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Grade:Upto college level

2 Answers

vikas askiitian expert
509 Points
10 years ago

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

vikas askiitian expert
509 Points
10 years ago

@1 = 90-@ , by mistake i have written @

correct this , then ans will be final ans ..

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