A square loop ABCD, carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be
A
2μ0IiL3π
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B
2μ0Ii3π
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C
μ0Ii2π
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D
μ0IiL2π
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Solution
The correct option is B2μ0Ii3π
The direction of currents in a long straight conductor XY and arm AB of a square loop ABCD are in the same direction.
So, there exists a force of attraction between the two conductors which is given by FAB as FAB=BiL=μ0IiL2π(L2)⎛⎜
⎜
⎜
⎜⎝∵BAB=μ0I2π(L2)⎞⎟
⎟
⎟
⎟⎠
The direction of currents in a long straight conductor XY and arm CD, the direction of currents are in the opposite direction. There exists a force of repulsion which will be experience by CD as FCD=BiL=μ0IiL2π(3L2)⎛⎜
⎜
⎜
⎜⎝∵BCD=μ0I2π(3L2)⎞⎟
⎟
⎟
⎟⎠
Now, As arms BC and CD are perpendicular to straight conductor XY and carry current in opposite directions, net force on both of them will cancel each other.
Net force on the loop ABCD is Floop=FBA−FCD=μ0IiL2π⎛⎜
⎜
⎜⎝1(L2)−1(3L2)⎞⎟
⎟
⎟⎠ Floop=2μ0Ii3π