Construction and Working of Nuclear Reactor or Atomic Pile

A nuclear reactor is a device within which a self-sustaining controlled chain reaction is produced by fissionable material. it is thus a source of control energy that is utilized for several useful purposes. The reactor has some important part which is given below:

1. Fuel: The fassionable material such as Uranium-235 and Plutonium-239 known as fuel. These materials play an important role in operating the nuclear reactor.


2. Moderator: It slows down the neutrons to thermal energies through the elastic collision between its nuclei and fission neutrons. Thermal neutrons have a very high probability of fissioning Uranium-235 nuclei. Examples: heavy water graphite beryllium oxide. Heavy water is the best moderator.


3. Control Rods: These rods are used to control the fission rate in the reactor. these Rods are fixed in the reactor walls. These rods are made up of the material of cadmium and Boron. These materials are good absorbers of slow neutrons. Therefore when the rods are pushed into the reactor, the fission rate decreases, and when they are pulled out, the fission grows.


4. Coolant: The coolant is used to remove the heat from the reactor which is produced inside the reactor. For this purpose air-water or CO₂ is passed through the reactor.


5. Shield: various type of intense rays are emitted from the reactor, which may be injurious to the people working near the reactor. To protect them from this radiation, thick concrete walls are erected around the reactor.


6. Safety device: In case of an accident or other emergency, a special set of control rods, called " shut-off rods" enter the reactor automatically. They immediately absorb the neutrons so that the chain reaction stops entirely.

Construction:

It is made up of a large number of uranium rods which are placed in calculated geometrical lattice between layers of pure graphite ( moderator) blocks. To prevent oxidation of Uranium, the rods are covered by close-fitting aluminum cylinders. The control rods are so inserted within the lattice that they will be raised or lowered between the Uranium rods whenever necessary. The whole reactor is encircled by a concrete shield.

Working:

The actual operation of the reactor is started by pulling out the control rods so that they do not absorb many neutrons. Then, the stray neutrons, which are always present in the tractor, start fissioning the U-235 nuclei. In each fission, two or three fast neutrons are produced. These neutrons repeatedly strike the moderator and slowed down. Then, These neutrons start fissioning the U-235 nuclei. So, a chain-reaction of fission starts. The number of neutrons, which is produced in fissioning, is controlled by pushing the cadmium rods into the reactor. These rods absorb a number of the neutrons. Thus, the energy produced in the reactor is kept under control to avoid any explosion. The coolant is pond pumped through the reactor to carry away The Heat generated by the fission of uranium nuclei. the hot CO₂ passes through the heat exchanger and convert cold water into steam. This steam is used to operate turbines for generating electricity.

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Einstein’s Mass Energy Relation Derivation

Einstein’s Mass-Energy Relation: Einstein's mass energy relation gives the relation between mass and energy. It is also knows as mass-energy equivalence principle.

According to Newtonian mechanics, Newton’s second law

$f=\frac{dP}{dt}$

Where $P$ is the momentum of the particle. So put $P=mv$ in above equation:

$f=\frac{d}{dt}\left ( mv \right )\quad\quad (1)$

According to theory of relativity, mass of the particle varies with velocity so above equation $(1)$ can be written as:

$f=m \frac{dv}{dt}+v\frac{dm}{dt}\quad\quad (2)$

When the particle is displaced through a distance $dx$ by the applied force $F$. Then the increase in kinetic energy $dk$ of the particle is given by

$dk= Fdx\quad\quad (3)$

Now substituting the value of force $F$ in equation $(3)$

$dk =m\frac{dv}{dt}\cdot dx+v\frac{dm}{dt}\cdot dx \quad (4) $

$dk=mv\cdot dv +v^{2}\cdot dm \:\: (5) \: \left \{ \because \frac{dx}{dt}=v \right \}$

The variation of mass with velocity equation

$m=\frac{m_{\circ}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\quad\quad (6) $

Square both sides in above equation:

