Page 655 - The Mechatronics Handbook

0066_Frame_C20.fm Page 125 Wednesday, January 9, 2002 1:47 PM

the terms are

– B exy z + B exz y 0
r × ( M ∇)B e = M x B exx zB exz x– and M × B e = M – B ez
⋅
x
−B exx y + B exy x B ey

Therefore,

T = M x∫ ( B exz – B exy z) vd
y
x
v
T = – M x∫ B ez v + M x∫ ( B exx zB exz x) v
d
–
d
y
v v
and
T = M x∫ B ey v + M x∫ ( B exy – B exx y) v
x
d
d
z
v v
The terms in the derived equations must be evaluated.
Let us find the analytic expression for the electromagnetic torque T . In particular, we have
x
∫
∫
∫ B exz y v = B xz y v + B xx( )z xy v +d B xy( )z y v +d B xz( )z ∫ zy vd
∫
2
d
d
v v v v v
where
∫ y v = 0, ∫ xy v = 0, ∫ zy v = 0
d
d
d
v v v
and
1 --l R 2 2
∫ y v = 2 1 ∫ ∫ ∫ R – z y ydzdx = 1 4 1 4
2
--plR =
2
d
--vR
d
v – --l −R 2 z 2 4 4
2 – R –
Therefore,
1
x ∫
M B exz yv = M --B xy( )z vR 4
d
v x 4
Furthermore,
x∫ B exy zv = 1 4
M d M --B xy( )z vR
x
v 4
Thus, for T , one has
x
T = M x∫ ( B exz yB exy z) v = M  1 )z vR – 1 )z vR 4  = 0
4
–
d
--B xy(
--B xy(
x x  4 4 
v
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