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76. We take the current (
i
= 50 A) to flow in the +
x
direction, and the electron to be at a
point
P
which is
r
= 0.050 m above the wire (where “up” is the +
y
direction). Thus, the
field produced by the current points in the +
z
direction at
P
. Then, combining Eq. 294
with Eq. 282, we obtain
G
G
Fe
i
r
v
e
=−
×
µ
0
2
π
b
g
e
j
±
.
k
(a) The electron is moving down:
G
vv
±
j (where
v
= 1.0
×
10
7
m/s is the speed) so
( )
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 Spring '08
 Any
 Physics, Current

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