A large straight current carrying conductor is bent in the form of Lshape. Find B at P

Explanation

Let us divide the conductor into two semi infinite segments 1 and 2.
Then, induction at P is
`vecB=vecB_(1)+vecB_(2)`
`vecB_(1)=(mu_(0)i)/(4pia)(sin(90-theta_(1))+sin90)hatk` ……(ii)
`vecB_(2)=(mu_(0)i)/(4pia)(sin(90-theta_(2))+sin90)hatk` ….(iii)
then `vecB=(mu_(0)i)/(4pia)(costheta_(1)+costheta_(2)+2)hatk`,
where `costheta_(1)=costheta_(2)=1/(sqrt2)`
Hence, `vecB=(2+sqrt2)(mu_(0)ihatk)/(4pia)`