Chapter 32, Problem 11P

To determine

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 $\text{rcB}$

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,:

$\mathrm{F}=\frac{m{v}^2}{r}$ (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

$F=eBv$ (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,

$mv=eBr$

$P=eBr$ (III)

• ‘P’ is the magnitude of momentum in kg.m/s².

In relativity the formula for magnitude of momentum is,

$P=\frac{\overline{E}}{c}$ (IV)

• ‘ $\overline{E}$’ is the energy in Joule(J).

• ‘c’ is the velocity of light (m/s).

Energy $\overline{E}$ in electron volt is

$\overline{E}=eE$ (V)

From equations (III),(IV)and(V), the energy of the particle

$\frac{eE}{c}=eBr$

$E=\text{rcB}$ (VI)