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`.

Answer

(i) 3 (ii) 13 (iii) 2.38

Explanation

(i) 3 (ii) 13 (iii) 2.38
(i) Direction of current at B should be perpendicular to paper outwards. Let current in this wire be . 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)=10/32 xxi_(A)=10/32xx9.6=3A`

(ii) 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^(-7)T`
`B_(2)=` Magnetic fiedl due to `I_(B), =((2xx10^(-7))/1.2)=5xx10^(-7)T`
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^(-7)T`
(iii) Force per unit length on wire B
`F/l=(mu_(0))/(2pi)(i_(A)i_(B))/r " "(r=AB=2m)`
`=((2xx10^(-7)(9.6xx3))/2=2.88xx10^(-6)N//m`