The resistance of wire `ABC` is double of resistance of wire `ADC`. The magnetic field at `O` is

Explanation

(a)
`R_(ABC)=R_(1)=2R_(0),R_(ADC)=R_(2)=R_(0)`
`i_(1)=(R_(0))/(R_(0)+2R_(0)).i=i/3,i_(2)=i-i_(1)=(2i)/3`
`B_(1)=(mu_(0)i_(1))/(2R).1/2=(mu_(0)i)/(12R), ox`
`B_(2)=(mu_(0)i_(2))/(2R).1/2=(mu_(0)i)/(6R), o.`
`B_(O)=B_(2)-B_(1)=(mu_(0)i)/(12R),o.`
(b)
`R_(ABC)=R_(1)=R_(0),R_(ADC)=R_(2)=R_(0)`
`i_(1)=(2R_(0))/(R_(0)+2R_(0))i=(2i)/3, i_(2)=i-i_(1)=i/3`
`B_(1)=(mu_(0)i_(1))/(4pid)(cos45^(@)+cos45^(@))=(mu_(0)i_(1))/(2sqrt(2)pid)`
`=(mu_(0)(2i//3))/(2sqrt(2)pi(a/2))=(2mu_(0)i)/(3sqrt(2)pia)=B_(2), ox`
`B_(3)=(mu_(0)i_(2))/(4pid)(cos45^(@)+cos45^(@))=(mu_(0)i_(2))/(2sqrt(2)pid)`
`=(mu_(0)(i//3))/(2sqrt(2)pi(a/2))=(mu_(0)i)/(3sqrt(2)pia)=B_(4), o.`
`B_(O)=(B_(1)+B_(2))-(B_(3)+B_(4))`
`(4mu_(0)i)/(3sqrt(2)pia)-(2mu_(0)i)/(3sqrt(2)pia)`
`=(sqrt(2)mu_(0)i)/(3pia), ox`
(c)
Magnetic field at `O`:
Due to the wire `AB:`
`d_(1)=L//4, cos alpha=(L//2)/(sqrt((L//2)^(2)+(L//4)^(2)))=2/(sqrt(5))`
`B_(1)=(mu_(0)(i//2))/(4pid_(1))[cos alpha+cos alpha]`
`=(mu_(0)(i//2))/(4pi(L//4))xx2xx2/(sqrt(5))=(2mu_(0)i)/(sqrt(5)piL),o.`
Due to the wire `DC`
`d_(2)=3L//4, cos beta=(L//2)/(sqrt((L//2)^(2)+(L//4)^(2)))=2/(sqrt(13))`
`B_(2)=(mu_(0)(i//2))/(4pid_(2))[cos beta+cos beta]`
`(mu_(0)(i//2))/(4pi.(3L//4))xx2ss2/(sqrt(13))=(2mu_(0)i)/(3sqrt(13)piL),ox`
Magnetic field due to `AD` is equal and opposite to `BC`
`B_(O)=B_(1)-B_(2)=(2mu_(0)i)/(piL)(1/(sqrt(5))-1/(2sqrt(13))), o.`