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Current carrying wire

  • Thread starter EvanQ
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
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:
  • #3
56
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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.
 
  • #4
rl.bhat
Homework Helper
4,433
7
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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