$m^{2}=\frac{m_\circ^{2}}{1-\frac{v^{2}}{c^{2}}}$

$m^{2}c^{2}-m^{2}v^{2}=m_\circ ^{2}c^{2}$

Differentiate the above equation which can be written as

$2mdm\cdot c^{2}- 2m \cdot dm \cdot v^{2}-2v\cdot dv\cdot m^{2}=0$

$c^{2}dm-v^{2}dm-vm\cdot dv$

$c^{2}dm=v^{2}dm+mv\cdot dv \quad\quad (7)$

Now substitute the value of $dk$ from equation $(5)$ in equation $(7)$. So above equation can be written as:

$dk = c^{2}dm$

Now consider that the particle is at rest initially and by the application of force it acquires a velocity $v$. The mass of body increase from ${m_{\circ}}$ to $m$. The total kinetic energy acquired by the particle is given by

$dk = \int_{m_\circ}^{m}c^{2}\cdot dm$

$k = c^{2}\left ( m-m_{\circ} \right )$

$k = mc^{2} - m_\circ c^{2}$

$k+m_\circ c^{2} = mc^{2}$

Where $k$ is the kinetic energy of the particle and $m_\circ c^{2}$ is the rest mass-energy of the particle. So The sum of these energies is equal to the total energy of the particle $E$. So

$E= m c^{2}$

Where $E$ is the total energy of the particle.

The above equation is called the mass energy equivalence equation.

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Concept of Simultaneity in Special Relativity

Concept of Simultaneity (Relative character of Time ): The interval aren't the same for two observes in relative motion. This cause an important incontrovertible fact that two events that appear to happen simultaneously to at least one observer are not simultaneous to another observer in relative motion.

Suppose two events occur (or two-time bombs explode) at different places $x_{1}$ and $x_{2}$ but at the same time $t_{0}$ with respect to an observer in a stationary frame (or on the ground). The situation of the different to an observer in moving frame $S'$ or to a pilot of a spaceship moving with velocity $v$ relative to stationary frame $S$ (or ground). To him, according to Lorentz transformation for time. The explosion at $x_{1}$ occurs at

$t'_{1} = \frac{t_{0} -x_{1}\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$ x_{2}$ occurs at

$t'_{2} = \frac{t_{0}- x_{2}\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Hence the two events (explosions) that occur simultaneously to one observer in the stationary frame are separated to another observer in a moving frame by an interval of

$t'_{2}-t'_{1}=\frac{\left ( x_{1}-x_{2} \right )\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Therefore, the principle of simultaneity in relativity is an absolute concept for two events. It depends on an observer or a frame of reference. The effect is not due to the time dilation. Hence, we conclude that there is no such thing as “absolute time” which is the same for all observers.

“Time is relative and it varies for all observers in relative motion.”

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Mean Value and Root Mean Square Value of Alternating Current

Derivation of the Mean Or Average Value of Alternating Current:

Let us consider alternating current $i$ propagating in a circuit, then the average value of the current.

$ i_{mean}=\frac{1}{\left ( \frac{T}{2} \right )}\int_{0}^{\frac{T}{2}}i \:dt \qquad (1)$

Where $\quad i = i_{0}sin \omega t \quad(2)$

Now substitute the value of current $i$ in above equation $(1)$

$ i_{mean}= \frac{2}{T}\int_{0}^{\frac{T}{2}}i_0. sin \omega t \cdot dt$

$ i_{mean}= \frac{2.i_{0}}{T}\int_{0}^{\frac{T}{2}}\sin \omega t \cdot dt$

$ i_{mean}= \frac{2 i_{0}}{T}[\frac{-cos\:\omega t}{\omega} ]_{0}^{\frac{T}{2}}$

The value of $\omega$ is $\frac{2 \pi}{T}$ i.e $\omega=\frac{2\pi}{T}$

$ i_{mean}= \frac{2 i_{0}}{T \left (\frac{2\pi}{T} \right )}\left [ -cos \left (\frac{2 \pi}{T} \right ) \left ( \frac{T}{2} \right ) \\ \qquad \qquad \qquad +cos0^\circ \right ] $

$ i_{mean}= \frac{i_{0}}{\pi}\left [ - cos\pi+cos0^{\circ} \right ]$

$ i_{mean}=\frac{i_{0}}{\pi}\left [1+1 \right ]$

$ i_{mean} =\frac{2 i_{0}}{\pi}$

$ i_{mean} = 0.637\: i_{0}$

From the above equation, we can conclude that the mean or average value of alternating current for one cycle is $0.637$ times or $63.7 \%$ of its peak value $i_{0}$.

