Biot-savart law

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This is a simple problem if you work with only magnitude and consider the symmetry of a ring. However, I am taking a detailed approach, and I am not getting the correct answer. Could you please look over my math???

a evenly distributed ring of charge on the positive Y-Z plane has a current, I, flowing counterclockwise or flowing from positive Z direction on top and -Z direction on bottom of ring. Figure out the magnetic field along the X axis of this ring of charge from the center of the ring.

[tex] d \vec{B} = \frac {\mu_{o} I}{4\pi} \frac {d\vec{l} \times \hat{r}}{r^2}[/tex]

below are pictures of the ring of charge. left picture is the sideway view with theta being constant. right picture is the frontal view with phi.

[tex] d\vec{l} = <0 , dl cos\phi , dl sin\phi>[/tex]
[tex] \hat{r} = <cos\theta , -w/r sin\phi , w/r cos\phi> [/tex]

[tex]{d\vec{l} \times \hat{r} = \frac{w}{r} cos^2\phi dl \cdot \hat{i} + \frac{w}{r} sin^2\phi dl \cdot \hat{i} + cos\theta sin\phi dl \cdot \hat{j} - cos\theta cos\phi dl \cdot \hat{k} [/tex]

[tex]\frac {d\vec{l} \times \hat{r}}{r^2} =\frac{w}{r^3} 1 dl \cdot \hat{i} +\frac{1}{r^2} cos\theta sin\phi dl \cdot \hat{j} - \frac{1}{r^2} cos\theta cos\phi dl \cdot \hat{k}[/tex]

[tex] dl = w d\phi [/tex]

[tex]\frac {d\vec{l} \times \hat{r}}{r^2} = \frac{w^2}{r^3} 1 d\phi \cdot \hat{i} +\frac{w}{r^2} cos\theta sin\phi d\phi \cdot \hat{j} - \frac{w}{r^2} cos\theta cos\phi d\phi \cdot \hat{k}[/tex]

ARE ALL MY STEPS CORRECT SO FAR????
 

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Answers and Replies

  • #2
gabbagabbahey
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a evenly distributed ring of charge on the positive Z plane has a current, I, flowing counterclockwise or flowing from positive Z direction on top and -Z direction on bottom of ring. Figure out the magnetic field along the X axis of this ring of charge from the center of the ring.

This problem description is not at all clear to me. For starters, what is 'the [itex]z[/itex] plane'? There is such a thing as a [itex]z[/itex] axis, but not a [itex]z[/itex] plane. Do you mean the plane defined by the equation [itex]z=0[/itex] (AKA the x-y plane)? Or do you mean the y-z plane (x=0) as your pictures would leave me to believe? Or perhaps you mean something else entirely?

Also, the 'top' and 'bottom' of the ring are not well defined. Which direction is up, and which direction is down?

I also see the variable [itex]w[/itex] in your calculations, but no explanation as to what it represents.
 
  • #3
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This problem description is not at all clear to me. For starters, what is 'the [itex]z[/itex] plane'? There is such a thing as a [itex]z[/itex] axis, but not a [itex]z[/itex] plane. Do you mean the plane defined by the equation [itex]z=0[/itex] (AKA the x-y plane)? Or do you mean the y-z plane (x=0) as your pictures would leave me to believe? Or perhaps you mean something else entirely?

Also, the 'top' and 'bottom' of the ring are not well defined. Which direction is up, and which direction is down?

I also see the variable [itex]w[/itex] in your calculations, but no explanation as to what it represents.

z plane refers to the z-y plane. w is the radius of the ring of charge. positive z direction current on top of ring and negative z direction current on bottom of ring (counterclockwise current flow)
 
  • #4
gabbagabbahey
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z plane refers to the z-y plane. w is the radius of the ring of charge. positive z direction current on top of ring and negative z direction current on bottom of ring (counterclockwise current flow)

Okay, and by 'top' you mean the point (0,w,0)?

[tex] d\vec{l} = <-dl sin\phi , dl cos \phi , 0>[/tex]

[itex]d\vec{l}[/itex] is an infinitesimal piece of the ring, right? If the ring is entirely in the y-z plane, why does your expression have a non-zero x-coordinate?

[tex] \hat{r} = <cos\theta , -w/r sin\phi , w/r cos\phi> [/tex]

Are you sure about the sign of the z-coordinate?

Also, I'm not sure why you would bother introducing a new variable [itex]\theta[/itex] here...why not just work with the variables given ion the problem statement [itex]x[/itex] and [itex]w[/itex]?

[tex]\vec{r}=<x,-w\sin\phi,-w\cos\phi>[/tex]

[tex]\implies r=\sqrt{x^2+w^2}[/tex]

[tex]\implies\hat{r}=<\frac{x}{\sqrt{x^2+w^2}},\frac{-w\sin\phi}{\sqrt{x^2+w^2}},\frac{-w\cos\phi}{\sqrt{x^2+w^2}}>[/tex]
 
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  • #5
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Okay, and by 'top' you mean the point (0,w,0)?

[itex]d\vec{l}[/itex] is an infinitesimal piece of the ring, right? If the ring is entirely in the y-z plane, why does your expression have a non-zero x-coordinate?
i think you're right



This doesn't look like a unit vector to me, its magnitude does not equal 1.

To properly find [itex]\hat{r}[/itex], Start by finding an expression for [itex]\vec{r}[/itex], and then divide by its magnitude.


it is a unit vector. try taking the absolute value of it...it comes out to 1
 
  • #6
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i've made the dl correction in the original post. THANKS. however, the math still comes out to be a bit off...i'm off by a half.
 
  • #7
gabbagabbahey
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it is a unit vector. try taking the absolute value of it...it comes out to 1

Yes, I realized after I posted that [itex]\cos^2\theta+\frac{w^2}{r^2}=1[/itex] from looking at your picture.
 
  • #8
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okay...it's been corrected...i'll check back in a couple of hours...need to be away from computer
 
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  • #9
gabbagabbahey
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Hmmm... looking at your picture it seems that you have defined [itex]\phi[/itex] such that [itex]z=w\cos\phi[/itex], but looking at you calculations it seems as though you are using [itex]z=-w\cos\phi[/itex]....which is it?
 
  • #10
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i got it....came out correctly...

thank you!
 

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