

∂

∂x

(iλ−Wm)=

∂

∂x

Wm′

whereWm′ is the coenergy, which is equal to the energyWm in structures that are

magnetically linear.

The forcefacting to pull the bar toward the fixed magnet core is given by

f=−fe=−

∂Wm

∂x

The stored energy in a linear magnetic structure is given by

Wm=

φF

2

=

φ^2 R(x)

2

whereφis the flux,Fis the mmf, andR(x)is the reluctance, which is a function of

displacement. Finally, we get

f=−

∂Wm

∂x

=−

φ^2

2

dR(x)

dx

(b) The reluctance of the air gaps in the magnetic structure is given by

R(x)=

2 x

μ 0 A

=

2 x

4 π× 10 −^7 × 0. 01

=

x

0. 6285 × 10 −^8

The magnitude of the force in the air gap is then given by