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gOlU g3n|[0]uS Grade: Upto college level
        

1568_21222_Untitled.jpg

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

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

6 years ago
vikas askiitian expert
510 Points
										

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


correct this , then ans will be final ans ..

6 years ago
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