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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.]

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Solution

(a)

Given: The speed of electron of monoenergetic electron beam is

The force exerted on the electron is given as,

Where, the velocity of the electron is

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

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

The value of

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

Thus, the radius of the circle traced by beam is

(b)

Given: The energy of the electron beam is

The energy of the electron is given as,

The value of

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

The velocity of electron is greater than speed of light

The mass of the particle is given as,

Where, the mass of the particle at rest is

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

Thus, the modified formula for radius of circular path is

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