Figure shows three cases: in all case the circular part has radius r and straight ones are infinitely long. For the same current the ratio of field B at centre P in the three case `B_1:B_2:B_3` is

Options

A. `(-pi/2):(pi/2):((3pi)/4-1/2)`
B. `(-pi/2+1):(pi/2+1):((3pi)/4+1/2)`
C. `-pi/2:pi/2:3 pi/4`
D. `(-pi/2-1):(pi/2-1/4):((3pi)/4+1/2)`

Answer

A

Explanation

case I:

`B_(A)=(mu_(0))/(4pi).i/r ox`
`B_(B)=(mu_(0))/(4pi).(pii)/r o.`
`B_(C)=(mu_(0))/(4pi).i/r o.`
So net magnetic field at the centre of case `1`
`B_(1)=B_(B)-B_(C)-B_(A) implies B_(1)=(mu_(0))/(4pi).(pii)/(r ) o.......(i)`
Case `2:` As we discussed before magnetic field at the centre `O` in this case

`B_(2)=(mu_(0))/(4pi).(pii)/r ox.........(ii)`
Case `3: B_(A)=0`

`B_(B)=(mu_(0))/(4pi).((2pi-pi//2)i)/r ox`
`B_(C)=(mu_(0))/(4pi).i/r o.`
`=(mu_(0))/(4pi).(3pii)/(2r) ox`
So net magnetic field at the centre of case `3`
`B_(3)=(mu_(0))/(4pi).i/r((3pi)/2-1) ox .........(iii)`
From equations (i),(ii) and (iii)
`B_(1):B_(2):B_(3) pi o.:piox:((3pi)/2-1) ox`
`=-pi/2:pi/2:((3pi)/4-1/2)`