Root mean square value of Alternating Current:

Let us consider a current $i$ propagating in a circuit, then the mean square value of alternating current.

$ \left (i_{mean} \right )^{2} = \frac{1}{T}\int_{0}^{T}i^{2} \cdot dt \quad(1)$

$ where\quad i= i_{0} \dot sin\omega t\quad (2)$

Now substitute the value of current $i$ in above equation $(1)$

$ \left ( i_{mean} \right )^{2} = \frac{1}{T}\int_{0}^{T} \left ( i_{0} \:sin\omega t\right )^{2} dt $

$ \left (i_{mean} \right )^{2} = \frac{i_{0}^2}{T}\quad \int_{0}^{T} sin^{2}\omega t \cdot dt$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{T}\int_{0}^{T} \frac{1-cos2\omega t}{2}\ dt$

$ \left (i_{mean} \right )^{2} = \frac{i_o^{2}}{2T}\int_{0}^{T}\left ( 1-cos2\omega t \right )dt$

$ \left (i_{mean} \right )^{2} = \frac{i_{0}^{2}}{2T}\left [ \left ( t \right )_{0}^{T} - \left ( \frac{sin2 \omega t}{2 \omega} \right )_{0}^{T} \right ] $

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ \left ( T-0 \right ) \\ \qquad\qquad\qquad -\frac{1}{2\omega}\left ( sin2\omega T-sin 0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ \left ( T-0 \right ) \\ \qquad\qquad\qquad -\frac{1}{2\omega}\left ( sin2\omega T-sin0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ T-\frac{1}{2\omega}\left ( sin4\pi \\ \qquad \qquad\qquad -sin0^{\circ} \right ) \right ]$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2T}\left [ T-\frac{1}{2\omega}\left ( 0-0 \right ) \right ]$

$ \left ( i_{mean} \right )^{2}=\frac{i_0^{2}}{2}$

$ \left (i_{mean} \right )^{2} = \frac{i_0^{2}}{2}$

So root mean square value of the above equation:

$ i_{rms} = \sqrt{i_{mean}^{2}}$

$i_{rms} = \frac{i_{0}}{\sqrt{2}}$

$i_{rms} = 0.707\:i_{0}$

Thus, the root mean square value of an alternating current is $0.707$ times or $70.7 \%$ of the peak value.

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Assumptions of Planck’s Radiation Law

Planck in 1900 suggested the correct explanation of the black body radiation curve. They gave the following assumption →

1. A chamber contains black body energy radiation and simple harmonic oscillators (atoms of Wall, i.e. Black lamp & Platinum coating inside wall, behave as oscillators or resonators) of molecular dimensions which can vibrate with all possible frequencies.


2. The frequency of energy radiation emitted by an oscillator is the same as the frequency of its vibration.


3. An oscillator cannot emit or absorb the energy in a continuous manner it can emit or absorb energy in a small unit (packet) called Quanta.

If an oscillator is vibrating with a frequency $ \nu $ it can only radiate in quanta of magnitude $h\nu $ i.e. “The oscillator can have only discrete energy value $E_{n}$ ” given by–

$E_{n}=nh\nu$

Where
$n$ – an integer
$h$– Planck ’s constant and the value is $6.626\times10^{-34} J-s$

The average energy of Planck’s oscillator of frequency $\nu$ -

$E_{\lambda}d\lambda = \frac{8\pi hc}{\lambda ^{5}} \frac{d\lambda }{(e^{\frac{hc}{\lambda kT}}-1)}$

$E_{\nu}d\nu= \frac{8\pi h\nu^{3}}{c^{3}}\frac{d\nu }{(e^{\frac{h\nu}{kt}}-1)}$

This assumption is most revolutionary in character. This implies that the exchange of energy between radiation and matter (Black lamp or platinum Coating ) cannot take place continuously but are limited to a discrete set of value $ 0, h\nu, 2h\nu, 3h\nu,------ nh \nu $.

