A circular coil of 20 turns and radius `10cm` is placed in a uniform magnetic field of `0.10T` normal to the place of the coil. If the current in the coil is `5.0A`, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) Given, `N=10^(29)m^-3`, `A=10^-5m^2`
Force on an electron of charge e, moving with drift velocity `v_d` in the magnetic field is given by
`F=Bev_d=Be(I)/(NeA) ( :'I=NeAv_d)`
`F=(BI)/(NA)=(0*10xx5*0)/(10^(29)xx10^-5)=5xx10^(-25)N`

Explanation

n=20, r=10 cm, B=0.10 T, I=5.0 A
Torque acting on the coil = `nBIA sin theta=0`