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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
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 ]
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