Magnetic field above the center of a square current loop

[astrocytosis]

Homework Statement

Find the exact magnetic field a distance z above the center of a square loop of side w, carrying a current I. Verify that it reduces to the field of a dipole, with the appropriate dipole moment, when z >> w.

Homework Equations

(1) dB = μ₀I/4πr² dl × rhat

(2) r = √((½w)²+z²)

The Attempt at a Solution

I treated the loop like four current-carrying wires of finite length and used Biot-Savart. I think due to symmetry, the magnetic field in x-y should cancel at the center so all the magnetic field is in z. It seems intuitive to me that dB should just be 4*(1).

But what's confusing me is the dl × rhat term; I'm not sure my approach to it is sound. If rhat is the unit vector pointing along the distance r from a point on the loop to the point on the z-axis, it must be at some angle to x-y plane.

I can define an angle φ such that sinφ = z/r and write r in terms of z and w. Can I say the contribution from dl × r to the magnetic field dB is dl sinφ and write:

dB = μ₀Iz/4πr³ dl

for each segment of the loop, where is r is as in (2) and dl = dw? Then my limits of integration would be 0 to w and I would multiply the result by 4. Does that make sense? Thank you.

 

[Chandra Prayaga]

A carefully drawn diagram will be of immense help.
Your approach of treating the loop like four current carrying wires is correct. certainly, the net magnetic field is along the z direction, and the total field is four times the field of one wire.
I am writing vectors in bold font. Calculating the distance from an infinitesimal portion dl of the loop to the point on the z-axis should be done carefully. Place the square loop in the xy plane, with sides parallel to the x and y axes, and the center of the loop at the origin.The point on the z axis then has coordinates (0, 0, z). Consider anyone side of the square loop, say one that is parallel to the y-axis. Take an infinitesimal part dy of this side, located at a distance y from the center of the side. The coordinates of that point are (w/2, y, 0). The vector r points from this point (w/2, y, 0) to (0, 0, z). That should help you calculate the distance r. The angle between dl and r is the angle between that side and r. It s not the angle defined by you.
Hope these remarks help.

 

[astrocytosis]

Ok, I found the length of r by subtracting (x, w/2, 0) - (0, 0, z) and taking the magnitude: r = √(x²+(w/2)²+z²). Then sinφ = z/r, so dB is

dB = μ₀I/4π * z/ √(x²+(w/2)²+z²) * dx

Integrating from 0 to w and multiplying by 4 gives μ₀Iz/π * zw/(z²+(w/2)²)^3/2. But this is still a little off from the solution.

 

[Chandra Prayaga]

1. In your case, dl is along the x axis, so you should find the angle between the x-axis and r. You will need that angle to calculate dl x r. That will give you the magnetic field dB. Then you can find its z component by finding the angle between dB and the z axis. You are using the angle between r and the z axis. That is not the same.
2. You should integrate from -w/2 to + w/2 to get th total magnetic field B.

 

[astrocytosis]

When I found these correct angles I was able to get to the right answer. Thank you!

 

[Chandra Prayaga]

You are welcome.