# Current Loop Generating a Magnetic Field

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1. Jan 19, 2015

### Wilfrid Somogyi

I'm trying to understand how to calculate the strength of the magnetic field generated by a current loop.

What I know so far is:
• When a current moves around a loop of wire it generates a magnetic field which looks a bit like that of a solenoid, only not so concentrated through the center of the loop.
• The magnetic field at the center of the current loop is calculated by:

• I understand what each of the terms mean except the term L and the term r (with the ^ above).
What I'd like to know is:
• How is this equation derived?
• What are the terms I don't recognise?
• Is there a way to calculate the magnetic field strength at any point on the plane that the loop lies in?

2. Jan 19, 2015

### blue_leaf77

Magnetic field strength caused by an elemental current $I \times dL$ is governed by Biot-Savart law $$d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{L}\times\vec{r}}{r^3}$$. Therefore dL and r (with hat) are elemental source and unit position vector of observation point relative to the elemental source, respectively. To get that equation just take the observation point at the center and integrate around the loop. With Biot-Savart law in hand surely it's possible to calculate magnetic field at any point in space.

3. Jan 19, 2015

### Wilfrid Somogyi

Thanks, I got most of that, but what do you mean by elemental source? (I'm only an A-level student you see)

4. Jan 19, 2015

### blue_leaf77

The source of magnetic field in our case is a loop in which some currents circulating. If you divide that loop into infinitesimal length (mathematically the length of such object is the limit toward zero), you will get elemental length/source. This is calculus stuff, if you sum infinitesimal elements whose length approaches zero you will end up with integral.
So if one were to picture this problem physically, he must think that the magnetic field at certain point in space must be the sum of magnetic field generated by all existing elemental sources. And again sum of elemental sources forms integral.

5. Jan 19, 2015

### Wilfrid Somogyi

Right. So in this case the integral S dL (apologies I have no integral sign on my phone) in the equation just becomes L? I'm assuming that as there is thing between the integral and and dL, so the one has been factored out? In which case why not just represent it as L? Or is it not simply the integral of 1 with respect to L?

6. Jan 20, 2015

### blue_leaf77

if your observation point is in the center and you calculated the cross product correctly you should find that the integrand is axially symmetric.
Try to work it out, it's not that hard.

7. Jan 20, 2015

### Wilfrid Somogyi

Will do, thanks for all the help!