Electromagnetism help: Find magnetic flux through a loop

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SUMMARY

The discussion focuses on calculating the magnetic flux through a loop surrounding a very large conductor carrying a direct current (DC) in a vacuum. The key equations used include the magnetic flux equation, \(\Phi=\int_S B\mathrm dS\), and the magnetic field generated by the conductor, \(B=\frac{\mu_0I}{2\pi x}\). The solution involves determining the differential area element \(dS\) and setting appropriate limits of integration based on the geometry of the problem. The final expression for the flux through the loop is \(\frac{\mu_0Ia}{\pi}\int_\alpha^{\pi} {(1+\cos\theta)}\mathrm d\theta\).

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with the right-hand rule for determining magnetic field direction
  • Knowledge of calculus, specifically integration techniques
  • Basic concepts of coordinate systems in physics
NEXT STEPS
  • Study the derivation of the differential area element \(dS\) in various coordinate systems
  • Learn about the application of the right-hand rule in electromagnetic contexts
  • Explore the implications of varying magnetic fields and their effects on loops
  • Investigate the relationship between magnetic flux and current in different geometries
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Students and educators in physics, particularly those focusing on electromagnetism, as well as engineers working with electromagnetic systems and applications.

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Homework Statement


Very large conductor with DC current is in vacuum. Find magnetic flux through a loop.
Given parameters: I,a,\alpha

Homework Equations


\Phi=\int_S B\mathrm dS - basic equation for magnetic flux
B=\frac{\mu_0I}{2\pi x} - electromagnetic induction created by very long
linear conductor at distance x

The Attempt at a Solution


After assuming the orientation of the loop, we can find direction of electromagnetic induction
by the right-hand rule. The flux through the loop is equal to the flux through some arbitrary shaped
surface which lies on the loop (see attachment). I don't know how to derive equation for dS.

Elementary flux through the infinitely small surface:
d\Phi=BdS\cos(B,n)=BdS

In my books solution it says that d\Phi=\frac{\mu_0I}{2\pi x}\cdot 2z\cos\theta dx, where
z=a\sin\theta, x=a(1-\cos\theta)

Flux through the loop is
\frac{\mu_0Ia}{\pi}\int_\alpha^{\pi} {(1+\cos\theta)}\mathrm d\theta=\frac{\mu_0Ia}{\pi}(\pi-\alpha-\sin\alpha)

Could someone explain how to derive equation for dS and how to set the limits of integration?
 

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Use your coordinate system to determine dS ... limits of integration are related.
ie. If dS lies in the x-y plane of a rectangular coordinate system you use dS = dx.dy
 
Since the center of your circle on the x-axis is not given, the problem can't be solved.
 
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Likes   Reactions: Simon Bridge
rude man said:
Since the center of your circle on the x-axis is not given, the problem can't be solved.

The radius of imaginary circle is a.
 
I know. That doesn't define the center of the "imaginary circle", just its radius.
 

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