Chapter 18, Problem 75GP

To determine

Number of protons are stored in the beam.

Answer to Problem 75GP

  $1.44\times {10}^{12}protons$

Explanation of Solution

Given info:

Speed of light is $3.0\times {10}^{8}m/s$ .

Circumference of the ring is $6300m$ .

Current is $11mA=11\times {10}^{-3}A$ .

Formula used:

Expression to calculate current:

  $I=nqA{v}_d$

Where,

  $v_d$ is drift speed.

  $I$ is the current,

  $A$ is the cross-sectional area

  $n$ is number of free electrons,

  $q$ is charge

Calculation:

It is known that $n$ is defined as the number of ions per unit volume. So, $n=\frac{N}{Volume}$ . It is also known that the volume of the accelerator pipe is $Volume=AS$ . Here, $A$ is the cross-sectional area of the pipe, $S$ is the length of the pipe. So, by the given formula

  $n=\frac{N}{AS}$

  $I=\frac{N}{AS}qA{v}_d$

  $I=\frac{Nq{v}_d}{S}$

  $N=\frac{IS}{q{v}_d}$

Substitute the given values:

  $N=\frac{\left(11\times {10}^{-3}\right)\left(6300\right)}{\left(1.6\times {10}^{-19}\right)\left(3.0\times {10}^8\right)}=1.44\times {10}^{12}protons$

Conclusion: The number is protons are $1.44\times {10}^{12}protons$ .