# Magnetic field in the open loop

1. Nov 3, 2009

### tomfrank

1. The problem statement, all variables and given/known data
I am trying to find the magnetic field at the point in the middle of the picture can someone give a hint on how to approach the problem?

2. Relevant equations

3. The attempt at a solution

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2. Nov 3, 2009

### ApexOfDE

Attachments Pending Approval...

3. Nov 4, 2009

any help?

4. Nov 4, 2009

### Staff: Mentor

The hint is to use the Biot-Savart law... (Hint: it often simplifies the integration to make your variable of integration theta...)

5. Nov 4, 2009

### ApexOfDE

you should divide the wire into some parts and find m.f created by these parts.

6. Nov 5, 2009

### tomfrank

would the left and the right side cancel out....

So I have to do it just for the bottom side???

7. Nov 5, 2009

### tomfrank

so thus this integral work :

B= ($$\mu$$*I)/(4*pi*x)$$\int$$ between(-pi/4) and (pi/4) of cos ($$\theta) d\theta$$

8. Nov 5, 2009

### fluidistic

I believe so (though I might be wrong). In all cases you should prove it via arithmetic so don't use this shortcut if you are not sure.

9. Nov 5, 2009

### ApexOfDE

No. Biot-Savart:

$$dB = \frac{\mu_0 I}{4 \pi} \frac{dl sin\theta}{r^2}$$

With $$\theta$$ is angle between dl and r. imo, you can use symmetry property to make life easier.

10. Nov 5, 2009

### tomfrank

so i stil do not quite get it....does the two side cancel out?

$$dB = \frac{\mu_0 I}{4\pi r^2} {dl sin\theta}$$

will the angle be (-pi/4) to (pi/4)

11. Nov 5, 2009

### ApexOfDE

No. r is changing with theta. Can you find relationship between sin(theta) and r? and between r and x?

About two sides, you can say that they cancel out.

12. Nov 5, 2009

### tomfrank

i see how you are looking at the picture.

basically the

sin (theta) dl = cos (theta') dl

and

1/r^2 = cos^2(theta')/ x^2

when i put all of them in the integral reduce down to

$$dB = \frac{\mu_0 I}{4\pi x} {dl cos\theta}$$

i do not know yet what boundary to use for the integral

13. Nov 5, 2009

### ApexOfDE

from this, i guess theta and theta' as shown in fig below:

http://img149.imageshack.us/img149/6568/88285645.jpg [Broken]

is that right?

Last edited by a moderator: May 4, 2017
14. Nov 5, 2009

### rl.bhat

dB = μo/4π*I*di*sinθ/r^2
where sinθ = x/r = x/sqrt( x^2 + l^2)
Take the limits of integration from -x/2 to +x/2.

15. Nov 5, 2009

### tomfrank

yes the picture with the angle is exactly what I was thinking....what are the limits for the integral angles???

16. Nov 5, 2009

### Staff: Mentor

The limits come from the geometry of the diagram. You really should be able to figure that part out.

And on your PM question to me, know, the B-field contributions from the two sides on the left and right do not cancel out.

17. Nov 5, 2009

### tomfrank

i did to -45 to 45....will that work?

18. Nov 5, 2009

### Staff: Mentor

Based on your initial drawing, yes.

19. Nov 5, 2009

thank you