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Magnetism
167
19.61
First, observe that
()
( )
(
2
22
5.00 cm
12.0 cm
13.0 cm
+= )
. Thus, the
triangle shown in dashed lines is a right triangle giving
1
sin
α
−
12.0 cm
67.4
13.0 cm
==
°
90.0
2
, and
2.6
β
=
°−
=
°
At point
P
, the field due to wire 1 is
( )( )
7
2
41
0
T
m
A
3
.
0
0
A
12.0 T
25
.
0
01
0
m
π
1
1
2
I
B
r
µ
−
−
×⋅
=
×
±
±
±²³´µ¶·
±
±
¸
±¸³´µ¶·
²
¹³´´µ¶·
²
and it is directed from
P
toward wire 2, or to the left and at
67.4° below the horizontal. The field due to wire 2 has
magnitude
( )( )
7
02
2
2
2
0
T
m
A
3
.
0
0
A
5.00 T
2
2
12.00
10
m
I
B
r
−
−
=
×
and at
P
is directed away from wire 1 or to the right and at
22.6° below the horizontal.
Thus,
BB
11
cos67.4
4.62 T
x
=−
°=−
sin67.4
11.1 T
y
=
−°
=
−
cos22.6
4.62 T
x
=°
=
+
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.
 Fall '10
 STAFF
 Physics, Magnetism

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