Page 255 - Engineering Electromagnetics, 8th Edition

CHAPTER 8 Magnetic Forces, Materials, and Inductance 237

Figure 8.3 Given P 1 (5, 2, 1), P 2 (1, 8, 5),
I 1 dL 1 =−3a y A · m, and I 2 dL 2 =−4a z A · m,
the force on I 2 dL 2 is 8.56 nN in the a y direction.

Solution. We have I 1 dL 1 =−3a y A · mat P 1 (5, 2, 1), and I 2 dL 2 =−4a z A · mat
P 2 (1, 8, 5). Thus, R 12 =−4a x +6a y +4a z , and we may substitute these data into (13),
4π10 −7 (−4a z ) × [(−3a y ) × (−4a x + 6a y + 4a z )]
d(dF 2 ) =
4π (16 + 36 + 16) 1.5
= 8.56a y nN

Many chapters ago, when we discussed the force exerted by one point charge on
another point charge, we found that the force on the first charge was the negative of
that on the second. That is, the total force on the system was zero. This is not the case
with the differential current elements, and d(dF 1 ) =−12.84a z nN in Example 8.2.
The reason for this different behavior lies with the nonphysical nature of the current
element. Whereas point charges may be approximated quite well by small charges,
the continuity of current demands that a complete circuit be considered. This we shall
now do.
The total force between two filamentary circuits is obtained by integrating twice:

I 1 I 2 dL 1 × a R12
F 2 = µ 0 dL 2 ×
4π R 2 12
(14)

I 1 I 2 a R12 × dL 1
= µ 0 × dL 2
4π R 2 12
Equation (14) is quite formidable, but the familiarity gained in Chapter 7 with
the magnetic field should enable us to recognize the inner integral as the integral
necessary to find the magnetic field at point 2 due to the current element at point 1.
Although we shall only give the result, it is not very difficult to use (14) to
find the force of repulsion between two infinitely long, straight, parallel, filamentary