A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the coil. If the current in the coil is` 5.0` A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field? (The coil is made of copper wire of cross sectional area `10^(-5) m^(2)` and the free electron density in copper is given to be about `10^(29) m^(-3)` )

Explanation

Torque acting on the coil of area A and a number on turns, placed in an external magnetic field B is given by
`tau = n BIA sin theta , " where" theta` is the angle between B and normal vector to the area A
Here, `theta = 0^(@)`
` n = 20 , r = 10 cm , B = 0.10 T , I = 5.0 A`
` :. tau = 0 `
(b) Net force on closed current loop in uniform magnetic field is zero.
(c ) Force on each electron (e) moving with a drift velocity v is given by
`F = Bev`
Also, `I = ne vA = ev = I/(nA)`
` :. " " F = (BI)/(nA)`
Where `n = N` Number of electrons per unit volume ` = 10^(29) m^(-3)`
And A = Area of cross section of wire ` = 10^(-5) m^(2)`
` F = (0.1 xx 5)/(10^(29) xx 10^(-5)) `
` = 5 xx 10^(-25) N`