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Heisenberg uncertainty principle

If the x-coordinate of the position of a particle is known to an accuracy of $\delta x$, then the x-component of momentum cannot be determined to an accuracy better than $\Delta P_{x}\approx \frac{\hbar }{\Delta x}$.

$\Delta P_{x}. \Delta x\approx \hbar$

The above inequality must be satisfied

$\Delta P_{x}. \Delta x\geqslant \hbar$

Where $\hbar $ - Planck’s Constant

This is the Uncertainty principle with macroscopic objects. Exact statement of the Uncertainty principle →

The product of the uncertainties in determining the position and momentum of the particle can never be smaller than the number of the order $\frac{\hbar }{2}$.

$\Delta P_{x}. \Delta x\geqslant \frac{\hbar}{2}$

Where $\delta x$ and $\delta P $ are defined as the root mean square deviation from their mean values.

The Uncertainty principle can also describe by the following formula →

$\Delta x.\Delta p_{x}\approx \frac{\hbar}{2}$

$\Delta x.\Delta p_{x}\geqslant \frac{\hbar}{2}$

$\Delta x.\Delta p_{x}\geqslant \frac{h}{4\pi }$

Expression for $y$ and $z$ component →

$\Delta y.\Delta p_{y}\geqslant \frac{h}{4\pi }$

$\Delta z.\Delta p_{z}\geqslant \frac{h}{4\pi }$

The uncertainty relation between energy and time →

$\Delta E.\Delta t\geqslant \frac{h}{4\pi }$

$\Delta E.\Delta t\geqslant \frac{\hbar }{2 }$

The uncertainty relation between momentum and Angular Position→

$\Delta L.\Delta \theta \geqslant \frac{h }{4\pi }$

$\Delta L.\Delta \theta \geqslant \frac{\hbar}{2}$

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The electric potential energy of a system of Charges

The Potential Energy of a system of two-point like charges→

When the system of two charged particles is configured, in which one charge is at rest of position and another is brought from infinity to near the first charge then the work done acquire by this charged particle is stored in the form of electric potential energy between these charges.

Derivation→

Let us consider, If two charge $q_{1}$ and $q_{2}$ in which one charge $q_{1}$ is at the rest of the position at point $P_{1}$ and another charge $q_{2}$ is brought from infinity to a point $P_{2}$ to configure the system then the electric potential at point $P_{2}$ due to charge particle $q_{1}$ →

$V=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}}{r}$

Where $r$ is the distance between the point $P_{1}$ and Point $P_{2}$

Here, Charge $q_{2}$ is moved in from infinity to point $P_{2}$ then the work required is →

$W=V q_{2}$

$W= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$

Since the electric potential at infinity is zero the work- done will also be zero. So total work-done from infinity to a point $P$ will be stored in the form of electric potential energy.

$U=W$

$ U= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$

The electric potential energy of a system of three-point-like charges→

To obtain the potential energy of a system of three charges. First, Obtain the work done between any two charges and then obtain the different work done for both those charges from the third charge, and then the total work done will be equal to electric potential energy.

