Question -
Answer -
(a)Speed of anelectron, v = 5.20 × 106 m/s
Magnetic field experienced by theelectron, B = 1.30 × 10−4 T
Specific charge of an electron, e/m =1.76 × 1011 C kg−1
Where,
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 theelectron is given as:
θ = Angle between themagnetic field and the beam velocity
Themagnetic field is normal to the direction of beam.
The beam traces a circular path of radius, r.It is the magnetic field, due to its bending nature, that provides thecentripetal force
for the beam.
Hence, equation (1)reduces to:
Therefore, the radius ofthe circular path is 22.7 cm.
(b) Energy of the electron beam, E =20 MeV
The energy of the electronis given as:
This result is incorrect because nothing canmove faster than light. In the above formula, the expression (mv2/2)for energy can only be used in the non-relativistic limit, i.e., for v << c
When very high speeds areconcerned, the relativistic domain comes into consideration.
In the relativisticdomain, mass is given as:
Where,
= Mass of the particle atrest
Hence, the radius of thecircular path is given as:
