Two very long straight wires P and Q carry currents of `10A` and `20A` respectively, and are at a distance `20cm` apart. If a third wire R (length `15cm`) having a current of `10A` is placed in middle between them, how much force will act on R? The direction of current in all the three wires is the same.

Answer

`3*0xx10^-5N`, towards Q

Explanation

Net force on conductor R towards conductor Q will be
`=F_2-F_1=(mu_0)/(4pi)(2I_2I_3)/(r)l-(mu_0)/(4pi)(2I_1I_3)/(r)l`
`=(mu_02I_3l)/(4pir)[I_2-I_1]`
`=(10^-7xx2xx10xx0*15(20-10))/(0*10)`
`=3*0xx10^5N` towards Q