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Magnetic Field

  1. Nov 21, 2005 #1
    a) A long wire is bent into shape as shown in the figure without he wire actually tocuhing itself. The radius of the circular sectio nis R. Determine the magnitude and direction of B at the cneter C of the circular section when the cirrent i is as indicated.

    Well i think that due to the cirular loop the magnetic field is giuven by

    [tex] B = \frac{\mu_{0} i }{4 \pi} \int \frac{ds \cross R}{R^3} [/tex]
    since R is perpendicular to ds, and constant valued
    [tex] B = \frac{\mu_{0} i }{4 \pi R^2} 2 \pi R = \frac{\mu_{0} i }{2R} [/tex]
    This field would point out of the page because of ds cross r.
    im not qute sure about the linear part though. Wouldnt some component of the linear part affect the magnetic field at C?

    Suppose the circular section of the wire is rotated without ditortion about hte indicated diameter until the plane of the circle is perpendicular to the straight section of the wire. The mangetic dipole moment associated with the circular section is now in the direction of the current in teh straight section of the wire. Determine B at C in this case

    I m not quite sure about this at all. For starters teh B field would point left.
    The linear part would affect the answer in this one wouldnt it ? WOuld it be the same answer as above minus the B field due to a lone straight wire that is
    [tex] B = \frac{\mu_{0} i }{2R} - \frac{\mu_{0} i}{2 \pi R} [/tex]
    is this correct? Please advise me on what i may have done wrong.

    Thank you for your help!
     

    Attached Files:

  2. jcsd
  3. Nov 22, 2005 #2
    im not quite sure what is going on from that picture, but recall this. for any point that lies along the direction of the flow of current (but not in the current) the vector quantity dl and the r vector (the two you have to take a cross product of to get the magnetic field) are either parallel or anit-parallel right? then whats the cross product and the contribution to the B field?
     
  4. Nov 22, 2005 #3
    so what does that mean ... the poin C is either parallel or perpendicular to the current i in the linear part of the wire?
     
  5. Nov 22, 2005 #4

    mezarashi

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    Homework Helper

    By the right hand rule, the magnetic field due to both the straight and the loop portions of the wire will be out of the page.

    You need to apply the Biot-Savart law at the loop section. For the infinite straight wire sectin, you may use Ampere's Law.

    In the second part, the magnetic field due to the loop will be towards the right, while due to the infinite wire portion remains out of the page. Using vector addition, you can add them together to find a resultant B.
     
  6. Nov 22, 2005 #5
    ok so for the loop section the magnetic field is [tex] \frac{\mu_{0} i}{2R} [/tex]
    but for the linear part Ampere's law yields [tex] \frac{\mu_{0} i}{2 \pi R} [/tex] simply adding these two is the magnetic field? I mena they both point in the same direction? Is this correct?

    For the rotated loop
    the loop would have the same vlaue but hte vector component would be j hat if we allowed j to point right and the magnitude of the linear part is the same magnitude but in the k hat direction if k points out of the page?
     
  7. Nov 22, 2005 #6

    mezarashi

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    What does your right hand rule say? You should be familiar with it enough to convince yourself.

    Yes.
     
  8. Nov 22, 2005 #7
    right hand rule for the linear part is straight out of teh page
    even fro the loop part the the magnetic field points out of the page

    thank you for your help so far... i know im kinda slow so thank you for your patience as well
     
  9. Nov 23, 2005 #8
    what you need is a sum of two fields:
    first field due to the straight wire and this field is perpendicular out of the page towards your eye which is your second term (mu,zero).I/2.pi.R.
    the field due to loop which is your first term
    that is (mu ,zero).I/2R.
    This second field is again perpendicular out of the page.
    Thats why you need their sum as they are vectors and directed out of the page.
    so the minus sign seems not o.k.
    much luck
    from the Netherlands
     
  10. Nov 23, 2005 #9
    what you need is a sum of two fields:
    first field due to the straight wire and this field is perpendicular out of the page towards your eye which is your second term (mu,zero).I/2.pi.R.
    the field due to loop which is your first term
    that is (mu ,zero).I/2R.
    This second field is again perpendicular out of the page.
    Thats why you need their sum as they are vectors and directed out of the page.
    so the minus sign seems not o.k.
    much luck
    from the Netherlands
     
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