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Electromagnetism help: Find magnetic flux through a loop

  • Thread starter gruba
  • Start date
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1. The problem statement, all variables and given/known data
Very large conductor with DC current is in vacuum. Find magnetic flux through a loop.
Given parameters: [itex]I,a,\alpha[/itex]

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

3. 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 [itex]dS[/itex].

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

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

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

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

Attachments

Simon Bridge

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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
 

rude man

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Since the center of your circle on the x axis is not given, the problem can't be solved.
 
208
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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 [itex]a[/itex].
 

rude man

Homework Helper
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I know. That doesnt define the center of the "imaginary circle", just its radius.
 

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