Let us consider a system is made up of three charges $q_{1}$, $q_{2}$ and $q_{3}$ which are placed at point $P_{1}$,$P_{2}$ and $P_{3}$. Now the work done between two charges $q_{1}$ and $q_{2}$ is

$W_{1}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r_{12}}$

Now, the charge $q_{3}$ is brought from infinity to the point $P_{3}$. Work has to be done against the forces exerted by $q_{1}$ and $q_{2}$ Therefore, The work done between charges of $q_{2}$ and $q_{3}$

$W_{2}=\frac{1}{4\pi\epsilon_{0}} \frac{q_{2}q_{3}}{r_{23}}$

Now, The work done between charges $q_{1}$ and $q_{3}$

$ W_{3}=\frac{1}{4\pi\epsilon_{0}} \frac{q_{1}q_{3}}{r_{13}}$

The total work done to make a system of three charges:

$ W=W_{1}+W_{2}++W_{3}$

Now substitute the value of $W_{1}$, $W_{2}$and $W_{2}$ in above equation i.e.

$ W=\frac{1}{4\pi\epsilon_{0}} \left[\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right]$

This work is stored in the form of electric potential energy in the system.

$U=W$

$ U=\frac{1}{4\pi\epsilon_{0}} \left[\frac{q_{1}q_{2}}{r_{12}}+ \frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right]$

Similarly, the Potential energy of a system of N point system i.e.

$ U=\frac{1}{4\pi\epsilon_{0}} \sum_{i=1}^{N}\sum_{j=1}^{N}\frac{q_{i}q_{j}}{r_{ij}}$

Here $i\neq j$

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Gauss's Law for Electric Flux and Derivation

Gauss's Law:

Gauss's law for electric flux is given by Carl Friedrich Gauss in 1813. He extended the work of Joseph-Louis Lagrange. This formula was first formulated in 1713 by Lagrange. Gauss's law stated that:

The electric flux passing normal through any closed hypothetical surface is always equal to the $\frac{1}{\epsilon_{0}}$ times of the total charge enclosed within that closed surface. This closed hypothetical surface is known as Gaussian surface.

Let us consider that a $+q$ coulomb charge is enclosed within the Gaussian's surface. Then according to Gauss's Law, the electric flux will be:

$\phi _{E}= \frac{q}{\epsilon_{0}}$

The electric flux of the electric field →

$\phi_{E}=\oint \overrightarrow{E}\cdot\overrightarrow{dA}$

Substitute this value of electric flux $\phi_{E}$ in the above formula so we get →

$\oint \overrightarrow{E}\cdot\overrightarrow{dA}=\frac{q}{\epsilon_{0}}$

Where $\epsilon_{0}$ → Permittivity of the free space

The above formula of Gauss's law is applicable only under the following two conditions:

1.) The electric field at every point on the surface is either perpendicular or tangential.
2.) The magnitude of the electric field at every point where it is perpendicular to the surface has a constant value.

Derivation of Gauss's law from Coulomb's law:

1.) When the charge is within the surface
2.) When the charge is outside the surface

1. When the charge is within the surface:

Let a charge $+q$ is placed at point $O$ within a closed surface of irregular shape. Consider a point $P$ on the surface which is at a distance $r$ from the point $O$. Now take a small element or area $\overrightarrow{dA}$ around the point $P$. If $\theta$ is the angle between $\overrightarrow{E}$ and $\overrightarrow{dA}$ then electric flux through small element or area $\overrightarrow{dA}$

$d\phi_{E}=\overrightarrow{E}\cdot\overrightarrow{dA}$

$d\phi_{E}=E\:dA\:cos\theta \qquad\quad\quad (1)$

When charge is inside the surface

According to Coulomb's law, the electric field intensity $E$  at point $P$.

$E=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}$

Now substitute the value of electric field intensity $E$ in equation $(1)$

$d\phi_{E}=\frac{q}{4\pi\epsilon_{0}}\frac{dA\:cos\theta}{r^{2}}$

but $\frac{dA\:cos\theta}{r^{2}}$ is the solid angle $d\omega$ subtended by $dA$ at point $O$. Hence the above equation can be written as

$d\phi_{E}=\frac{q}{4\pi\epsilon_{0}}d\omega$

So, The total flux $\phi_{E}$ over the entire surface can be found by integrating the above equation

$\oint d\phi_{E}= \frac{q}{4\pi\epsilon_{0}}\oint d\omega$

For entire surface solid angle $d\omega$ will be equal to $4\pi$ i.e. $d\omega=4\pi$

$\phi _{E}= \frac{q}{\epsilon_{0}}$

If the closed surface enclosed with several charges like $q_{1},q_{2},q_{3},.....-q_{1},-q_{2},-q_{3},.....$. Now each charge will contribute to the total electric flux $\phi_{E}$.

