Question

An electron (charge \[ = 1.6 \times {10^{ - 19}}C\] ) is moving in a circle of radius \[5.1 \times {10^{ - 11}}m\;\] at a frequency of \[6.8 \times {10^{15}}\;revolutions/sec\] . The equivalent current is approximately.
A. \[5.1 \times {10^{ - 3}}amp\]
B. \[6.8 \times {10^{ - 3}}amp\]
C. \[1.1 \times {10^{ - 3}}amp\]
D. \[2.2 \times {10^{ - 3}}amp\]

Answer 1

Hint: In the given question we are given the value of the radius of the circular path and the value of the frequency of the electron. Using these values we can find the amount of electric current flowing in the given circular path by using the relationship between the current and the total charge flowing per unit time through the given path.

Formula used:
Current, $I = \dfrac{q}{t}$
where $q$ is the charge flowing in $t$ period of time.
Total charge, $q = ne$
where $n$ is the number of electrons and $e$ is the charge on a single electron.
And $\text{time} = \dfrac{1}{\text{frequency}}$

Complete step by step solution:
Given: Radius of the circular path, $r = 5.1 \times {10^{ - 11}}m\;$
Frequency of the electron, $\nu = 6.8 \times {10^{15}}\;revolutions/sec$
Since, $\text{time} = \dfrac{1}{\text{frequency}}$ therefore,
$t = \dfrac{1}{{6.8 \times {{10}^{15}}}}$
Also, the number of electrons given is, $n = 1$

Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or tightly depending on the situation. Electrons can move freely inside the confines of the body when the nucleus is only lightly holding them. Because electrons are negatively charged particles, they cause a number of charges to flow when they move, and we refer to this movement of charges as an electric current.

We know that, $I = \dfrac{q}{t}$ and $q = ne$. In this formula, $n$ should always have a positive integral value because electrons (or any other charge) cannot exist in fraction. Therefore, we get,
$I = \dfrac{{ne}}{t}$
Putting the known values in the above expression, we get,
$I = \dfrac{{1 \times \left( {1.6 \times {{10}^{ - 19}}} \right)}}{{\left( {\dfrac{1}{{6.8 \times {{10}^{15}}}}} \right)}} \\ $
Simplifying this, we get,
$I = \left( {6.8 \times {{10}^{15}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}} \right) \\ $
$\therefore I = 1.1 \times {10^{ - 3}}A$

Hence, option C is the correct answer.

Note: An important point to notice in this question is that a single electron is given $\left( {n = 1} \right)$ which is indicated by the term “an electron” given in the question. Also, one might make a mistake by taking the frequency as time in the formula used. We first need to find the value of time in the SI unit from the frequency given in the question.