Posted by Zom on Monday, October 29, 2012 at 3:31pm.
i1 =10A, i2 =1A sde a=10cm=0.1 m, distance between the straight cureent and the side 12 of the loop is b=10 cm =0.1 m
Assume that the directions of the currents are following :
i1↑I ⃞ ↓i2
Points 1,2,3, and 4 are located clockwise starting from the left bottom corner of the square
Magnetic field of the current i1
at the side 12 locations is
B12 =μ₀i1/2πb =4π10⁻⁷10/2π0.1 = ..
Magnetic field of the current i1 at the side 34 location is
B34 =μ₀i1/2π(a+b) = 4π10⁻⁷10/2π0.2 = ..
Amperes force on the side 12 is
F12=i2aB12 =
=10.1 4π10⁻⁷10/2π0.1=...
Amperes force on the side 34 is
F12=i2aB34 =
=10.1 4π10⁻⁷10/2π0.2=...
F23=F41. These forces are directed in opposite directions, =>
The net force is directed to the left (to the straight current) and is
F= F12 F34 =
.
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