Homework Help: 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

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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