Page 259 - Engineering Electromagnetics, 8th Edition

CHAPTER 8 Magnetic Forces, Materials, and Inductance 241

Since the loop is of differential size, the value of B at all points on the loop may be
taken as B 0 . (Why was this not possible in the discussion of curl and divergence?)
The total force on the loop is therefore zero, and we are free to choose the origin for
the torque at the center of the loop.
The vector force on side 1 is

dF 1 = Idx a x × B 0
or
dF 1 = Idx(B 0y a z − B 0z a y )
For this side of the loop the lever arm R extends from the origin to the midpoint
1
of the side, R 1 =− dy a y , and the contribution to the total torque is
2
dT 1 = R 1 × dF 1
1
=− dy a y × Idx(B 0y a z − B 0z a y )
2
1
=− dx dy I B 0y a x
2
The torque contribution on side 3 is found to be the same,
1
dT 3 = R 3 × dF 3 = dy a y × (−Idx a x × B 0 )
2
1
=− dx dy IB 0y a x = dT 1
2
and
dT 1 + dT 3 =−dx dy IB 0y a x
Evaluating the torque on sides 2 and 4, we find

dT 2 + dT 4 = dx dy IB 0x a y
and the total torque is then
dT = Idx dy(B 0x a y − B 0y a x )

The quantity within the parentheses may be represented by a cross product,
dT = Idx dy(a z × B 0 )
or

dT = IdS × B (15)

where dS is the vector area of the differential current loop and the subscript on B 0
has been dropped.
We now define the product of the loop current and the vector area of the loop as
2
the differential magnetic dipole moment dm, with units of A · m . Thus
dm = IdS (16)