Question

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 1

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 = $50.215.10^{-7}$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

$B=\frac{\mu_{0}.I}{2.\pi.r}$

in which

$\mu_{0}$ is permeability of free space and is $4.\pi.10^{-7}$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 = $\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}$

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 = $\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.082}$

F₁ = $278.975*10^{-7}$N

Current in F₃ is flowing thoruhg the negative side of the referential, so:

F₃ = $-\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}$

F₃ = $-228.76*10^{-7}$N

Net force is total force:

$F_{net} = F_{1}+F_{3}$

$F_{net}=(278.975-228.76).10^{-7}$

$F_{net}=50.22.10^{-7}$

The total force acting on the square loop is $F_{net}=50.22.10^{-7}$N.