The question is incomplete. Here is the complete question.
A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.
Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.
Answer: Net Force = [tex]50.215.10^{-7}[/tex]N
Explanation: Force and Magnetic field are related through the following formula:
F = I.L.B.sinθ
Magnetic field (B) in a straight long wire is given by
[tex]B=\frac{\mu_{0}.I}{2.\pi.r}[/tex]
in which
[tex]\mu_{0}[/tex] is permeability of free space and is [tex]4.\pi.10^{-7}[/tex]T.m/A
I is current in the wire;
r is distance to the wire;
Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.
So, for the net force, the relevant forces will be on the sides parallel to the wire.
For the other forces, angle is 90°, sin(90°) = 1, then:
F = I.L.B
Replacing magnetic field:
F = [tex]\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}[/tex]
Note: The side closest to the wire is F₁, while the farthest is F₃.
Note2: As the constant unit is in meters, distance and length of side of the square loop are also in meters.
Calculating forces:
F₁ = [tex]\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.082}[/tex]
F₁ = [tex]278.975*10^{-7}[/tex]N
Current in F₃ is flowing thoruhg the negative side of the referential, so:
F₃ = [tex]-\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}[/tex]
F₃ = [tex]-228.76*10^{-7}[/tex]N
Net force is total force:
[tex]F_{net} = F_{1}+F_{3}[/tex]
[tex]F_{net}=(278.975-228.76).10^{-7}[/tex]
[tex]F_{net}=50.22.10^{-7}[/tex]
The total force acting on the square loop is [tex]F_{net}=50.22.10^{-7}[/tex]N.
