Page 355 - Schaum's Outline of Theory and Problems of Applied Physics

340 ELECTROMAGNETIC INDUCTION [CHAP. 28

N turns
Cross-sectional
area A

I
I
Fig. 28-4. (From Arthur Beiser, Schaum’s Outline of Basic Mathematics for Electricity and Electronics, 2nd Ed.,
c 1993, The McGraw-Hill Companies. Reproduced with permission of The McGraw-Hill Companies.)

SOLVED PROBLEM 28.8
2
Find the inductance in air of a 500-turn solenoid 10 cm long whose cross-sectional area is 20 cm .
2
2
2
µ 0 N A (1.26 × 10 −6 H/m)(500) (2 × 10 −3 m ) −3
L = = = 6.3 × 10 H = 6.3mH
l 10 −1 m
SOLVED PROBLEM 28.9
A solenoid 20 cm long and 2 cm in diameter has an inductance of 0.178 mH. How many turns of wire
does it have?
The radius of the solenoid is r = 1cm = 0.01 m, and so its cross-sectional area is
2
2
A = πr = (π)(0.01 m) = 3.14 × 10 −4 m 2
2
Since L = µ 0 N A/l in air,

Ll (0.178 × 10 −3 H)(0.2m)
N = = = 300 turns
µ 0 A (4π × 10 −7 T·m/A)(3.14 × 10 −4 m )
2
SOLVED PROBLEM 28.10
2
An inductor consists of an iron ring 5 cm in diameter and 1 cm in cross-sectional area that is wound
with 1000 turns of wire. If the permeability of the iron is constant at 400 times that of free space at the
magnetic intensities at which the inductor will be used, find its inductance.

An inductor of this kind is essentially a solenoid bent into a circle. The length of the equivalent solenoid is
therefore
l = πd = (π)(0.05 m) = 0.157 m

2
2
2
and its cross-sectional area is A = 1cm /(100 cm/m) = 10 −4 m . The permeability of the core is µ = 400 µ 0 . The
inductance of this inductor is therefore
3 2
2
2
µN A (400)(4π × 10 −7 T·m/A)(10 ) (10 −4 m )
L = = = 0.32 H
l 0.157 m
SOLVED PROBLEM 28.11
The current in a circuit falls from 5 to 1 A in 0.2 s. If an average emf of2Vis induced in the circuit while
this is happening, find the inductance of the circuit.
Since V e =−L I/ t, we have (disregarding the minus sign)
V e t (2V)(0.1s)
L = = = 0.05 H
I 5A − 1A