Question

SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 68.0 m in diameter and has a 0.37 A beam, how many electrons are in the beam

Answer 1

Answer:

The no. of electron in the beam = $1.64\times10^{12}$

Explanation:

Given :

The diameter of circular ring = 68 m.

The current flowing in the beam = 0.37 A

Speed of light = $3\times10^{8} ms^{-1}$

We know that the current is equal to the charge per unit time.

⇒    $I = \frac{Q}{t}$

∴    $Q=It$

Here given in the question, electron revolving in a circle with the diameter

$d = 68$m

⇒ Time take to complete one round $(t) =$ $\frac{\pi d }{v}$

∴    $Q = \frac{I\pi d }{v}$

     $Q = \frac{0.37 \times 3.14 \times 68}{3 \times 10^{8} }$

     $Q = 26.33 \times 10^{-8}$

Now, for finding the no. of electron we have to divide $Q$ to the charge of the electron  $q = 1.6 \times 10^{-19}$

∴     $n$ =  $\frac{26.33 \times 10^{-8} }{1.6 \times 10^{-19} }$

      $n = 1.64 \times 10^{12}$

Thus, the no. of electron in the beam is $1.64 \times 10^{12}$.