Magnetic Field above a Rectangular Circuit

[Contingency]

Homework Statement

On the XY there lies a conducting wire-rectangle with sides parallel to the axis.
The current is given and constant.
What is the magnitude of the magnetic field along an axis parallel to the z axis, going through the intersection of the rectangle's diagonals?

Homework Equations

[tex] \vec { dB } =\frac { I }{ c{ \left| \vec { r } \right| }^{ 3 } } \vec { dl } \times \vec { r } [/tex]

The Attempt at a Solution

Due to symmetry, I can expect the field along this axis to be in the z direction. I can integrate along just one of each pair of the circuit's sides and double the result.
What I'm having trouble with is the cross product in Biot-Savart's Law - it seems to me that the angle [itex] { \theta }_{ \vec { dl } ,\vec { r } } [/itex]changes throughout the integral. If this is true then it seems the calculation of the field is not too easy..
I'd just like to make sure I'm correct before I get into it - does [itex] { \theta }_{ \vec { dl } ,\vec { r } } [/itex]change throughout the integral?

 

[VantagePoint72]

Remind yourself how the calculation goes for an infinite line of current using the Biot-Savart law. The same calculation will apply to each piece of the loop, just with finite integration bounds.

 

[Contingency]

I haven't calculated that field with Biot-Savart.. But I presume the implied answer is that the angle changes and this is not negligible.. Right?

 

[VantagePoint72]

Yes, the angle changes. It is usually one of the first calculations done in any EM text after introducing the Biot-Savart law. You can take at look at "Introduction to Electrodynamics" by Griffiths, for instance.

 

[Nickie]

Your understanding of the problem and approach to solving it is correct. The angle between the current element and the distance vector does indeed change throughout the integral. However, this can be taken into account by breaking the integral into smaller segments and considering the contributions from each segment separately. As long as the angle is properly accounted for in each segment, the overall result will be correct. Additionally, the symmetry of the problem can be used to simplify the calculation.