Square loop carrying current near a wire

[Reshma]

This one is again from Griffiths.
a) Find the force on a square loop of side 'a' placed at a distance 's' from an infinite wire. Both the loop and the wire carry a current 'I'.

I found the magnitude of the magentic field using Biot-Savart's law:
B = \frac{\mu_0 I}{2\pi s}

The force is given by:
\vec F_{mag} = I\int\left(d\vec l \times \vec B\right)
"dl" is a wire element.

So, when I consider only the magnitude:
F = \frac{\mu_0 I^2}{2\pi s}\int dl
Here the wire is infinite, so how is it possible to integrate over the length of the wire?

[siddharth]

The B you first find is due to the infinite wire. So, when you integrate to find the force acting on the loop, dl will be an element on the loop. So you should integrate over the square loop.

[Reshma]

The B you first find is due to the infinite wire. So, when you integrate to find the force acting on the loop, dl will be an element on the loop. So you should integrate over the square loop.
I'm sorry, I don't get it. Here we have to take the direction into consideration. But the current flows in the wire as well as the loop. So if we take the current on the square loop, what will \int dl look like?

[Reshma]

Help me out here someone!!

[siddharth]

Help me out here someone!!

I have my exams going on, so I can't reply immediately.

Anyway, you find the magnetic field B due to the infinite wire as,
B = \frac{\mu_0 I}{2\pi s}.

The force on a small element 'dl' of conductor in a magnetic field is
F = i \vec{dl} \times \vec{B}
where, B is the external magnetic field, 'i' is the current on the conductor.

So, in this case, the external magnetic field is due to the infinte wire and 'dl' and i are with respect to the square loop. Therefore, when you integrate, you do so along the square loop, and not the infinite wire.

Did you get the answer in the book?

[Gokul43201]

Anyway, you find the magnetic field B due to the infinite wire as,
B = \frac{\mu_0 I}{2\pi s}.Since 's' is given as a specific distance in the question, you should use the more general equation :

B = \frac{\mu_0 I}{2\pi d}
where d can be either s or s+a or s+x (0<x<a), depending on where in the square you take your elemental length.

[Reshma]

Since 's' is given as a specific distance in the question, you should use the more general equation :

B = \frac{\mu_0 I}{2\pi d}
where d can be either s or s+a or s+x (0<x<a), depending on where in the square you take your elemental length.

Thanks. I suppose the forces at the sides of the square should cancel. Then effective magnetic force operates at the bottom and at the top of the square loop.
At the bottom: d=s
F = \frac{\mu_0 I^2}{2\pi s}\int dl

F = \frac{\mu_0 I^2}{2\pi s}s

F = \frac{\mu_0 I^2}{2\pi }

At the top: d=s+a
B = \frac{\mu_0 I}{2\pi (s+a)}

F = \frac{\mu_0 I^2}{2\pi (s+a)}\int_0^{s+a} dl

Are my boundary conditions for "dl" correct? If so the force obtained will be the same as for the bottom segment.

[Gokul43201]

F = \frac{\mu_0 I^2}{2\pi (s+a)}\int_0^{s+a} dl

Are my boundary conditions for "dl" correct? If so the force obtained will be the same as for the bottom segment.Everything is right except for the limits on that last integral. dl refers to the length of an element on the bottom segment. Let this element be at some general position 'x' along the segment. What are the limits for x (ie : what values can x take as you move the element through various positions along the bottom segment) ?

[Reshma]

Let this element be at some general position 'x' along the segment. What are the limits for x (ie : what values can x take as you move the element through various positions along the bottom segment) ?
Limits for x would be 0 to a, right?

[Gokul43201]

Yes. And make sure you understand why.

[Reshma]

Yes. And make sure you understand why.
I stumbled upon a glitch here. :bugeye:
For the bottom segment:
F = \frac{\mu_0 I^2}{2\pi s}\int dl
F = \frac{\mu_0 I^2}{2\pi s}s
I should have taken the limits on 'dl' as 0 to a here too, right?