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Magnetic field of a bent wire

  1. Nov 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A straight wire carrying current I goes down the positive y-axis ( I is also in this direction) from infinite y to the origin. At the origin it changes direction 90 degrees and goes along positive x to infinity. Find B at (0,0,z).
    I'm given that
    ##\int_{0}^{\infty} \frac{dx}{(x^2+y^2)^{\frac{3}{2}}} = \frac{1}{z^2}##

    2. Relevant equations
    dB = ##\frac{\mu_0 I}{4 \pi} \frac{dl sin(\theta)}{r^2}##
    This has been simplified from the Biot Savart law by subbing in unit vector r = r/r, and taking the magnitude of the cross product. ##\theta## is the angle between dl and r.
    3. The attempt at a solution

    I'm having trouble working out the direction. I split the wire into two segments, calculating B for the segment on the y axis first, then for the segment on the x axis, then adding them.
    For segment along y axis:
    dl is -dy (or is it just dy?), sin(##\theta##) = ##\frac{z}{\sqrt{y^2+z^2}}## and r = ##\sqrt{y^2+z^2}##:
    dB = ##- \frac{\mu_0 I}{4 \pi} \frac{z dy}{(y^2+z^2)^{\frac{3}{2}}}##
    = ##- \frac{\mu_0 I z}{4 \pi} \frac{dy}{(y^2+z^2)^{\frac{3}{2}}}##
    Integrate dB between 0 and infinity using the given result for the integral:

    B = ##- \frac{\mu_0 I }{4 \pi} \frac{1}{z}##
    Happy with the magnitude, half that of an infinite wire as expected.

    Ignoring the other segment for now, what do I do about the direction?? From the right hand rule, it's in the -x direction, but if I multiply by the negative unit vector in x then the negatives cancel, and it's actually in the positive x direction, which is wrong! Isn't it?
    Last edited: Nov 11, 2015
  2. jcsd
  3. Nov 11, 2015 #2
    You are right about the field direction being in the - x direction. Your mistake is in taking the negative sign at the end of your integration seriously. When you write the magnitude of a cross product;

    C = AB sinθ, that is the magnitude. It is always positive. The direction of the vector is completely determined by the right hand rule.

    What you should say is:
    Even though the integral gave you a negative sign, we want only the magnitude of B from that integration. The direction is determined to be - x from the right hand rule
  4. Nov 11, 2015 #3
    Brilliant, thanks!
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