Question

(a) A monoenergetic electron beam with electron speed of 5.20 × 106m/s is subject to a magnetic field of 1.30 × 10–4 Tnormal to the beam velocity. What is the radius of the circle tracedby the beam, given e/m for electron equals 1.76 × 1011C /kg (b) Is the formula you employ in (a) valid for calculating radius ofthe path of a 20 MeV electron beam? If not, in what way is itmodified?[Note: Exercises 11.20(b) and 11.21 (b) take you to relativisticmechanics which is beyond the scope of this book. They have beeninserted here simply to emphasise the point that the formulas youuse in part (a) of the exercises are not valid at very high speeds orenergies. See answers at the end to know what ‘very high speed orenergy’ means.]

Answer 1

(a)

Given: The speed of electron of monoenergetic electron beam is 5.20× 10 6   ms -1 , the magnetic field is 1.30× 10 −4  T normal to the beam velocity.

The force exerted on the electron is given as,

F=evBsinθ(1)

Where, the velocity of the electron is v, magnetic field experienced by the electron is B, angle between velocity of electron and direction of magnetic field is θ, charge on the electron is e , and mass of the electron is m.

The magnetic force provides centripetal force to the beam due to which it traces a circular path.

F c , the centripetal force and r, the radius of circular path is given by,

F c = m v 2 r (2)

For equilibrium, equation (1) is equal to (2),

evBsinθ= m v 2 r r= v ( e m )Bsinθ (3)

The value of e/m is 1.76× 10 11   Ckg −1 .

By substituting the given values in the above expression, we get,

r= 5.20× 10 6 ( 1.76× 10 11 )×1.30× 10 −4 ×sin 90 0 =0.227 m

Thus, the radius of the circle traced by beam is 0.227 m.

(b)

Given: The energy of the electron beam is 20 MeV.

The energy of the electron is given as,

E= m v 2 2 v= ( 2E m ) 1 2

The value of m is 9.1× 10 −34  kg.

By substituting the given values in the above expression, we get,

v= ( 2×20×1.6× 10 −13 9.1× 10 −34 ) 1 2 =2.652× 10 9   ms -1

The velocity of electron is greater than speed of light ( 3× 10 8   ms -1 ), which is not possible. So, this formula can’t be used. When very high speed is considered, the relativistic domain comes into consideration.

The mass of the particle is given as,

m= m 0 [ 1− v 2 c 2 ] 1 2 (4)

Where, the mass of the particle at rest is m 0 and speed of light is c.

By substituting the given values in the above expression, we get,

r= m 0 vc eBsinθ [ c 2 − v 2 ] 1 2

Thus, the modified formula for radius of circular path is r= m 0 vc eBsinθ [ c 2 − v 2 ] 1 2 .