Page 287 - Engineering Electromagnetics, 8th Edition

CHAPTER 8 Magnetic Forces, Materials, and Inductance 269

Thus, for this uniform field
12 = µ 0 n 1 I 1 πR 1 2

and
M 12 = µ 0 n 1 n 2 πR 1 2
Similarly,
21 = µ 0 n 2 I 2 πR 1 2
2
M 21 = µ 0 n 1 n 2 πR = M 12
1
If n 1 = 50 turns/cm, n 2 = 80 turns/cm, R 1 = 2 cm, and R 2 = 3 cm, then
2
−7
M 12 = M 21 = 4π × 10 (5000)(8000)π(0.02 ) = 63.2 mH/m
The self-inductances are easily found. The flux produced in coil 1 by I 1 is
11 = µ 0 n 1 I 1 πR 1 2
and thus
2
L 1 = µ 0 n S 1 d H
1
The inductance per unit length is therefore
2
L 1 = µ 0 n S 1 H/m
1
or
L 1 = 39.5 mH/m
Similarly,
2
L 2 = µ 0 n S 2 = 22.7 mH/m
2
We see, therefore, that there are many methods available for the calculation of
self-inductance and mutual inductance. Unfortunately, even problems possessing a
high degree of symmetry present very challenging integrals for evaluation, and only
afew problems are available for us to try our skill on.
Inductance will be discussed in circuit terms in Chapter 10.

D8.12. Calculate the self-inductance of: (a) 3.5 m of coaxial cable with a =
0.8mmand b = 4 mm, filled with a material for which µ r = 50; (b)a toroidal
coil of 500 turns, wound on a fiberglass form having a 2.5 × 2.5cm square
cross section and an inner radius of 2 cm; (c)a solenoid having 500 turns about
acylindrical core of 2 cm radius in which µ r = 50 for 0 <ρ < 0.5cm and
µ r = 1 for 0.5 <ρ < 2 cm; the length of the solenoid is 50 cm.

Ans. 56.3 µH; 1.01 mH; 3.2 mH