Two very long, straight parallel wires P and Q carry currents `5A` and `10A` respectively and are at a distance of `10cm` apart, as shown in figure. If a third wire R (length `10cm`) having a current of `5A` is placed in middle between them, how much force will be acting on R? The direction of current in all the wires is the same.

Answer

`10^-5N` towards Q

Explanation

Force on R due to wire P is
`F_1=(mu_0)/(4pi)(2I_1I_2)/(r_1)l=(10^-7xx2xx5xx5xx(0*10))/((5xx10^-2))`
`=10^-5N` towards P
Force on R due to wire Q is
`F_2=(mu_0)/(4pi)(2I_3I_2l)/(r_2)=(10^-7xx2xx10xx5xx(0*10))/((5xx10^-2))`
`=2xx10^-5N` towards Q
Net force on `R=F_2-F_1=2xx10^-5-10^5`
`=10^-5N` towards Q