The difference is the force on the loop.

[Alouette]

1. An infinite straight wire on the z-axis carries current I2=3.6A in the +z direction, upward. A rectangular loop is placed in the xz plane with its nearest side parallel to the wire a distance d=0.9m away. The loop has height (z-length) h=1.5m, width (x-length) w=1m, and carries a current I1=1.2.



2. Equations
B = Iμ/2∏r

F = Il X B

F = I₂(μI₁/2∏d)

The Attempt at a Solution

I tried using the equations, first by solving for B and plugging in μ, I1, and r=0.9

Then I used second equation and multipled I2 times that B, I got as an incorrect answer 9.6e-7

So I've drawn the diagram, but I'm missing where the height and length of the loop come into the equation, I'm certain they are part of the problem solving since I omitted them from my attempt.

I also know the perpendicular-to-wire sides of the square loop cancel out, so I need to focus on the parallel parts?...

FINAL EDIT:
Ah now I know, I needed to use this:
(I2h)(μ₀I/2π(d)) - (I2h)(μ₀I/2π(d+h))

My question now is why is the current of the infinite wire multiplied by the height of the square loop in this equation? Because the height is the length I assume? Just want to double check.

And also why is it necessary to subtract one F from the other?

 

[tiny-tim]

Hi Alouette!

F = Il X B
…
I also know the perpendicular-to-wire sides of the square loop cancel out, so I need to focus on the parallel parts?...
…
… why is the current of the infinite wire multiplied by the height of the square loop in this equation? Because the height is the length I assume? Just want to double check.

And also why is it necessary to subtract one F from the other?

Your equation F = Il X B has the length (of the wire) as a vector.

The vector points in the same direction as the current, so it points up on one side of the loop, and down on the other side, so the cross products are opposite (and unequal for the vertical sides … for the horizontal sides, as you say, they're opposite and equal, so they cancel ).