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Homework Help: Infinitely Long Wire with Loop (Magnetic Field)

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data

    The wires below are infinitely long and some of them are with loops and semi loops. The current I is constant. What is the magnetic field in point O, when R is the distance from it?

    In F, E and H, the magnetic field exerted by the terminal infinitely long wire is 0 because it is 180 degrees?
    If the loop is a semi-circle, I use B/2? If the loop is a quarter of a circle, then B/4 and so on?

    How to solve for G?

    2. Relevant equations

    B= μI/2R

    3. The attempt at a solution

    I included in my calculations the magnetic fields of both the infinite wire and the loop.

    A: μI/2R Outwards
    B: μI/2R + 2μI/4piR Outwards
    C: μI/2R - 2μI/4piR Outwards
    D: μI/2R - μI/2R +2μI/8piR = 2μI/8piR Inwards
    E: 2μI/8piR Inwards
    F: μI/2R + 2μI/16piR Inwards
    G: ?
    H: μI/2R + 2μI/8piR + μI/2R Inwards

    Attached Files:

  2. jcsd
  3. Mar 23, 2014 #2

    Simon Bridge

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    Homework Helper

    You are asked to find the field at point O right?
    Then yes: EF and H have sections of wire that do not contribute.
    If B is the field of a complete loop, then an 1/nth of a loop contributes B/n.

    You have equations:
    BL = μI/2R <---<<< for the field center of a current loop radius R
    BW =2μI/4piR=μI/(2piR) <---<<< the field radial distance R from infinite straight wire

    This makes for a nice way to summarize your answers to make double-checking easy:

    If we make BW=B, then BL=pi.B

    Using + to indicate "into the page"
    Your answers are:

    A: -B
    B: -(pi+1)B (what is the total current at the place the loop and the wire meet?)
    C: -(pi-1)B (there is another difference between C and B)
    D: +(pi-pi+1/2)B = B/2 (are you saying the two wires cancel out? Check by RH rule.)
    E: +B/2 (which is half that due to a wire - does that make sense?)
    F: +(pi+1/4)B
    G: ?
    H: +(pi + 1/2 +pi)

    ... I think one of us got confused between the contributions of the wire and that of the loop-section.


    $$B_L=\frac{\mu_0 I}{2R}=\pi\frac{\mu_0 I}{2\pi R}=\pi B_W$$

    For some of them, only half the wire contributes to the field.
    Did you check that the final direction you gave is that for the net field - i.e. a negative magnitude into the page is a positive magnitude out of the page - so the net direction is "outwards".

    For G - it's the same as the others: what's the problem?
    Last edited: Mar 23, 2014
  4. Mar 24, 2014 #3
    Thank you for your help!
    I confused the two equations, sorry.

    My answers now:
    A: -B
    B: -(pi+1)B The current is NI, which is 2I? So the loop’s magnetic field is μ2I/2R= μI/R.
    C: +(+pi-1)B The wires don’t meet, so the current stays the same?
    D: +(2+1/2pi)B
    E: +pi.B/2
    F: +(1/4pi+1)B
    G: +(2+1/2pi)B
    H: +(1 +1/2pi)B

    yes, these are all net directions, assuming I'm correct.
    Last edited: Mar 24, 2014
  5. Mar 25, 2014 #4

    Simon Bridge

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    It's interesting about the gap in C isn't it?
    It may just be that the gap is supposed to be invisibly small and shown large for emphasis.
    I think you have the right of it: one subtracts the wire while the other adds it.

    Sometimes you have to account for gaps by subtracting the field that would otherwise be present.

    You seem to have a handle on what's required enough to check your own math - well done.
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