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the magnetic field B at the centre of circular coil of radius r is π times that due to a long straight wire at a distance r from it, for equal currents. The adjoining diagram shown three cases in all cases the circular part has radius r and straight one are infinitely long. For the same current the field B at the centre P in cases 1,2,3 has the ratio

Asked by Prashant DIGHE 7th December 2019, 10:15 PM
Answered by Expert
In the above configuration (1) , magnetic field at P has three contributions,
(i) due to semi infinite wire left of p, (ii) due to semi circle, (iii) due to semi-infinite wire at right side of p.
magnetic field due to left side semi-infinite wire and magnetic field due to right side semi-infinite wire cancel each other.
Hence net field due to semi circle = μo i / (4 r )
(Let us consider counterclockwise direction of current in loop gives +ve direction magnetic field )
In the above configuration, magenetic field at P due to infinite wires is zero.
Semi-circle part of wire gives magnetic field -(μo i) / (4 r )
In the above configuration, semi-finite wire at left side of P does not contribute magnetic field.
magnetic field at P due to three-quarter-circle gives magnetic field  -(3/4)(μo i) / (2r)
magnetic field at P due to semi-infinite wire at right side = (μo i) / (4πr)
Hence net magnetic field at P = - [ (3/4)(μo i) / (2r) ]+  [ (μo i) / (4πr) ]
Hence ratio of magnetic field of three configuration = μo i / (4 r ) : -(μo i) / (4 r ) : - [ (3/4)(μo i) / (2r) ]+  [ (μo i) / (4πr) ]
Above ratio can be simplified to get ,  ( -π/2 ) : (π/2) : (3π/4) - (1/2)
Answered by Expert 8th December 2019, 11:28 AM
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