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drop perpendicular from P to line which is below OP ...let this perpendicular cut this line at Q ....
now we have right angled triangle POQ in which ,
POQ = @ , PQO = 90o & OPQ = 90-@
length of this perpendicular = xsin@
now , we have formula for magnetic field due to current carrieng straight wire (B) = (uoI/4pia)[sin@1+sin@2]
here , a = perpendicular distance
@1 , @2 are angles substended with perpendicular line to end points of wire ...
@1 = @ (from left end it is making @)
@2 = 90 (right end is at infinity )
now , B = (uoI/4pi(xsin@) ) . [sin@+1]
direction of this field is in -ve Z axis ...
in same manner , B due to another straight wire is same ...
total magnetic field = 2B
= (uoI/2pi(xsin@))[1+sin@] (-k)
this is the required ans
@1 = 90-@ , by mistake i have written @
correct this , then ans will be final ans ..
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