Page 657 - The Mechatronics Handbook

0066_Frame_C20.fm Page 127 Wednesday, January 9, 2002 1:48 PM

Case 2: Magnetization Perpendicular to the Axis of Symmetry
For orientation of the magnetization vector perpendicular to the axis of symmetry, the following equation
is used to find the electromagnetic torque:

⋅
T = ∫ (M × B e + r × ( M ∇)B e ) v
d
v
where

– B eyz z + B ezz y
B exz
⋅
⋅
( M ∇)B e = [ ∂B e ]M = r × ( M ∇)B e =
z M z B exz zB ezz x–
M B eyz
B exz y + B eyz x
B ezz
and

– B ey
M × B e = M z B ex
0

Thus,

∫
∫
T = – M z B ey v + M z ( B exz yB eyz z) v
d
d
–
x
v v
∫
∫
T = M z B ex v + M z ( B ezz zB ezz x) v
d
–
d
y
v v
∫
T z = M z ( B eyz xB exz y) v
–
d
v
Expressing the fluxes and performing the integration, we have the following expressions for the torque
components as the function of the magnetic field:

--R +
1 2
T = – vM z B y + B xx( -----l + B yy( 1 2 B yz( 1 2 
--R
x  )y 24 )y 8 )z 8 
1 2 
1 2
T = vM z B x + B xz(  3 2 -----l + B xx( -----l + B xy( 1 2
--R –
--R
y )y 8  12  )x 24 )y 8
T z = vM z B xy( )y   -----l – 1 2  
1 2
--R
4
12
The electromagnetic forces are found to be
∫
F = M z B exz dv = vM z B xz
x
v
∫
F = M z B eyz dv = vM z B yz
y
v
∫
F z = M z B ezz dv = vM z B zz
v
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