Two long straight parallel wieres are `2m` apart, perpendicular to the plane of the paper. The wire A carries a current of `9.6 A`, directed into the plane of the paper. The wire `B` carries a current such that the magnetic field of induction at the point `P`, at a distance of `10/11` m from the wire B, is zero. find
a. the magnitude and directiion of the current in B.
b. the magnitude of the magnetic field of induction of the pont `S`.
c. the force per unit length on the wire `B`.

Explanation

(a) Direction of current at B should be perpendicular to paper
outwards. Let current in this wire be `i_B`. Then
`(mu_0)/(2pi) (i_A)/(2+10/11)=(mu_0)/(2pi) (i_b)/(10//11)`
or `(i_B)/(i_A)=10/32`
or `i_B=10/32xxi_A=10/32xx9.6=3A`
(b) Since `AS^2+BS^2=AB^2`
`:. -ASB=90^@`
At `S:B_1`=Magnetic field due to `i_A`
`=(mu_0)/(2pi) (i_A)/1.6=((2xx10^-7)(9.6))/1.6=12xx10^-7T`
`B_2`= Magnetic field due to `i_B=(mu_0)/(2pi) (i_B)/1.2=((2xx10^-7)(3))/1.2`
`=5xx10^-7T`
Since `B_1 and B_2` are mutually perpendicular, net magnetic
field at S would be:
`B=sqrt(B_1^2+B_2^2)=sqrt((12xx10^-7)^2+(5xx10^-7)^2)`
`=13xx10^-7T`

(c) Force per unit length on wire B:
`F/l=(mu_0)/(2pi) (i_Ai_B)/r (r=AB=2m)`
`=((2xx10^-7)(9.6xx3))/2=2.88xx10^-6 Nm^-1`