Question

What is the net force on the square coil?

A. `25xx10^(-7)N` moving towards wire
B. `35xx10^(-7)N` moving away from wire
C. `35xx10^(-7)N` moving towards wire
D. `35xx10^(-7)N` moving away from wire

Answer 1

Correct Answer - A
Force on side `BC` and `AD` are equal but opposite so their net will be zero.

But `F_(AB)=10^(-7)xx(2xx2xx1)/(2xx10^(-2))xx15xx10^(-2)=3xx10^(-6)N`
and `F_(CD)=10^(-7)xx(2xx2xx1)/((12xx10^(-2)))xx15xx10^(-2)`
`=0.5xx10^(-6)N`
` implies F_(net)=F_(AB)-F_(CD)=2.5xx10^(-6)N`
`25xx10^(-7)N` , towards the wire.