Two long parallel wires are carrying currents as shown

Find magnitude and direction of magnetic field at `P,Q` and `R`.
(b) (i)
(ii)
At what distance from left wire, magnetic field is zero on the line joining the wires.
Find magnitude of magnetic field at `P`.
(d) Two straight infinitely long and thin parallel wires are spaced `d` distance apart and carry a current `i` each. find the magnetic field at a point distance `d` from both wires when the currents are in the `(i)` same and `(ii)` opposite directions.

Explanation

(a)
At `P:`
Due to wire `①`, `B_(1)=(mu_(0)i)/(2pid)`, upward
Due to wire `②`, `B_(2)=(mu_(0)i)/(2pi(3d))` downward
`B_(P)=B_(1)-B_(2)=(mu_(0)i)/(2pid)(1-1/3)=(mu_(0)i)/(3pid)`, upward
At `Q:`
Due to wire `①` ,`B_(1)=(mu_(0)i)/(2pi(3d))`, downward
Due to wire `②`, `B_(2)=(mu_(0)i)/(2pid)`, downward
`B_(Q)=B_(1)+B_(2)=(mu_(0)i)/(pid)`, downward
At `R:`
Due to wire `①` , `B_(1)=((mu_(0)i)/(2pi(3d)))`, downward
Due to wire `②` , `B_(2)=(mu_(0)i)/(2pid)`, upward
`B_(R) =B_(2)-B_(1)=(mu_(0)i)/(2pid)(1-(1)/(3))=(mu_(0)i)/(3pid)`, upward
(b) (i)
Let magnetic field is zero at distance `x`, left wire
At `P:`
`B_(1)=B_(2)`
`(mu_(0)i)/(2pix)=(mu_(0)(i))/(2pi(d-x)) implies 4/x =1/(d-x) implies x=(4d)/5`
(ii)

The magnetic field won't be zero between the wires because due to both wires magnetic field will be upward. The magnetic field will be zero nearer the smaller current outside `AB`. Let it be zero at distance `x` from wire `B` as shown.
At `P:`
Due to `①`, `B_(1)=(mu_(0)(4i))/(2pi(d+x))`, upward
Due to `②` `B_(2)=(mu_(0)(i))/(2pix)`, downward
`B_(P)=0 i.e., B_(1)=B_(2)`
`=(mu_(0)(4i))/(2pi(d+x))=(mu_(0)(i))/(2pix) implies 4/(d+x)=1/x`
`x=d/3`
Magnetic field is zero at distance `d=d//3=4d//3` from left wire.
(iii)
`angleAPB=90^(@)`
At `P:`
`B_(1)=(mu_(0)(3i))/(4pi(3d))=(mu_(0)i)/(2pid)`, along `PB`
`B_(2)=(mu_(0)(4i))/(4pi(4d))=(mu_(0)i)/(2pid)`, along `PA`

`B_(1)` and `B_(2)` are perpendicular
`B_(P)=sqrt(B_(1)^(2)+B_(2)^(2))=(sqrt(2)mu_(0)i)/(2pid)=(mu_(0)i)/(sqrt(2)pid)`
(d) (i)

`B_(1)=B_(2)=(mu_(0)i)/(2pid)=B`
`B_(P)=2Bcos30^(@)=sqrt(3)B=(sqrt(3)mu_(0)i)/(2pid)`
`B_(1)=B_(2)=(mu_(0)i)/(2pid)=B`
`B_(P)=2Bcos60^(@)=B=(mu_(0)i)/(2pid)`