$\phi_{E}= \frac{1}{\epsilon_{0}}\left [ q_{1}+q_{2}+q_{3}...-q_{1}-q_{2}-q_{3}... \right ]$

Here $\quad q=q_{i}-q_{j}$

$\phi_{E}= \frac{1}{\epsilon_{0}}\sum_{i=1,j=1}^{n}(q_{i}-q_{j})$

$\phi_{E}= \frac{1}{\epsilon_{0}}\sum q$

Where $\sum q$ → Algebraic Sum of all the charges

2. When the charge is outside the surface:

Let a point charge $+q$ be situated at point $O$ outside the closed surface. Now a cone of solid angle $d\omega$ from point $O$ cuts the surface area $dA_{1}$, $dA_{2}$ at point $P$ and $Q$ respectively. The electric flux for an outward normal is positive while for inward normal is negative so

The electric flux at point $P$ through an area

$d\phi_{1}$= $-\left (\frac{q}{4\pi \epsilon_{0}} \right )d\omega$

The electric flux at point $Q$ through area

$d\phi_{2}$= $+\left (\frac{q}{4\pi \epsilon_{0}} \right )d\omega$

The Total electric flux will be sum of all the electric flux passing through areas of surface →

$\phi_{E}=d\phi_{1}+d\phi_{2}$

$\phi_{E}=-\left ( \frac{q}{4\pi \epsilon_{0}} \right )d\omega+\left ( \frac{q}{4\pi \epsilon_{0}} \right )d\omega $

$\phi_{E}=0$

The above equation is true for all cones from point $O$ through any surface, however irregular it may be-

The total electric flux over the entire surface due to an external charge is zero.

This verifies Gauss's law.

Application of Gauss's law:

There are following some important application given below:

1. Electric field intensity due to a point charge


2. Electric field intensity due to uniformly charged spherical Shell (for Thin and Thick)


3. Electric field intensity due to a uniformly charged solid sphere (Conducting and Non-conducting)


4. Electric field intensity due to uniformly charged infinite plane sheet (for Thin and Thick)


5. Electric field intensity due to uniformly charged parallel sheet


6. Electric field intensity due to charged infinite length wire

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Vector Form of Coulomb's Law

Derivation of vector form of Coulomb's law:

Let us consider, Two-point charges $+q_{1}$ and $+q_{2}$ are separated at a distance $r$ (magnitude only) in a vacuum as shown in the figure given below.

Let $\overrightarrow{F_{12}}$ is the force on charge $+q_{1}$ due to charge $+q_{2}$ and $\overrightarrow{F_{21}}$ is the force on charge $+q_{2}$ due to charge $+q_{1}$. Then

$\overrightarrow{F_{12}}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{21}}\qquad(1)$

Where $\widehat{r}_{21}$ ➝ Unit Vector Pointing from charge $+q_{2}$ to charge $+q_{1}$

$\overrightarrow{F_{21}}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{12}}\qquad(2)$

Where$\widehat{r}_{12}$ ➝ Unit Vector Pointing from charge $+q_{1}$ to charge $+q_{2}$

From the above figure, we can conclude that the direction of unit vector $\widehat{r}_{12}$ and $\widehat{r}_{21}$ is opposite. i.e.

