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Magnetic Field of rectangular current loop

  • Thread starter cconfused
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  • #1
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Homework Statement



Find the magnetic field of a rectangular current loop lying symmetrically on the xy-plane. Find the magnetic field at (0,0,z)

Homework Equations



Biot-Savart law or derived formula (Mu/4*Pi) * sin(theta2)-sin(theta1)/s

The Attempt at a Solution


I am NOT good with getting my trianges correctly. I know that B1=B3 and B2=B4 but I cannot figure out the values of the thetas!
 

Answers and Replies

  • #2
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Try and show a little more work than what you've got.
 
  • #3
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Well I know how to solve it for a square loop theta 1 =-theta2=45 degrees
= sqrt2*Mu*I/Pi*R
Now for a rectangle I think that theta1=-theta2 (for sides B1=B3) = -b/(sqrt(a^2+z^2)
and theta1=theta2 (for sides B2=B4) = a/(sqrt(b^2+z^2) and than I add them together but I'm not sure
 
  • #4
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Noone can help me with this question?
 
  • #5
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This is actually pretty tough. You're going to have to use the Biot-Savart law from scratch. The formula you're trying to use is for a point in the same plane as the wire, and isn't going to work here.

[tex]\mathbf{B}(\mathbf{r})=\frac{\mu_0 I}{4 \pi} \int \frac{d \mathbf{l}\times \hat{\mathbf{r}}}{r^2}[/tex]

You should realize some amount of symmetry. I remember I worked this out once, because I was making rectangular coils for a magneto-optical trap, and it took me a bit. What level of physics is this?
 
Last edited:
  • #6
fluidistic
Gold Member
3,654
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This is actually pretty tough. You're going to have to use the Biot-Savart law from scratch. The formula you're trying to use is for a point in the same plane as the wire, and isn't going to work here.

[tex]\mathbf{B}(\mathbf{r})=\frac{\mu_0 I}{4 \pi} \int \frac{d \mathbf{l}\times \mathbf{r}}{r^2}[/tex]

You should realize some amount of symmetry. I remember I worked this out once, because I was making rectangular coils for a magneto-optical trap, and it took me a bit. What level of physics is this?
Slight modification: notice that it should read [tex]\mathbf{B}(\mathbf{r})=\frac{\mu_0 I}{4 \pi} \int \frac{d \mathbf{l}\times \mathbf{r}}{r^3}[/tex].
 
  • #7
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Oops, yeah, I forgot the hat, I'll fix mine and let yours be an alternate. :)

Edit: Nvm, the hat on the r vector isn't working... so the later version is the best.
 
  • #8
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It's third year...E&M 2..any help? I do know there is symmetry of the horizontal and vertical components, and they all add up to give the total magnetic field
 
  • #9
1,860
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Actually, I've been thinking about this. You can use the derived result that you mentioned earlier, or start from scratch from Biot-Savart. Either way, it's your homework and not mine. You have to make the effort. If you have, then show use you've made the effort by posting some work.
 

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