P30_025.PDF
(
58 KB
)
Pobierz
Chapter 30 - 30.25
has length
√
2
a
,so
r
=
a/
√
2and
B
=
µ
0
i/
√
2
πa
. The fields due to the wires at the upper left and lower
right corners both point toward the upper right corner of the square. The fields due to the wires at the
upper right and lower left corners both point toward the upper left corner. The horizontal components
cancel and the vertical components sum to
B
total
=4
µ
0
i
√
2
πa
cos45
◦
=
2
µ
0
i
πa
=
2(4
π
×
m
/
A)(20A)
π
(0
.
20m)
·
=8
.
0
×
10
−
5
T
.
In the calculation cos45
◦
was replaced with 1
/
√
2. The total field points upward.
25. Each wire produces a field with magnitude given by
B
=
µ
0
i/
2
πr
,where
r
is the distance from the
corner of the square to the
c
enter. Accordin
g
to the Pythagorean theorem, the diagonal of the square
10
−
7
T
Plik z chomika:
kf.mtsw
Inne pliki z tego folderu:
P30_011.PDF
(64 KB)
P30_012.PDF
(56 KB)
P30_009.PDF
(60 KB)
P30_002.PDF
(56 KB)
P30_001.PDF
(57 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
Zgłoś jeśli
naruszono regulamin