# Current carrying wire

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

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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rl.bhat
Homework Helper
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.

rl.bhat
Homework Helper
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