There are two parallel current carrying wires X and Y as shown in figure. Find the magnitude and direction of the magnetic field at points, P, Q and R.

Explanation

Here, `I_1=10, I_2=15A, r=5cm=5xx10^-2m` Let `hatn` be the unit vector perpendicular to the plane of paper unwards.
The magnetic field at P due to current in wire X is
`vecB_1=(mu_0)/(4pi)(2I_1)/(r)hatn=10^-7(2xx10)/(5xx10^-2)hatn=4xx10^-5hatnT`
The magnetic field at P due to current in wire Y is
`vecB_2=(mu_0)/(4pi)(2I_2)/((r+r+r))(-hatn)=10^-7=(2xx15)/(15xx10^-2)(-hatn)=-2xx10^-5T`
The resultant magnetic field at P is
`vecB_P=vecB_1+vecB_2=(4xx10^-5)hatn-(2xx10^-5)hatn=2xx10^-5Thatn`
`=2xx10^-5T` acting normally outwards.
At point Q, both `vecB_1` and `vecB_2` will be acting normally inwards. So
`vecB_Q=vecB_1+vecB_2=(mu_0)/(4pi)(2xx10)/(5xx10^-2)(-hatn)+(mu_0)/(4pi)(2xx15)/(5xx10^-2)(-hatn)`
`=[10^-7xx(20)/(5xx10^-2)+10^-7xx(30)/(5xx10^-2)](-hatn)=(4xx10^-5+6xx10^-5)(-hatn)`
`=10^-4T` acting normally inwards.
At point R, `vecB_1` will be acting normally inwards and `vecB_2` will be acting normally outwards. So,
`vecB_R=vecB_1+vecB_2=10^-7xx(2xx10)/((5+5+5)xx10^-2)(-hatn)+10^-7xx(2xx20)/((5xx10^-2))(hatn)`
`=(-4/3xx10^-5+8xx10^-5)(hatn)=20/3xx10^-5T`, acting normally outwards