a. Direction of current at `B` should be perpendicular to paper outwards. Let curent 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=10/32x9.6=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^I-7)(9.6))/1.6`
`=12xx10^-7T`
`B_2=`magnetic field to due to `i_B`
`=mu_0/(2pi) i_B/1.2`
`=((2xx10^i-7)(3))/1.2`
`=5xx10^-7T`
Since `B_1 and B_2` are mutually perpedicular.Net magnetic field at S would be
`b=sqrt(B_1^2+B_2^2)`
`=sqrt((12xx106i-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^I-6N/m`
