An electron having kinetic energy 10eV is circulating in a path of radius 0.1 m in an external magnetic field of intensity `10^(-4)`T. The speed of the electron will be

Options

A. 11 cm
B. 18 cm
C. 12 cm
D. 16 cm

Answer

A

Explanation

Kinetic energy of electron `((1)/(2) xx mv^(2)) = 10` magnetic induction `B = 10^(-4) Wb//m^(2)`
Therefore `(1)/(2) (9.1 xx 10^(-31)) v^(2) = 10 xx (1.6 xx 10^(-29))`
Or, `v^(2) = (2 xx 10 xx (1.6 xx 10^(29)))/(9.1 xx 10^(-31)) = 3.52 xx 10^(12)`
Or, `v = 1.876 xx 10^(6)` m
Centripetal force = `(mv^(2))/(r) = Bev`
Therefore `r = (mv)/(Be) = ((9.1 xx 10^(-31))xx(1.876 xx 10^(6)))/(10^(-4) xx (1.6 xx 10^(-29)))`
`= 11 xx 10^(-2)` m = 11 cm