Find the magnetic field at the center of a rectangle

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SUMMARY

The discussion focuses on calculating the magnetic field |B| at the center of a rectangular loop of wire with sides a and b carrying a current I. The formula B = μI/2πR is initially derived for a straight wire, and the user explores its application to the rectangular loop by considering contributions from each side of the rectangle. The approach involves integrating the magnetic field contributions from each wire segment, utilizing the Biot-Savart Law, and leveraging symmetry to simplify the calculations.

PREREQUISITES
  • Understanding of the Biot-Savart Law
  • Familiarity with magnetic field calculations
  • Knowledge of vector calculus, particularly cross products
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the application of the Biot-Savart Law in different geometries
  • Learn about magnetic field integration techniques for complex shapes
  • Explore the concept of magnetic field symmetry in current-carrying loops
  • Investigate the derivation of magnetic fields for various wire configurations
USEFUL FOR

Students of physics, electrical engineers, and anyone interested in electromagnetism and magnetic field calculations in current-carrying conductors.

vande060
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Homework Statement


find |B| at the center
of a rectangular loop of wire of sides a and b carrying current I



Homework Equations



B = uI/2piR



The Attempt at a Solution


phyprobbox.jpg


- current flowing counterclockwise

- original art work :D

I did the proof for biot savart law as a homework problem, and I got B = uI/2piR for straight wire, but the proof I used was based of positive and negative infinity for bounds of the integral. I was wondering however, if the same formula (B = uI/2piR ) could be used to tackle the above problem like this

B = 2(uI/2pi(b/2)) + 2(uI/2pi(a/2))

of course I can do simplification on my own, but I am curious if this is the correct route?
 
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For a straight current carrying length of wire, the differential element for the magnetic field at a point is given by

dB = \frac{\mu_o I \vec{dL} \times \hat{r}}{4 \pi R^2}

where dL is the differential line element of the wire and \widehat{r} is a unit vector in the direction of r, the vector from the point of interest to the line element. R is the magnitude of r.

The cross product will simplify to dLsin(θ), where θ is the angle between dL and \widehat{r}.

To find B you should integrate over the length of each wire section. Symmetry will allow you to boil this down to a couple of integrals, and if you look carefully, both of them have the same form with just a slight change of which constants are plugged where.
 
gneill said:
For a straight current carrying length of wire, the differential element for the magnetic field at a point is given by

dB = \frac{\mu_o I \vec{dL} \times \hat{r}}{4 \pi R^2}

where dL is the differential line element of the wire and \widehat{r} is a unit vector in the direction of r, the vector from the point of interest to the line element. R is the magnitude of r.

The cross product will simplify to dLsin(θ), where θ is the angle between dL and \widehat{r}.

To find B you should integrate over the length of each wire section. Symmetry will allow you to boil this down to a couple of integrals, and if you look carefully, both of them have the same form with just a slight change of which constants are plugged where.

Ive got in now thanks
 

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