$\hat{r_{12}}=-\hat{r_{21}}\qquad(3)$

So from equation $(2)$ and equation $(3)$, we can write as

$\overrightarrow{F_{21}}=-\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{21}}\qquad(4)$

Now, Put the value of equation $(1)$ in equation $(4)$. So equation $(4)$, we can write as

$\overrightarrow{F_{21}}=-\overrightarrow{F_{12}}\qquad (5)$

The above equation $(5)$ shows that " The Coulomb's force is Action and Reaction Pair. This force acts on different bodies." If

$\overrightarrow{F_{12}}=\overrightarrow{F_{21}}=\overrightarrow{F}$

And

$ \hat{r_{12}}=\hat{r_{21}}=\hat{r}$

Then generalized vector form of Coulomb's Law$\overrightarrow{F}=\frac{1}{4\pi\varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\hat{r}$

Where $\hat{r}=\frac{\overrightarrow{r}}{r}$

$ \overrightarrow{F}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^3}\:\overrightarrow{r}$

Where $\overrightarrow{r}$ is displacement vector

This is a generalized vector form of Coulomb's law.

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Conversion of Galvanometer into an Ammeter

What is Ammeter?

An Ammeter is an instrument that is used to measure the electric current in the electric circuits directly in Ampere. The instrument which measures the current of the order of milliampere $(mA)$ is called the milliammeter. The internal resistance of the ideal ammeter is always zero.

What is Galvanometer?

The galvanometer is an instrument that is used to measure the very small amount of the electric charge passing through the circuit. The internal resistance of the Galvanometer is not zero.

Galvanometer used as Ammeter: To use the galvanometer as an ammeter in the circuit, The resistance of the galvanometer should be very small or almost zero as compared to the other resistance of the circuit. Because the internal resistance of an ideal ammeter is zero.

So a low resistance is connected in parallel to the galvanometer which is known as a shunt.

When a low resistance is connected in parallel to the galvanometer then the resultant resistance decreases as compared to the other resistance of the circuit and it can be easily used as an ammeter and the actual current can be measured through it.

Mathematical Analysis:

Let us consider, $G$ is the resistance of the coil of the galvanometer and the $i_{g}$ current, passing through it, produces full scale deflection. If $i$ is the maximum current of the circuit then a part of current $i_{g}$ passes through the galvanometer and the remaining current $(i-i_{g})$ passes through the shunt $S$. Since $S$ and $G$ are parallel, the potential difference across them will be the same:

$i_{g} \times G = \left( i- i_{g} \right) \times S \qquad(1)$

$\frac{i_{g}}{i}=\frac{S}{S+G}$

i.e. only $\frac{S}{S+G}$th part of the total current will flow in the coil of the Galvanometer. Again from equation $(1)$:

$S=\left(\frac{i_{g}}{i-i_{g}}\right)G \qquad(2)$

If the current $i_{g}$ passes through the coil of the galvanometer and produces a full-scale deflection on the meter scale of a galvanometer, then the current $i$ in the circuit corresponds to the full-scale deflection. Thus, with a shunt $S$ of the above value, the galvanometer will be an ammeter in the range $0$ to $i$ ampere.

Example: Let a current of $1 A$ in the coil of a galvanometer produce a full-scale deflection. To convert it into an ammeter of range $10A$, a shunt is required such that when the current in the circuit is $10A$, only $1A$ flows in the coil remaining passes through the shunt. Now From substitute the $i_{g}=1A$ and $i=10A$ in the above equation $(2)$:

$\frac{S}{G}= \frac{1}{\left( 10 -1 \right)}$

$\frac{S}{G}= \frac{1}{9}$

The resistance of the shunt should be only $\frac{1}{9}$th the resistance of the galvanometer coil.

Note: As the shunt resistance value is very small so the combined resistance of the galvanometer and the shunt also becomes very small and hence the ammeter has a much smaller resistance than the galvanometer.

Resistance of Ammeter:

$\frac{1}{R_{A}}=\frac{1}{G}+\frac{1}{S}$

$R=\frac{G\:S}{G+S}$

When a Galvanometer is used in the circuit and connected in the series to measure the electric current:

The galvanometer is used in series to measure the electric current of the circuit so that the whole amount of the current passes through it. but the galvanometer will have some resistance due to the resultant resistance of the circuit increasing and the current in the circuit somewhat decreasing. Therefore the current read by the Galvanometer is less than the actual current.

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