Magnetic field of a bent wire and Biot Savart law

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SUMMARY

The discussion focuses on calculating the magnetic field (B) at the origin (0,0,z) due to a bent wire carrying current I, using the Biot-Savart law. The wire consists of two segments: one along the positive y-axis and the other along the positive x-axis. The magnetic field due to the y-axis segment is derived as B = - (μ₀ I) / (4 π z), indicating a magnitude half that of an infinite wire. The direction of the magnetic field is confirmed to be in the -x direction, as determined by the right-hand rule, despite initial confusion regarding the negative sign from the integration.

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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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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
 
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Brilliant, thanks!
 

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