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A square loop ABC carrying a current i i...

A square loop ABC 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

`(mu_(0)Ii)/(2pi)`

B

`(2mu_(0)IiL)/(3pi)`

C

`(mu_(0)IiL)/(2pi)`

D

`(2mu_(0)Ii)/(3pi)`

Text Solution

Verified by Experts

The force of attraction on aprt AB of the loop by conductor XY
`=(mu_(0)IiL)/(2pixx(L)/(2))=(mu_(0)Ii)/(pi)`
Similarly, the repulsive force on part CD of the loop by condcutor XY
`=(mu_(0)IL)/(2pixx3(L)/(2))=(mu_(0)Ii)/(3pi)`
The magnetic force on AD and BC by the conductor XY will be zero. Therefore, the net atractive force on the loop
`ABCD=(mu_(0)Ii)/(pi)-(mu_(0)Ii)/(3pi)=(2pi_(0)Ii)/(3pi)`
The option D is correct.
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