Correct Answer - `3*33cm` on the left hand side of wire X
The resultant magnetic field will be zero at a point P on the left hand side of wire X. Let x be the distance of point P from wire x. Distance of point P from wire Y is (x+r). Magnetic field at P due to current in wire X is
`B_1=(mu_0)/(4pi)(2I_1)/(x)=(10^-7xx2xx4)/(x)`
acting perpendicular upwards.
Magnetic field at P due to current in wire Y is
`B_2=(mu_0)/(4pi)(2I_2)/((x+r))=(10^-7xx2xx10)/((x+5xx10^-2))` acting perpendicular downwards
Resultant magnetic field will be zero at P if
`B_1=B_2`
or `(10^-7xx2xx4)/(x)=(10^-7xx2x10)/((x+5xx10^-2))`
or `4/x=(10)/((x+5xx10^-2))`
or `4x+20xx10^-2=10x` or `6x=20xx10^-2`
or `x=(20xx10^-2)/(6)m=20/6cm=10/3cm=3*33cm`