A long wire carrying a current `i` is bent to form a plane angle `theta`. The magnetic field at a point on the bisector of this angle situated at a distance `d` forms vertex is

Explanation

Two straight parts of the system will create the same magnetic field in the same direction and hence we can multiply the result for one half of the system by two.

Magnetic field intensity due to a straight wire of finite length can be written as follows:
` B = (mu_(0)I)/(4 pi d) (sin alpha + sin beta) ` ....(i)
In the given situation as per figure `alpha = 90^(@) - theta " and " beta = 90^(@)` due to infinite length on one side.
Here, the value of d can be calculated from triangle OPQ.
`sin theta = d/x rArr d = x sin theta`
We can substitute these values in equation (i) and multiply the result with 2 to compensate for the second half of the system. Hence, the net magnetic field intensity can be written as follows:
`B = 2 xx (mu_(0)I)/(4 pi x sin theta) [sin (90^(@) - theta) + sin 90^(@)]`
`rArr" " B = (mu_(0)I)/(2 pi x sin theta) [ cos theta + 1] = (mu_(0)I)/(2 pi x) ((1 + cos theta)/(sin theta))`
`rArr" " B = (mu_(0)I)/(2 pi x) (2 cos^(2) . theta/2)/(2 sin. theta/2 cos. theta/2) = ( mu_(0)I)/ (2 pi x ) cos . theta/2 `