(i) A very long wire carrying a current I is bent at right angles .Find magnetic field at a point lying on a perpendicular to the wire , drawn through the point of bending at a distance d from it
(ii) Three long wires carrying same currect are placed as shown Find magnetic field at O.

Explanation

Magnetic field at point `P` at height `d` above point `O`
Due to wire `①`, `B_(1)=(mu_(0)i)/(4pid)`, towards `+x`-axis
Due to wire `②`, `B_(2)=(mu_(0)i)/(4pid)`, towards `-y`-axis
`B_(1) is bot^(ar)` to `B_(2)`

`B_(P)=sqrt(2) B=(sqrt(2)mu_(0)i)/(4pid)`
`=(mu_(0)i)/(2sqrt(2) pid)`
(b)
At `P:`
Due to wire `①`, `B_(1)=(mu_(0)i)/(2pisqrt(2)d)`
Due to wire `②`, `B_(2)=(mu_(0)i)/(2pid)`
Due to wire `③`, `B_(3)=(mu_(0)i)/(2pisqrt(2)d)=B_(1)`

Resultant of `B_(1)` and `B_(3)` is
`B'=2B_(1)cos 45^(@)`
`=(mu_(0)i)/(2pid)`

`B'=B_(2)`
`B_(P)=0`
(c)
At point `B:`

At `B:`
Due to wire `A,B_(A)=(mu_(0)i)/(2pi(3d))`, along `BC`
Due to wire `C,B_(C)=(mu_(0)i)/(2pi(4d))`, along `AB`
Resultant magnetic field at `B`,
`B=sqrt(B_(A)^(2)+B_(C)^(2))=(mu_(0)i)/(2pid) sqrt(1/(3^(2))+1/(4^(2)))`
`=(mu_(0)i)/(2pid). 5/(3xx4)=5/24 (mu_(0)i)/(pid)`
`tan theta =(B_(C))/(B_(A))=3/4 implies theta=tan^(-1)(3//4)`
At point `C:`

At `C:`
Due to wire, `A,B_(A)=(mu_(0)i)/(4pi(5d))`
Due to wire, `A,B_(B)=(mu_(0)i)/(2pi(4d))`
Resulatant magnetic field at `C`
`B=sqrt(B_(A)^(2)+B_(B)^(2)+2.B_(A).B_(B) cos alpha)`
`=(mu_(0)i)/(2pid) sqrt(1/(5^(2))+1/(4^(2))+2xx1/5xx1/4xx4/5)`
`=(sqrt(73)mu_(0)i)/(40 pid)`
`tan theta=(B_(B) sin alpha)/(B_(A)+B_(B)cos alpha)=(1/4xx3/5)/(1/5+1/4xx4/5)=3/8`
`theta=tan^(-1) (3//8)`