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axis. We take
G
B
to be in the same direction as that of the current flow in the hypotenuse.
Then, with
BB
==
G
0 0750
.T
,
cos
0.0692T ,
sin
0.0288T.
xy
B
B
θ
=−
=
=
(a) Eq. 2826 produces zero force when
G
G
LB

so there is no force exerted on the
hypotenuse of length 130 cm.
(b) On the 50 cm side, the
B
x
component produces a force
iB
yx
A
±
k, and there is no
contribution from the
B
y
component. Using SI units, the magnitude of the force on the
A
y
side is therefore
4 00
0500
0 0692
0138
..
.
.
Am
T
N
.
bg
b
g
b
g
=
(c) On the 120 cm side, the
B
y
component produces a force
A
±
k, and there is no
contribution from the
B
x
component. The magnitude of the force on the
A
x
side is also
4 00
120
0 0288
.
.
T
N
.
b
g
=
(d) The net force is
iB iB
AA
±±
,
kk
+=
0
keeping in mind that
B
x
< 0 due to our initial assumptions. If we had instead assumed
G
B
went the opposite direction of the current flow in the hypotenuse, then
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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