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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 :
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2μ0Ii 3π -
2μ0Ii 2π -
2μ0IiL 3π -
2μ0IiL 2π
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Correct Option: A
The direction of current in conductor 
XY and AB is same
∴ FAB = iℓB (attractive)
| FAB = i(L). | (←) | (←) | ||||||
| 2π |  |  | π | |||||
| 2 | ||||||||
FBC opposite to FAD
FBC (↑) and FAD (↓)
⇒ cancels each other
FCD = iℓB (repulsive)
| FCD = i(L). | (→) = | (→) | ||||||
| 2π | | | 3π | |||||
| 2 | ||||||||
Therefore the net force on the loop
Fnet = FAB + FBC + FCD + FAD
| Fnet = | - | = | ||||
| π | 3π | 3π |
