A circular coil of 20 hours and radius 10cm is placed in an uniform magnetic field of `0.17` normal to the plane of the coil if the current in the coil is `5.0A`. What is the
average force on each electron in the coil due to the magnetic field ? The coil is made of copper wire of cross - (a) sectional area `10^(-5) m ^(+2) ` and the free clectron density in copper is given to be about `10^(29)//m^(3) `.

Explanation

Number of turns, `N = 20`
Radius of circular coil , ` r = 10 cm = 0.1 m`
Magnetic field , `B = 0.1 ` T
Angle between area vector and magnetic field, `theta = 0^(@)`
Current , `I = 5.0 A " " (A = pi r^(2) = "Area")`
Number density of electrons, ` n = 10^(29)//m^(3) `
Area , `A = 10^(-5) m ^(2) ` ltbr. Magnitude of force , `F = e (v_(d) xx B) `
`F = BeV_(d) sin theta rArr I = n Ae v_(d) `
` v_(d) = I/(n Ae) `
` F = e (I/(nAe)) B sin 90^(@) `
` = (IB sin 90^(@))/(nA) `
` F = (0.1 xx 5 xx 1)/ (10^(-5) xx 10^(29)) = 5 xx 10^(-25) N `