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

Number of turns on the circular coil, `n = 20`
Radius of the coil, `r = 10 cm = 0.1m`
Magnetic field strength, `B = 0.10 T`
Current in the coil, `I = 5.0A`
(a) The total torque on the coil is zero because the field is uniform.
(b) The total force on the coil is zero because the field is uniform.
(c) Cross-sectional area of copper coil, `A = 10^(-5)m^(2)`
Number of free electrons per cube meter in copper, `N = 10^(29)//m^(3)`
Charge on the electron, `e = 1.6 xx 10^(-19)C`
Magnetic force, `F =Bev_(d)`
Where,
`v_(d)=` Drift velocity of electrons
`=(I)/(NeA)`
`:. F = (BeI)/(NeA)`
`=(0.10 xx 5.0)/(10^(29) xx 10^(-5)) = 5 xx 10^(-25)N`
Hence, the average force on each electrons is `5 xx 10^(-25)N`.