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whatisreality
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Homework Statement
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}##
Homework 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?
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