How Does a Bent Wire Affect the Magnetic Field at a Point?

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SUMMARY

The discussion focuses on calculating the magnetic field produced by two segments of wire carrying a 27.0A current, each measuring 2.00mm, positioned 3.00cm from a bend. The relevant equation for this calculation is ΔB = (μ₀/4π)(IΔLsinθ)/R², where μ₀ is the permeability of free space, I is the current, ΔL is the length of the wire segment, θ is the angle, and R is the distance to the point of interest. The final computed magnetic field at point P, located midway between the two wire segments, is approximately 16.968μT.

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Homework Statement



A wire carrying a 27.0A current bends through a right angle. Consider two 2.00mm segments of wire, each 3.00cm from the bend.

YF-28-13.jpg


Find the magnitude of the magnetic field these two segments produce at point P, which is midway between them.


Homework Equations



F = IL x B


The Attempt at a Solution



B = F / IL
= F / 27 x 0.004 m (both 2mm segments combined)

once again stuck on how to find F, and unsure if i am even using the correct equation. where does the 3cm from the bend bit come into play if at all?
 
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Relevant equations

F = IL x B
It is the force acting on a current carrying conductor in a magnetic field.
The correct equation is . deltaB = (mu)o/4pi{I*delta L*sin(theta)}/R^2
where (theta ) is the angle between (delta L) and line joining the point P and mid point of delta L and R is the distance between delta L and P.
 
Last edited:
ΔB = μo/4π(IΔLsinθ)/R^2

so I = 27
θ = 45
ΔL = 2mm?
and R can be found using trig.

so what is μ and o??

soz just really lost. any help would be great.
 
Field due to two elements =ΔB = 2[μo/4π(IΔLsinθ)/R^2]
= 2[10^-7*27*2x10^-3*0.707/(1.414*1.5*10^-2)^2]
= 16.968x10^-6T
 

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