Consider two straight, parallel, long, current carrying conductors AB and CD carrying currents `I_(1)` and `I_(2)` respectively in same direction and let these be separated by a distance d. Now magnetic field B1 developed at a point Q on 2nd conductor due to current `I_(1)` flowing in 1st conductor is
`B_(1)=(mu_(0)I_(1))/(2pid)`
As per right hand rule `B_(1)` is acting normal to the plane of the paper pointing inward. Thus, conductor CD carrying current `I_(2)` is in a magnetic field which is perpendicular to its length. Therefore, force experienced by 2nd conductor CD due to `B_(1)`.
`F_(21)=B_(1)I_(2)l`,
where l = length of the 2nd conductor
or `F_(21)=(mu_(0)I_(1))/(2pid)I_(2)l=(mu_(0)I_(1)I_(2)l)/(2pid)`
and force per unit length
`(F_(21))/l=(mu_(0)I_(1)I_(2))/(2pid)=(mu_(0))/(4pi).(2I_(1)I_(2))/d`
The force `F_(21)` in accordance with Fleming.s left hand rule is directed towards the conductor AB. In the same way, it is found that force experienced per unit length of wire AB is
`(F_(12))/l=(mu_(0))/(4pi).(2I_(1)I_(2))/d`
and is directed towards CD.
IF `I_(1)=I_(2)=1A` and d=1 m, then `(F_(12))/p=(mu_(0))/(4pi)xx(2xx1xx1)/1=(mu_(0))//(2pi)=2xx10^(-7)Nm^(-1)`. Hence, SI base unit of current i.e., ampere is defined as the value of that steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross-section and placed 1 m apart in vacuum would produce on each of these conductors a force equal to `2xx10^(-7)Nm^(-1)`.
(b) Here `I_(1)=I_(2)=I=3A`, distance between two parallel conductors d = 12 cm = 0.12 m. So the normal distance of point P midway between them from either conductor r`=d/2=0.06m`
Then `B_(1)=B_(2)=(mu_(0)l)/(2pir)=(4pixx10^(-7)xx3)/(2pixx0.06)=1xx10^(-5)T`
