This preview shows page 1. Sign up to view the full content.
60. (a) Our notation is as follows:
h
is the height of the toroid,
a
its inner radius, and
b
its
outer radius. Since it has a square cross section,
h = b – a
= 0.12 m – 0.10 m = 0.02 m.
We derive the flux using Eq. 2924 and the selfinductance using Eq. 3033:
Φ
B
a
b
a
b
BdA
Ni
r
hdr
Nih
b
a
==
F
H
G
I
K
J
=
F
H
G
I
K
J
z
z
µµ
00
22
ππ
ln
and
2
0
ln
2
B
Nh
N
b
L
ia
µ
Φ
⎛⎞
⎜⎟
π
⎝⎠
.
Now, since the inner circumference of the toroid is
l
= 2
π
a
= 2
π
(10 cm)
≈
62.8 cm, the
number of turns of the toroid is roughly
N
≈
62.8 cm/1.0 mm = 628. Thus
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Inductance

Click to edit the document details