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

1016_21222_12.jpg  

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

ans of  figure 1


magnetic field due to any arc = uoI@/4pia   .............1


@ is angle substended at the point , a = radus of arc ..


 


magnetic field due to straight wire = (uoI/4pia).[sin@1+sin@2]    ...........2


 


angle @ substended by circular arc = (2pi - 2@)


Barc (field due to arc) = uoI(2pi-2@)/4piR


                             = uoI(pi-@)/2piR                ...................3


 


Bline(field due to straight wire) = (uoI/4piRcos@). [ 2sin@ ]           ...................4


perpendicula distance = Rcos@


total magnetic field will be summation of these two ,


B(total) = uoI/2piR. [ pi-@ + tan@]


 


this is the required ans


 

6 years ago
vikas askiitian expert
510 Points
										

in figure 2


magnetic field due to arc whose radius is a  will be


Ba = uoI(2pi-@)/4pi(a)   


 magnetic field due to arc whose radius is b will be


Bb = uoI(@)/4pib


 


net magnetic field = B = sum of both


         B = uoI/4pi .  [ (2pi-@)/a + @/b ]     


 


this is the required ans

6 years ago
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