1.
Fadx = force on segment "ad" = (/4) (2I1 I2 H/L)
Fadx = (10-7) (2 x 0.575 x 0.344 x 0.26/0.20)
Fadx = 5.14 x 10-8 N
2.
Fbcx = force on segment "bc" = (/4) (2I1 I2 H/(L+ W))
Fbcx = (10-7) (2 x 0.575 x 0.344 x 0.26/(0.20 + 0.79))
Fbcx = 1.04 x 10-8 N
3)
FnetY = 0
since force on the upper section of wire is cancelled by the force on lower section of wire.
4)
the net force by the left wire on the loop is in positive x-direction toward right, hence the net force by the wire on right must be towards left and equal. hence the current in wire on right must be down direction
so along negative Y-direction
5)
force by the wire on left = force by wire on right
(/4) (2I1 I2 H/L) - (/4) (2I1 I2 H/(L+ W)) = (/4) (2I3 I2 H/(2L)) - (/4) (2I3 I2 H/(2L+ W))
(I1/L) - (I1/(L + W)) = (I3/(2L)) - (I3/(2L + W))
(0.575/0.20) - ((0.575)/(0.20 + 0.79)) = (I3/(0.40)) - (I3/(0.40 + 0.79))
I3 = 1.4 A
A rectangular loop of wire with sides H-26 cm and w-79 cm carries current . 0.344...
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