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Magnetic field in the open loop

  • Thread starter tomfrank
  • Start date
38
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1. Homework Statement
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. Homework Equations



3. The Attempt at a Solution
 

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Attachments Pending Approval...
 
38
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any help?
 

berkeman

Mentor
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1. Homework Statement
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. Homework Equations



3. The Attempt at a Solution
The hint is to use the Biot-Savart law... (Hint: it often simplifies the integration to make your variable of integration theta...)
 
122
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you should divide the wire into some parts and find m.f created by these parts.
 
38
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would the left and the right side cancel out....

So I have to do it just for the bottom side???
 
38
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so thus this integral work :

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

fluidistic

Gold Member
3,625
96
would the left and the right side cancel out....

So I have to do it just for the bottom side???
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.
 
122
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so thus this integral work :

B= ([tex]\mu[/tex]*I)/(4*pi*x)[tex]\int[/tex] between(-pi/4) and (pi/4) of cos ([tex]\theta) d\theta[/tex]
No. Biot-Savart:

[tex]
dB = \frac{\mu_0 I}{4 \pi} \frac{dl sin\theta}{r^2}
[/tex]

With [tex]\theta[/tex] is angle between dl and r. imo, you can use symmetry property to make life easier.
 
38
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so i stil do not quite get it....does the two side cancel out?

[tex]

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

[/tex]

will the angle be (-pi/4) to (pi/4)
 
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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.
 
38
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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

[tex]


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

[/tex]

i do not know yet what boundary to use for the integral
 
122
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1/r^2 = cos^2(theta')/ x^2
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:

rl.bhat

Homework Helper
4,433
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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.
 
38
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yes the picture with the angle is exactly what I was thinking....what are the limits for the integral angles???
 

berkeman

Mentor
55,971
6,013
yes the picture with the angle is exactly what I was thinking....what are the limits for the integral angles???
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.
 
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i did to -45 to 45....will that work?
 
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thank you
 

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