An infinte wire place along z-axis has current `I_(1)` in positive z-direction A conducting rod placed in xy plane parallel to y-axis has current `I_(2)` in positive y-direction The ends of the rod subtend `+30^(@)` and `-60^(@)` at the origin with positive x direction The rod is at a distance a from the origin. Find net force on the rod .

Answer

`F=-(mu_(0)I_(1)I_(2))/(4pi)log_(e)(3)`

Explanation

The arrangement is shown here.

Please not that the direction of `I_(1)` and the `+z-` axis, both are outwards towards the reader. Let us consider an infinitesimal elemne to length by at a distance `y` from `x-` axis. If `vec(B)` be he field due to the wire 1 at the element is
`B_(1)=(mu_(0)I_(1))/(2pir)=(mu_(0)I_(1))/(2pisqrt(a^(2)+y^(2)))`
The force on this element due to the field of wire 1 is
`d vec(F)=I(dvec(l)xxvec(B))" "impliesdvec(F)=I_(2)(dvec(y)xxvec(B)_(1))`
`implies|dvec(F)|=dF=(I_(2)dy)((mu_(0)I_(1))/(2pisqrt(a^(2)+u^(2))))sintheta`
`impliesdF=(mu_(0)I_(1)I_(2))/(2pi)((sintheta)/(sqrt(a^(2)+u^(2))))dy`
[along positive `z-` direction, using right - hand thumb rule]
But `sin theta =(y)/(sqrt(a^(2)+y^(2)))`
`impliesdF=(mu_(0)I_(1)I_(2))/(2pi)((ydy)/(a^(2)+y^(2)))`
`F=(mu_(0)I_(1)I_(2))/(2pi)int_(-sqrt(3)a)^((a)/(sqrt(3)))(y)/(a^(2)+y^(2))`
Since, `int (ydy)/(a^(2)+y^(2))=(1)/(2)log_(e)(a^(2)+y^(2))`

`impliesF=(mu_(0)I_(1)I_(2))/(4pi)log_(e)[log_(e)(a^(2)+y^(2))|_(-sqrt(3)a)^((a)/(sqrt(3)))]`
`impliesF=(mu_(0)I_(1)I_(2))/(4pi)log_(e)[(a^(2)+(a^(2))/(2))/(a^(2)+3a^(2))]`
`impliesF=(mu_(0)I_(1)I_(2))/(4pi)log_(e)((1)/(3))impliesF=-(mu_(0)I_(1)I_(2))/(4pi)log_(e)(3)`