Two semi infinitely long straight current carrying conductors are held in the form as shown in figure. One common end of them is at the origin. If both the conductors carry same current I, find the value of the magnetic field induction at a point (a, b).

Answer

`(mu_0I)/(4piab)[(a+b)+sqrt(a^2+b^2)]`

Explanation

`B_1=(mu_0)/(4pi)I/b[sin theta_1+sinpi/2]=(mu_0I)/(4pib)[(a)/(sqrt(a^2+b^2))+1]`
Its direction is normally into the plane of paper.
Magnetic field at point P due to conductor II is
`B_2=(mu_0)/(4pi)I/a[sin theta_2+sin pi/2]=(mu_0I)/(4pia)[(b)/(sqrt(a^2+b^2)+1)]`
It is also acting normally into the plane of paper:
Resultant magnetic field at P.
`B=B_1+B_2`
`=(mu_0I)/(4pi)[(1)/(sqrt(a^2+b^2))(a/b+b/a)+(1/a+1/b)]`
`=(mu_0I)/(4pi)[(sqrt(a^2+b^2))/(ab)+(a+b)/(ab)]`
`=(mu_0I)/(4piab)[sqrt(a^2+b^2)+(a+b)]`