Find the net force acting on the wire shown in the figure.
[Take, B=1T, i = 2 A, R=0.5 m]

Explanation

Force on wire `AB=|F_(1)|= i(R )B = iRB`
Force on wire `BC=|F_(2)|= i((R )/(2))B=(iRB)/(2)`
Force on wire `CD=|F_(3)|= i(R )B=iRB`
The direction of forces are shown in figure

`F_(X)=` components of forces along + X-axis
`=F_(2)+F_(3) cos 45^(@) -F_(1) cos 45^(@)`
`=F_(2)" " [ :. F_(1)=F_(3)]`
`F_(Y)=`components of forces along + Y-axis
`=F_(1) sin 45^(@) +F_(3) sin 45^(@)`
`=2F_(1) sin 45^(@) =sqrt(2)F_(1)`
Resultant force or net force on wire
`=sqrt((F_(x))^(2)+(F_(Y))^(2))=sqrt(F_(2)^(2) +(sqrt(2)F_(1))^(2))`
`=sqrt(((iRB)/(2))^(2)+(sqrt(2)iRB)^(2))`
`=sqrt(((iRB)^(2))/(4)+2(iRB)^(2))=sqrt((9(iRB)^(2))/(4))`
`=(3)/(2) iRB=(3)/(2) xx 2 xx 0.5 xx 1 = 1.5N`