Square loop carrying current near a wire

In summary, we are trying to find the force on a square loop of side 'a' placed at a distance 's' from an infinite wire, where both the loop and the wire carry a current 'I'. To find this force, we first use Biot-Savart's law to find the magnitude of the magnetic field, which is given by B = (μ0I)/(2πs). Then, we use the equation F = i(dl x B) to find the force on a small element 'dl' of conductor in a magnetic field. In this case, the external magnetic field is due to the infinite wire and 'dl' and i are with respect to the square loop. Therefore, when we integrate to find
  • #1
Reshma
749
6
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:
[tex]B = \frac{\mu_0 I}{2\pi s}[/tex]

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

So, when I consider only the magnitude:
[tex]F = \frac{\mu_0 I^2}{2\pi s}\int dl[/tex]
Here the wire is infinite, so how is it possible to integrate over the length of the wire?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
siddharth said:
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 [itex]\int dl[/itex] look like?
 
  • #4
Help me out here someone!
 
  • #5
Reshma said:
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,
[tex]B = \frac{\mu_0 I}{2\pi s}[/tex].

The force on a small element 'dl' of conductor in a magnetic field is
[tex] F = i \vec{dl} \times \vec{B}[/tex]
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 infinite 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?
 
Last edited:
  • #6
siddharth said:
Anyway, you find the magnetic field B due to the infinite wire as,
[tex]B = \frac{\mu_0 I}{2\pi s}[/tex].
Since 's' is given as a specific distance in the question, you should use the more general equation :

[tex]B = \frac{\mu_0 I}{2\pi d}[/tex]
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.
 
  • #7
Gokul43201 said:
Since 's' is given as a specific distance in the question, you should use the more general equation :

[tex]B = \frac{\mu_0 I}{2\pi d}[/tex]
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
[tex]F = \frac{\mu_0 I^2}{2\pi s}\int dl[/tex]

[tex]F = \frac{\mu_0 I^2}{2\pi s}s[/tex]

[tex]F = \frac{\mu_0 I^2}{2\pi }[/tex]

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

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

Are my boundary conditions for "dl" correct? If so the force obtained will be the same as for the bottom segment.
 
  • #8
Reshma said:
[tex]F = \frac{\mu_0 I^2}{2\pi (s+a)}\int_0^{s+a} dl[/tex]

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) ?
 
  • #9
Gokul43201 said:
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?
 
  • #10
Yes. And make sure you understand why.
 
  • #11
Gokul43201 said:
Yes. And make sure you understand why.
I stumbled upon a glitch here. :bugeye:
Reshma said:
For the bottom segment:
[tex]F = \frac{\mu_0 I^2}{2\pi s}\int dl[/tex]
[tex]F = \frac{\mu_0 I^2}{2\pi s}s[/tex]
I should have taken the limits on 'dl' as 0 to a here too, right?
 

1. How does a square loop carrying current near a wire interact with the wire?

The square loop carrying current creates a magnetic field around itself, and this field interacts with the magnetic field created by the wire. This interaction can cause the loop to experience a force or torque, depending on the relative orientations of the loop and the wire.

2. What factors affect the strength of the interaction between the loop and the wire?

The strength of the interaction depends on the current flowing through the loop and the wire, the distance between them, and the orientation of the loop with respect to the wire. Additionally, the shape and size of the loop can also affect the strength of the interaction.

3. Can the interaction between the loop and the wire be attractive or repulsive?

Yes, the interaction can be either attractive or repulsive depending on the direction of the currents in the loop and the wire. If the currents are flowing in the same direction, the interaction will be attractive, while if they are flowing in opposite directions, the interaction will be repulsive.

4. How does the loop's motion affect the interaction with the wire?

If the loop is stationary, the interaction with the wire will cause it to experience a force or torque. However, if the loop is in motion, the magnetic fields will also interact, causing the loop and wire to create electromagnetic waves and potentially inducing currents in each other. This can lead to more complex interactions between the loop and the wire.

5. What are some real-world applications of this phenomenon?

This phenomenon is commonly used in electromechanical devices, such as motors and generators, where the interaction between a loop and a wire is used to convert electrical energy into mechanical energy or vice versa. It is also utilized in magnetic levitation systems, where the repulsive force between a loop and a wire is used to suspend an object in mid-air.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
836
Replies
1
Views
268
  • Advanced Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
948
  • Introductory Physics Homework Help
Replies
3
Views
136
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
274
  • Advanced Physics Homework Help
Replies
1
Views
4K
  • Advanced Physics Homework Help
Replies
6
Views
3K
  • Advanced Physics Homework Help
Replies
5
Views
2K
Back
Top