Magnetic flux thorugh a loop of wire

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SUMMARY

The discussion centers on calculating the magnetic flux through a square loop of wire positioned at a distance from a long straight wire carrying current I, as outlined in Griffiths Problem 7.8. The magnetic field B at a distance r from the wire is defined by the equation B = (μ₀ I) / (2π r). The correct expression for the area element in the flux calculation is a ds, where ds represents an infinitesimal distance along r, rather than . This clarification resolves the confusion regarding the area element used in the flux integral.

PREREQUISITES
  • Understanding of magnetic fields, specifically B = (μ₀ I) / (2π r)
  • Familiarity with the concept of magnetic flux and the equation Φₐ = ∫ B · da
  • Knowledge of calculus, particularly integration techniques
  • Basic principles of electromagnetism as outlined in Griffiths' textbook
NEXT STEPS
  • Study the derivation of magnetic fields around current-carrying wires
  • Learn about the application of the Biot-Savart Law in calculating magnetic fields
  • Explore advanced integration techniques for calculating flux in varying magnetic fields
  • Review Griffiths' Electrodynamics, focusing on problems related to magnetic flux
USEFUL FOR

Students of electromagnetism, physics educators, and anyone seeking to deepen their understanding of magnetic flux calculations in the context of current-carrying conductors.

stunner5000pt
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Homework Statement


Griffiths Problem 7.8
A square loop of wire (side a) lies on a table dsiantce s from a very long straight wire, which carries a current I, as shown in the figure.

Find the flux of B through the loop

Homework Equations


Flux of B is given by
[tex]\Phi_{B} = \int \vec{B} \dot d\vec{a}[/tex]


The Attempt at a Solution


THe area element is constant
but hte magnetic field is not
For a wire, B at a dsitance r is given by
[tex]\vec{B} = \frac{\mu_{0} I}{2\pi r} \hat{phi}[/tex]

Flux is then
[tex]\Phi_{B} = \int_{s}^{s+a} \frac{\mu_{0} I}{2\pi r} a^2 ds[/tex]

but hte solution says that the area element should be just a, and not a^2 .. why is that?

thanks for help!
 

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stunner5000pt said:
but hte solution says that the area element should be just a, and not a^2 .. why is that?
The element of area is "a ds", where "ds" is distance along r: you are treating the area as infinitesimal rectangular strips, parallel to the wire.
 
Doc Al said:
The element of area is "a ds", where "ds" is distance along r: you are treating the area as infinitesimal rectangular strips, parallel to the wire.

thank you
i get it now
 

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