Question

An electron revolves \[6 \times {10^{15}}\,times/sec\] in a circular loop. Current in the loop is
A. \[0.96\,mA\]
B. \[0.96\,A\]
C. \[28.8\,A\]
D. None

Answer 1

Hint: In order to solve this question we need to establish and use the relationship between charge and current. Since, we know that current is charge per unit time or we can say that current is the product of charge and frequency. We also need to quantize charge by the relation q=ne and in this question, n=1.

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 $time = \dfrac{1}{{frequency}}$

Complete step by step solution:
Given: Frequency of the revolution, $\nu = 6 \times {10^{15}}\,times/sec$
Since, $time = \dfrac{1}{{frequency}}$ therefore,
$t = \dfrac{1}{{6 \times {{10}^{15}}}}$
Also, the number of electrons given is, $n = 1$

While some electrons are bound in an atom, they are all similarly attached to protons. The atom's electrons that are farthest from the nucleus can be extracted. Because there are less electrons in the atom when some electrons are taken out, there are more protons than electrons. The electrically neutral body becomes positively charged after the elimination of electrons. The body can also get electrons from an outside source, which is the opposite of the first scenario. In this instance, the body's electron count rises and it acquires a negative charge

We know that, the current in the loop can be written as,
$I = \dfrac{q}{t}$ and $q = ne$
Where $n$ 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,
$I = \dfrac{{1 \times \left( {1.6 \times {{10}^{ - 19}}} \right)}}{{\left( {\dfrac{1}{{6 \times {{10}^{15}}}}} \right)}}$
Simplifying this, we get,
$I = \left( {6 \times {{10}^{15}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}} \right)$
Thus, $I = 9.6\, \times {10^{ - 4}}A$
That is, $I = 0.96\,mA$

Hence, option A. is the answer.

Note: You should not confuse frequency and time. Frequency is the reciprocal of time. In this question, you were given with n=1 but in some cases there will be more than one electron and do not forget to use the relation of quantization in such cases and put the value of n according to the number of electrons.