
An electron of mass $m$ moves around the nucleus in a circular orbit of radius $r$ under the action of centripetal force $F$. The equivalent electric current is:
A) $\dfrac{e}{2\pi }\sqrt{\dfrac{F}{mr}}$
B) $2\pi e\sqrt{\dfrac{F}{mr}}$
C) $\dfrac{e}{\pi }\sqrt{\dfrac{F}{mr}}$
D) $\dfrac{e}{2\pi }\sqrt{\dfrac{mr}{F}}$
Answer
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Hint: The equivalent electric current can be defined as the charge of an electron divided by the time taken to move unit distance. Now, we can calculate $v$ from the centripetal force equation $F=\dfrac{m{{v}^{2}}}{r}$ . Then, putting the value of $v$ in the formula of electric current, we can find $i$ .
Complete step by step solution:
Given, the mass of an electron is $m$ , the radius of a circular orbit is $r$ , and the value of centripetal force is $F$ .
Now, let, the velocity of the electron is $v$ , and the equivalent electric current is $i$ .
We know, electric current can be defined as $i=\dfrac{q}{t}$ , where $q$ is the charge, and $t$ is the time
So, $i=\dfrac{e}{\dfrac{2\pi r}{v}}$
Or, $i=\dfrac{ev}{2\pi r}$ ………… (i)
Now, we know, centripetal force, $F=\dfrac{m{{v}^{2}}}{r}$
Therefore, $v=\sqrt{\dfrac{Fr}{m}}$ ………..(ii)
Now, putting the value of $v$ from equation (ii) in equation (i), we can have,
Equivalent electric current,
$\Rightarrow i=\dfrac{e\sqrt{\dfrac{Fr}{m}}}{2\pi r}$
$\Rightarrow i=\frac{e}{2\pi }\sqrt{\frac{F}{mr}}$
Therefore, the correct answer is (A), $\dfrac{e}{2\pi }\sqrt{\dfrac{F}{mr}}$.
Additional Information:
Electrons spin not only on their own axis but also in a constant circular motion around the nucleus. In spite of moving in such a way, the electrons are very stable. The electrostatic forces of attraction between the electrons and the nucleus cause the stability of electrons mainly. Another name of the electrostatic forces is Coulombic forces of attraction. This attraction between the nucleus and the electrons causes the Centripetal force for the electrons. It is an inward force to the center of the circle from the circumference.
Note: The nucleus contains a positive charge and the charge of an electron negative. So, there exists an electrostatic force acting between the nucleus and electrons. This electrostatic force is the source of the centripetal force of a revolving electron around the nucleus.
Complete step by step solution:
Given, the mass of an electron is $m$ , the radius of a circular orbit is $r$ , and the value of centripetal force is $F$ .
Now, let, the velocity of the electron is $v$ , and the equivalent electric current is $i$ .
We know, electric current can be defined as $i=\dfrac{q}{t}$ , where $q$ is the charge, and $t$ is the time
So, $i=\dfrac{e}{\dfrac{2\pi r}{v}}$
Or, $i=\dfrac{ev}{2\pi r}$ ………… (i)
Now, we know, centripetal force, $F=\dfrac{m{{v}^{2}}}{r}$
Therefore, $v=\sqrt{\dfrac{Fr}{m}}$ ………..(ii)
Now, putting the value of $v$ from equation (ii) in equation (i), we can have,
Equivalent electric current,
$\Rightarrow i=\dfrac{e\sqrt{\dfrac{Fr}{m}}}{2\pi r}$
$\Rightarrow i=\frac{e}{2\pi }\sqrt{\frac{F}{mr}}$
Therefore, the correct answer is (A), $\dfrac{e}{2\pi }\sqrt{\dfrac{F}{mr}}$.
Additional Information:
Electrons spin not only on their own axis but also in a constant circular motion around the nucleus. In spite of moving in such a way, the electrons are very stable. The electrostatic forces of attraction between the electrons and the nucleus cause the stability of electrons mainly. Another name of the electrostatic forces is Coulombic forces of attraction. This attraction between the nucleus and the electrons causes the Centripetal force for the electrons. It is an inward force to the center of the circle from the circumference.
Note: The nucleus contains a positive charge and the charge of an electron negative. So, there exists an electrostatic force acting between the nucleus and electrons. This electrostatic force is the source of the centripetal force of a revolving electron around the nucleus.
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