Magnetic force with 2 Current Carrying Wires

[mike_302]

Homework Statement

Two parallel loops are parallel, coaxial, and almost in contact, 1.29 mm apart. Each loop is .117 m in radius, the top loop carrying 137 A clockwise, and the bottom look carrying 137 A counter-clockwise. Calculate the force that the bottom loop exerts on the top.

Homework Equations

dB = (u0/4pi)(I*ds x r^)/r^2 where r^ is the unit vector.
For a long wire: B = (u0*I)/(2pi*r) where r is the distance from the center of the wire
Fb = Int(I*ds x B) for a current carrying wire

The Attempt at a Solution

So the answer works out to Fb= I^2*u0*r/d where d is the distance between the two loops.
To get this, I simply said Fb = ILB, where L is hte circumference of the loops, and then I had to say B was the magnetic field around a long straight wire... But I don't quite understand why I approximate the circle as a long straight wire? IS it because the two loops are stated as being so close together that one side doesn't affect anything except that which is directly above it?

Can someone explain this, and possibly even the math that I might have done directly from that dB equation to figure that out?

Thanks!

 

[anathema]

Thank you for your post and for your attempt at solving this problem. Your approach is on the right track, but there are a few things that can be clarified to help you better understand the math and the concept behind it.

First, let's start with the equation dB = (u0/4pi)(I*ds x r^)/r^2. This is the equation for the magnetic field produced by a small segment of current-carrying wire, where dB is the magnetic field produced by that segment, u0 is the permeability of free space, I is the current in the wire, ds is the length of the segment, r^ is the unit vector pointing from the segment to the point where the magnetic field is being measured, and r is the distance between the segment and the point.

Now, let's apply this equation to the two parallel loops in your problem. Since the loops are coaxial, we can consider them to be made up of many small current-carrying segments, each producing its own magnetic field. The total magnetic field at any point due to both loops will be the sum of the magnetic fields produced by each segment. However, since the loops are very close together, we can simplify our calculation by assuming that the distance between any point and the two loops is the same. This allows us to combine the two loops into one long straight wire, with a current of 137 A in one direction and 137 A in the opposite direction. This is why we use the equation for the magnetic field around a long straight wire, B = (u0*I)/(2pi*r), to calculate the magnetic field at the point where the two loops are almost in contact.

Now, to calculate the force exerted by the bottom loop on the top loop, we use the equation Fb = ILB, where I is the current in the top loop, L is its circumference, and B is the magnetic field at the point where the two loops are almost in contact. Substituting the values given in the problem, we get Fb = (137 A)(2*0.117 m)(u0*137 A)/(2*pi*1.29*10^-3 m) = 7.59*10^-4 N.

I hope this explanation helps you understand the concept and the math behind it. Keep up the good work and don't hesitate to ask for further clarification if needed. Good luck with your