Concept explainers
The energy of a particle in a synchrotron in the relativistic limit (E).
Answer to Problem 11P
Solution:
The energy of a particle in a synchrotron in the relativistic limit (E)is
Explanation of Solution
Given Info:
The magnetic field is ‘B’.
The radius of the orbit is ‘r’.
Formula Used:
The magnitude of force due to circular motion in a synchrotron is,:
(I)
• ‘F’ is the magnitude of force in Newton(N).
• ‘v’ is the velocity of charge particle ‘e’ (m/c)
• ‘r’ is the radius of the orbit.
The magnitude of force is experienced by the charge particle due to magnetic field is
(II)
• ‘F’ is the magnitude of force in Newton(N).
• ‘B’ is the magnetic field n Tesla (T)
Calculation:
From equations (I)and (II)we get,
(III)
• ‘P’ is the magnitude of momentum in kg.m/s2.
In relativity the formula for magnitude of momentum is,
(IV)
• ‘ ’ is the energy in Joule(J).
• ‘c’ is the velocity of light (m/s).
Energy in electron volt is
(V)
From equations (III),(IV)and(V), the energy of the particle
(VI)
Chapter 32 Solutions
Physics: Principles with Applications
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