In a chamber, a uniform magnetic field of `6*5G(1G=10^-4T)` is maintained. An electron is shot into the field with a speed of `4*8xx10^6ms^-1` normal to the field. (i) Explain why the path of the electron is a circle. Determine the radius of the circular orbit. `(e=1*6xx10^(-19)C, m_e=9*1xx10^(-31)kg)`.

Explanation

Magnetic field strength, `B = 6.5 G = 6.5 xx 10^(-4)T`
Speed of the electron, `v = 4.8 xx 10^(6)m//s`
Charge on the electron, `e = 1.6 xx 10^(-19)C`
Mass of the electron, `m_(e) =9.1 xx 10^(-31)kg`
Angle between the shot electron and magnetic field, `theta = 90^(@)`
Magnetic force exerted on the electron in the magnetic field is given as:
`F = eV sin theta`
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
`F_(e) = (mv^(2))/(r)`
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
`F_(e) = F`
`(mv^(2))/(r) - evB sin theta`
`r = (mv)/(B esin theta)`
`=(9.1 xx 10^(-31)xx4.8 10^(6))/(6.5 xx 10^(4) xx 1.6 xx 10^(-19)sin 90^(@))`
`=4.2 xx 10^(-2)m = 4.2 cm`
Hence, the radius of the circular orbit of the electron is `4.2 cm`.