Electrons in a synchrotron

Given that the velocity of the electrons is $v\approx c\approx3\times10^8 m/s$ and the distance travelled by the electrons = circumference of the synchroton = $L=300m$ , then the time required for the electrons to flow through the synchroton once is $t=\frac{L}{v}=\frac{300}{3\times10^8}=1\times10^{-6}$ seconds. This means that once every $10^{-6}$ seconds $10^{11}$ electrons flow through the synchroton. Now, the total amount of charge on the electrons is $C=10^{11}\times 1.6\times 10^{-19}=1.6\times10^{-8}$ Coulombs. Therefore, the electric current flowing through the synchroton is $I=\frac{C}{t}=\frac{1.6\times10^{-8}}{10^{-6}}=1.6\times10^{-2}=\boxed{1.6E-2}$ Amps.

In one second, each electron orbited the circle 3 * 10^8/300 =10^6 times. There were 10^11 electron, each with 1.6 10^-19 Coulombs of charge. The current is 10^6 * 10^11 * 1.6 10^-19 = 1.6*10^-2 amperes.

First determine the total charge (Q) by multiplying the number of electrons (10^11) by the charge of electrons (1.61*10^-19)

Q= 1.61E-8 C

Then determine the time (t) by using v=d/t. The d is 300 m and the v 3x10^8.

t= .00001 sec

Substitute both values into the equation I=Q/t and you get I = .0161 Amps

We know that current is the rate of flow of electric charge, that is I=Q/t where I represents the current, Q is the electric charge in coulomb and t is the time in second. First find the total electric charge produce by the electrons, that is the product of the total number of electrons(100 billion electrons) and their respective charge(1.6 \times 10^{-19} Coulomb), then we will get the value 1.6 \times 10^{-8} Coulomb.

Next ,take the speed of the accelerated electrons and divide it by the circumference of 300m. We will get the value 1 \times 10^6 , which indicates the number of times the electrons circulate the synchroton in 1 second. After that take 1 divide by the value 1 \times 10^6 to get the time of 1 circulation done by the accelerated electrons. The time will be 1 \times 10^{-6} second.

Finally, take the value of total electric charge and divide by the time. The electric current produce is 0.016 Ampere.

Time taken by each electron to move round the synchotron once = distance/speed of an electron = (3E2/3E8) s =1E-6 s. Now, by the given assumption, the electron distribution is homogeneous throughout the synchotron, i.e., we can say that all electrons cross the same cross sectional part of the synchotron in same part of time. And we find out Total charge of all the electrons(Q) =ne (where n denotes the no. of electrons in the synchotron and e the charge of an electron) =1E11*1.6E-19 coulombs =1.6E-8 coulombs. And, from above assumption, we say that these charges cross a cross sectional part of the synchotron in the same time as that of an electron, i.e it crosses the whole synchotron in 1E-6 s. Therefore, current flow in the synchotron(I) = Q/t =1.6E-8/IE- 6 amperes =1.6E-2 amperes =0.016 amperes Hence current flow in the synchotron will be 0.016 amperes. (ANS)

By definition, the magnitude of the electric current equals the rate at which electric charge (in this case electrons) passes through a conductor. Let us choose any point P on the circle. Let $\Delta N$ be the number of electrons that pass through that point in a time interval $\Delta t$ . Since the $N=10^{11}$ electrons are uniformly distributed over the circle (of circumference $L=300~\mbox{m}$ ) , we can write $\frac{\Delta N}{N}= \frac{c {\Delta t}}{L}$ which implies $\frac{ \Delta{N}}{\Delta t}=\frac{c N}{L}.$ That is to say, the above equations gives us the number of electrons passing through point P per unit time. The current is simply $I=\frac{\Delta q}{\Delta t}= \frac{ e \Delta N}{\Delta t}=\frac{c e N }{L}= 1.6\times 10^{-2}~\mbox{A}.$