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

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AI Thread Summary
The discussion focuses on calculating the magnetic field at point P due to two segments of a bent wire carrying a 27.0A current. The relevant equation for the magnetic field produced by a current-carrying conductor is provided, specifically ΔB = (μo/4π)(IΔLsinθ)/R^2. Participants express confusion about determining the force (F) and the significance of the distance from the bend, which is 3cm. The calculation involves using trigonometry to find the distance R and the angle θ, leading to a final magnetic field value of approximately 16.968μT. Clarification on the symbols used in the equations, particularly μ and o, is also sought to aid understanding.
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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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