## Guest

Khimraj
3007 Points
4 years ago
Magnetic field due to a wire of finite length is given by
B = (uoI/4$\pi$r)(sin$\Theta 1$ + sin$\Theta 2$)
where $\Theta 2$ is angle QOM
and $\Theta 1$ is angle POM
r is perpendicular distance
So B = (uoI/4$\pi$a)(5a/3 + 5a/3)
So B = (5/6)uoI/$\pi$a.
Khimraj
3007 Points
4 years ago
Magnetic field due to a wire of finite length is given by
B = (uoI/4$\pi$r)(sin$\Theta 1$ + sin$\Theta 2$)
where $\Theta 2$ is angle QOM
and $\Theta 1$ is angle POM
r is perpendicular distance
So B = (uoI/4$\pi$a)(3/5 + 4/5)
So B = (7/20)uoI/$\pi$a.