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What is the net force on the square coil...

What is the net force on the square coil

A

`25 xx 10^(-7) N` moving towards wire

B

`25 xx 10^(-7)N` moving away from wire

C

`35 xx 10^(-7)N` moving towards wire

D

`35 xx 10^(-7) N` moving away from wire

Text Solution

Verified by Experts

The correct Answer is:
A
`F_(1) = (mu_(0))/(2pi xx 2 xx 10^(-2)) xx 2 xx 1 xx 15 xx 10^(-2)`
`F_(2) = (mu_(0) xx 2 xx 1 xx 15 xx 10^(-2))/(2pi xx 12 xx 10^(-2))`
now `F_("net") = F_(1) - F_(2)`
`= (4pi xx 10^(-7) xx 2 xx 1 xx 15 xx 10^(-2))/(2pi xx 2 xx 10^(-2)) (1- (1)/(6))`
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