# Electromagnetism help: Find magnetic flux through a loop

1. Nov 7, 2015

### gruba

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: $I,a,\alpha$

2. Relevant 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 $x$

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 $dS$.

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?

File size:
31.6 KB
Views:
48
2. Nov 7, 2015

### Simon Bridge

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

3. Nov 8, 2015

### rude man

Since the center of your circle on the x axis is not given, the problem can't be solved.

4. Nov 8, 2015

### gruba

The radius of imaginary circle is $a$.

5. Nov 8, 2015

### rude man

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

6. Nov 9, 2015