In figure, a long wire AB is placed on a table. A wire CD `10cm` long, is just above AB. It can slide up and down on two vertical wires. If the wires carry a current of `20A`, then how much above AB, will the wire CD settle? The mass of the wire CD is `0*50` gram.

Answer

`1*63mm`

Explanation

Let the wire CD settle at distance r above the wire AB. The magnetic force on CD of length l due to current in AB is
`F=(mu_0)/(4pi)(2i_1i_2)/(r)l`
Here, `l=10cm.=0*1m, i_1=i_2=20A`
`:. F=(10^-7xx2xx20xx20xx0*1)/(r)=(8xx10^-6)/(r)N`
Since the currents in AB and CD are in opposite directions, therefore force on CD will be upwards. The gravitional force on CD acting vetically downwards will be
`F_1=mg=0*5xx10^-3xx9*8N`
In the state of equibrium, `F=F_1`
or `(8xx10^-6)/(r)=0*5xx10^-3xx9*8`
or `r=(8xx10^-6)/(0*5xx9*8xx10^-3)=1*63xx10^-3m`