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Grade: Upto college level

                        

9 years ago

Answers : (2)

vikas askiitian expert
509 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

 

9 years ago
vikas askiitian expert
509 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

9 years ago
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