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Question regarding the Biot-Savart Law and a Circular Loop

  1. Nov 19, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    The Biot-Savart Law states that the direction and magnitude of the magnetic field B
    produced at a point P by a current length element I dl is given by:

    [tex] d\vec{B} = \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \hat{r}}{r^2} [/tex]

    Where [tex] \hat{r} = \frac{\vec{r}}{r} [/tex]

    where r is the position vector from the current length element to the point P.

    (i)

    A circular loop of wire of radius R lies in the xy-plane and is centred at the origin.
    The wire carries a steady current I, which is seen to be flowing clock-wise round the
    loop when looking along the z-direction from the negative z towards the positive z.
    Calculate the magnetic field at a distance Z on the axis of the circular loop and show
    that the z-component of the field is given by:

    [tex] B_z = \frac{\mu_oIR^2}{2(Z^2+R^2)^{3/2}} [/tex]

    (ii)

    A single-turn, circular loop of radius 10 cm is to produce a field at its centre that
    will just cancel the earth’s magnetic field at the equator, which is 0.7 G directed
    north. Find the current in the loop (1 T ≡ 104 Gauss).

    (iii)

    Make a sketch, showing the orientation of the loop and the current for part (ii).

    2. Relevant equations

    Given in question

    3. The attempt at a solution

    Okay I'm doing part (i)

    so far I have inserted R hat to give:

    [tex] dB = \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times\vec{r}}{r^3} [/tex]

    to get B I need to integrate:

    [tex] B = \int \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times\vec{r}}{r^3} [/tex]

    and I can take out a few items:

    [tex] B = \frac{\mu_0}{4\pi} \int \frac{Id\vec{l}\times\vec{r}}{r^3} [/tex]

    But I am not sure where to go now.

    Could anyone give me any suggestions?

    ???

    TFM
     
  2. jcsd
  3. Nov 19, 2008 #2

    gabbagabbahey

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    Find relevant expressions for [itex]\vec{r}[/itex], [itex]r^3[/itex] and [itex]\vec{dl}[/itex]
     
  4. Nov 19, 2008 #3

    TFM

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    Okay. Well, firstly:

    [tex] d\vec{l} [/tex] is a line integral

    [tex] r^3 [/tex] is the distance from the point on the wire to where the Magnetic field is being measured

    [tex] \vec{r} [/tex] is a vector, but I am not sure of what it represents...?

    ???

    TFM
     
  5. Nov 19, 2008 #4

    gabbagabbahey

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    You said it yourself in your first post:

    draw a picture, and try to express this in terms of cylyndrical coordinates and unit vectors....remember, since you are trying to find the field at "a distance Z on the axis of the circular loop", your point P is at [itex](r, \phi, z)=(0,0,Z)[/itex]
     
  6. Nov 19, 2008 #5

    TFM

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    Is this the right diagram:

    I think it should be the second one, since you stated:

    ???

    TFM
     

    Attached Files:

  7. Nov 19, 2008 #6

    gabbagabbahey

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    The second one is close, but you have mislabeled 'r'...the radius of the circular path of the wire is capital 'R', little 'r' should represent the distance form P to an infinitesimal length of the wire dl....you should shade in a short section of the wire and call it 'dl'...then find an expression for the vector from 'dl' to 'P'...
     
  8. Nov 20, 2008 #7

    TFM

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    Okay so firstlty I have the midifed attached image (Its a JPEG, hence the reason the filling in is poor)

    so I need an expression for the vector from dl to P. would it not be [tex] \hat{z} [/tex], since the point P is on the Z axis, and so the vector would go from the centre of the loop to P, which is just up along the Z-Axis

    ???

    TFM
     

    Attached Files:

  9. Nov 20, 2008 #8

    gabbagabbahey

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    P is on the z-axis, but dl is not!...So how could the vector [itex]\hat{z}[/itex] possibly go from dl to P?
     
  10. Nov 20, 2008 #9

    TFM

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    I am slightly confused...Where is dl from?

    TFM
     
  11. Nov 20, 2008 #10

    gabbagabbahey

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    dl is an infinitesimal length of the wire.....where is the wire? Pick a tiny section of the wire and call it dl....what are it's coordinates?
     
  12. Nov 20, 2008 #11

    TFM

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    I see now, the wire is a loop around the x-axis, not up down it. so a point on the z axis would be [tex] R^2 + Z^2 [\tex] from the wire?

    TFM
     
  13. Nov 20, 2008 #12

    gabbagabbahey

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    [tex]\sqrt{R^2+Z^2}[/tex] gives you the length of the vector [itex]\vec{r}[/itex], but you also need to know its direction...draw an arrow from dl to P...which way does it point?
     
  14. Nov 20, 2008 #13

    TFM

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    Its pointing towards the Z-axis, but at an angle towards it, not parallel

    ???

    TFM
     
  15. Nov 21, 2008 #14

    gabbagabbahey

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    Okay, call that angle [itex]\alpha[/itex]; then the direction of [itex]\vec{r}[/itex] is [itex]-\cos \alpha \hat{r}+\sin \alpha \hat{z}[/itex] Is it not?...now look at the triangle, what are [itex]\cos \alpha[/itex] and [itex]\sin \alpha[/itex] in terms of R and Z?
     
  16. Nov 21, 2008 #15

    TFM

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    [tex] cos \alpha =\frac{Adjacent}{Hypotenuse} [/tex]

    [tex] sin \alpha =\frac{Opposite}{Hypotenuse} [/tex]

    Hypotenuse will be the lkine from I to P, and the adjacent will be R, opposite will be Z

    TFM
     
  17. Nov 21, 2008 #16

    gabbagabbahey

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    Yes, but isn't the length of the line from dl to P [itex]\sqrt{Z^2+R^2}[/itex]?...what does that give you for [itex]\vec{r}[/itex]?
     
  18. Nov 21, 2008 #17

    TFM

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    Well:

    [tex] cos\alpha = \frac{R}{\sqrt{Z^2 + R^2}} [/tex]

    and

    [tex] sin\alpha = \frac{Z}{\sqrt{Z^2 + R^2}} [/tex]

    ???

    TFM
     
  19. Nov 22, 2008 #18

    gabbagabbahey

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    Yes, and what does that make [itex]\vec{r}[/itex]?
     
  20. Nov 22, 2008 #19

    TFM

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    Well Since:

    [tex] \hat{r} = -\cos \alpha \hat{r}+\sin \alpha \hat{z} [/tex]

    that would make:

    [tex] \hat{r} = -\frac{R}{\sqrt{R^2 + Z^2}} \hat{r} + \frac{Z}{\sqrt{R^2 + Z^2}} \hat{z} [/tex]

    ???

    TFM
     
  21. Nov 22, 2008 #20

    gabbagabbahey

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    Uggh...let's switch to [itex]\hat{s}[/itex] for the cylindrical unit vector so that yo don't confuse it with [itex]\hat{r}[/itex] which is the direction of the vector from dl to P....then you have:

    [tex]\hat{r} = -\frac{R}{\sqrt{R^2 + Z^2}} \hat{s} + \frac{Z}{\sqrt{R^2 + Z^2}} \hat{z}[/tex]

    And since you've already found the magnitude of [itex]\vec{r}[/itex] to be [tex]{\sqrt{R^2 + Z^2}[/tex]...that means the vector [itex]\vec{r}[/itex] is....?
     
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