A monoenergetic electron beam with electron speed of 5.20 × 106 m s−1 is subject to a magnetic field of 1.30 × 10−4 T normal to the beam velocity. What is the a radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg−1.
Speed of an electron, v = 5.20 × 106 m/s
Magnetic field experienced by the electron, B = 1.30 × 10−4 T
Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
e = Charge on the electron = 1.6 × 10−19 C
m = Mass of the electron = 9.1 × 10−31 kg−1
The force exerted on the electron is given as:
`F = e|vecv xx vecB|`
`= evBsin theta`
θ = Angle between the magnetic field and the beam velocity
The magnetic field is normal to the direction of beam.
`:. theta = 90^@`
F = eVB ...(1)
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force `(F = (mv^2)/r)` for the beam.
Hence, equation (1) reduces to:
`evB = (mv^2)/r`
`:. r = mv/(eB) = v/((e/m)B)`
`= (5.20 xx 10^6) /(1.76 xx 10^11) xx 1.30 xx 10^(-4) = 0.227 m = 22.7 cm`
Therefore, the radius of the circular path is 22.7 cm.