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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# SPEAR, a storage ring about $${\bf{72}}.{\bf{0}}\;{\bf{m}}$$ in diameter at the Stanford Linear Accelerator (closed in 2009), has a $${\bf{20}}.{\bf{0}} - {\bf{A}}$$ circulating beam of electrons that are moving at nearly the speed of light. (See Figure 20.39.) How many electrons are in the beam?

About $$9.42 \times {10^{13}}\;{\rm{electron}}$$ number of electrons are there.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

• The diameter of the storage ring is $$d = 72.0\;{\rm{m}}$$.
• The current carried by the circulating beam of electrons is $$I = 20.0\;{\rm{A}}$$.

## Step 2: Significance of the charge of a ring

The charge of a ring is described as the amount of current carried out by a ring in unit time. The charge carried is directly proportional to the current and the diameter of an object and inversely proportional to the velocity of that object.

## Step 3: Determination of the number of electrons

The equation of the number of electrons inside the beam is expressed as:

$$n = \frac{{\pi Id}}{{c{q_e}}}$$

Here, $$n$$ is the number of electrons inside the beam, $$I$$ is the current carried by the circulating beam of electrons, $$d$$is the diameter of the storage ring,$$c$$ is the velocity of light and $${q_e}$$ is the charge of one electron.

Substitute the values in the above equation.

\begin{align}n &= \frac{{\left( {3.14} \right)\left( {20.0\;{\rm{A}}} \right)\left( {72.0\;{\rm{m}}} \right)}}{{\left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {62.8\;{\rm{A}}} \right)\left( {72.0\;{\rm{m}}} \right)}}{{\left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {4521.6\;{\rm{A}} \cdot {\rm{m}}} \right)}}{{\left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{A}} \cdot {\rm{s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\end{align}

Hence, further as:

\begin{align}n &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{A}} \cdot {\rm{s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{A}} \cdot {\rm{s}} \times \frac{{1\;{\rm{C}}}}{{1\;{\rm{A}} \cdot {\rm{s}}}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= \frac{{\left( {1.507 \times {{10}^{ - 5}}\;{\rm{C}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C/electron}}} \right)}}\\ &= 9.42 \times {10^{13}}\;{\rm{electron}}\end{align}

Thus, about $$9.42 \times {10^{13}}\;{\rm{electron}}$$ number of electrons are