Page 656 - The Mechatronics Handbook

0066_Frame_C20.fm Page 126 Wednesday, January 9, 2002 1:48 PM

Then, for T , we obtain
y
T = – M x∫ B ez dv + M x∫ ( B exx – B exz x) dv
z
y
v v
1 2 
= M –  B z + -----l + 1 2 1 2   1 2 -----l v
--R +
--R v +
1 2
--R –
x  B zx( )x 24 B zy( )y 8 B zz( )z 8  B xx( )z 4  12 
= – vM B z + B xx(  1 2 1 2  + B yy( 1 --R + B zz( 1 2
2
--R –
--l
--R
x )z 4  8  )z 8 )z 4
Finally, we obtain the expression for T z as
T = M x∫ B ey dv + M x∫ ( B exy xB exx y) dv
–
z
v v

= vM B y + B xx(  1 2 1 2  B yy( 1 2 – 1 2
--R
--l –
--R –
--R B yz(
x )y 8  4  )y 8 )z 8
Thus, the following electromagnetic torque equations result:
T = 0
x
2
--R –
T = – vM B z + B xx(  1 2 1 2  + B yy( 1 --R + B zz( 1 2
--R
--l
y x )z 4  8  )z 8 )z 4
T = vM B y +  1 2 1 --R 2  – 1 2 1 2
--R –
--l –
--R
z x B xx( )y 8  4  B yy( )y 8 B yz( )z 8
The electromagnetic forces are found as well. In particular, from
F = M x∫ B exx v
d
x
v
F = M x∫ B exy v
d
y
v
and
F = M x∫ B exz v
d
z
v
using the expressions for the expanded magnetic fluxes, e.g.,

∫ B exx v = ∫ ( B xx + B xx( )x x + B xx( )y y + B xx( )z z) vd
d
v v

and performing the integration, one has the following expressions for the electromagnetic forces as the
function of the magnetic field:
F = vM B xx , F = vM B xy , F =
x
x x y x z vM B xz

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