(a) (i)
At `O:`
Wire `①` is in `x-y` plane.
Magnetic field due to this wire be `bot^(ar)` to plane `x-y` at `O`.
`B_(1)=(mu_(0)i)/(4piR)`, along `z`-axis
Wire `②` is in `y-z` plane.
`B_(2)=(mu_(0)i)/(4R)`, along `-x`-axis
`B_(3)=(mu_(0)i)/(4piR)`, along `-z`-axis
`vec(B)_(O)=-(B_(1)+B_(3)) hatk-B_(2)hati`
`=-(mu_(0)i)/(4R) hati-(mu_(0)i)/(2 pi R) hat(k)`
`|vec(B)|= B_(o) = (mu_(0)i)/(4R) sqrt(1 + ((2)/(pi))^(2))`
(ii)
`B_(1)=(mu_(0)i)/(4piR)`, along `-z` axis
`B_(2)=(mu_(0)i)/(4R)`, along `-x` axis
`B_(3)=(mu_(0)i)/(4piR)`, along `-x` axis
`vec(B)_(O)=-(B_(2)+B_(3)) hati-B_(3)hatk`
`=-((mu_(0)i)/(4R)+(mu_(0)i)/(4piR)) hati-(mu_(0))/(4piR) hatk`
`|vec(B)_(O)|=B_(O)=(mu_(0)i)/(4R) sqrt((1+1/(pi))^(2)+(1/(pi))^(2))`
(iii)
`B_(1)=(mu_(0)i)/(4piR)`, along `-z`direction
`B_(2)=0`, current is divided in two parts and magnetic field due to these parts is zero.
`B_(3)=(mu_(0)i)/(4piR)`, along `-y` direction
`vec(B)_(O)=-B_(3)hatj-B_(1)hatk`
`=-(mu_(0)i)/(4piR) hatj-(mu_(0)i)/(4piR) hat k`
`|vec(B)_(O)|=B_(O)=(sqrt(2)mu_(0)i)/(4piR)=(mu_(0)i)/(2sqrt(2)piR)`
Due to long wire, magnetic field at `P` is
`B=(mu_(0)I)/(2piR)`
`vec(B)=B_(x)hati-B_(y) hatj`
`=B sin thetahati-B cos theta hatj`
`=B(y/r hati-x/r hatj)`
`=(mu_(0)I)/(2pir)(y/r hati-x/r hatj)`
`=(mu_(0)I)/(2pir^(2))(y hati-x hatj)`
`=(mu_(0)I)/(2pi(x^(2)+y^(2)))(yhati-xhatj)`
(c) There can be two closed loops.
Magnetic field at `P` due to loop, since is anticlockwise, hence magnetic field will be towards the observer i.e. along `=x` axis
`vec(B)_(1)=Bhatk`
where `B`: magnitude of magnetic field due to each loop.
`vec(B)_(P)=vec(B)_(1)+vec(B)_(2)=Bhati+Bhatk`
Unit vector along magnetic field at `P`
`hat(B)_(P)=(vec(B)_(P))/(|vec(B)_(P)|)=(Bhati+Bhatk)/(sqrt(2)B)=1/(sqrt(2))(hati+hatk)`.