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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Magnetic dipole moment of a revolving electron

The magnetic dipole moment of a revolving electron (Or Magnetic Moment due to Orbital Angular Momentum):

An electron revolving in an orbit about the nucleus of an atom behaves like a current carrying loop. It is called a minute current-loop and produces a magnetic field. Every current loop is associated with a magnetic moment.

Magnetic Dipole Moment of a Revolving Electron

Let us consider, that the magnetic moment associated with a loop carrying current $i$ and having area $A$ is:

$\mu_{L}= i.A \qquad(1)$

The current due to a revolving electron is

$i=\frac{e}{T}$

Where

$T$- The period of revolution of electron motion around the nucleus i.e $T=\frac{2 \pi r}{v}$
$e$- Charge on an electron So from the above equation

$i=\frac{e}{\frac{2 \pi r}{v}}$

$i=\frac{ev}{2 \pi r} \qquad(2)$

The area of the current loop is:

$A=\pi r^{2} \qquad(3)$

Now put the value of $i$ and $A$ in equation $(1)$

$\mu_{L}= \left( \frac{ev}{2 \pi r} \right) \left( \pi r^{2} \right)$

$\mu_{L}= \frac{evr}{2} \qquad(4)$

$\mu_{L}= \left(\frac{evr}{2}\right) \left( \frac{m}{m} \right)$

$\mu_{L}= \left(\frac{e}{2m}\right) \left( mvr \right)$

$\mu_{L}= \left(\frac{e}{2m}\right) L \qquad(5) \qquad (\because L=mvr) $

Where $L$- The orbital angular momentum of the electron and another value of $L$ is

$L=\frac{nh}{2\pi} \qquad(6)$

$\mu_{L}= \left(\frac{e}{2m}\right) \frac{nh}{2\pi}$

$\mu_{L}= n \: \left(\frac{eh}{4m \pi}\right)$

Where $n=1,2,3......$ is the principle quantum number.

This equation gives the magnetic moment associated with the orbital motion of the electron.

Bohr Magneton:

Bohr Magneton is defined as the angular momentum of an eletcron in ground state.

We know that:

Principle Quantum Number$(n)=1$

Charge of a electron $(e)=1.6\times10^{-19} C$

Planck Constant $(h)= 6.623 \times 10^{-34} J-sec$

Mass of electron $(m)= 9.1 \times 10^{-31} Kg$

So the magnetic moment of an electron in the ground state:

$\mu_{B}= n \: \left(\frac{eh}{4m \pi}\right)$

Now subtitute the value of $n,e,h,m$ in above equation:

$\mu_{B}= \frac{1 \times 1.6\times10^{-19} \times 6.623 \times 10^{-34} }{4 \times 3.14 \times 9.1 \times 10^{-31} }$

$\mu_{B}= 9.274 \times 10^{-24} A-m^{2}$

The magnetic moments due to orbital motion of electrons in higher orbits are multiples of the Bohr magneton value.

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Magnetic potential energy of current-loop in a magnetic field

Magnetic potential energy:

When a current carrying loop is placed in an external magnetic field the torque is acted upon the current loop which tends to rotate the current loop in a magnetic field. Therefore the work is done to change the orientation of the current loop against the torque. This work is stored in the form of magnetic potential energy in the current loop. This is known as the magnetic potential energy of the current loop.

Note: The current loop has magnetic potential energy depending upon its orientation in the magnetic field.

Derivation of Potential energy of current-loop in a magnetic field:

Let us consider, A current loop of magnetic moment $\overrightarrow{m}$ is held with its axis at an angle $\theta$ with the direction of a uniform magnetic field $\overrightarrow{B}$. The magnitude of the torque acting on the current loop or magnetic dipole is

$\tau=m \: B \: sin\theta \qquad(1)$

Now, the current loop is rotated through an infinitesimally small angle $d\theta$ against the torque. The work done to rotate the current loop

$dW=\tau \: d\theta$

$dW=m \: B \: sin\theta \: d\theta $ {from equation $(1)$}

If the current loop is rotated from an angle (or orientation) $\theta_{1}$ to $\theta_{2}$ then the work done

$W=\int_{\theta_{1}}^{\theta_{2}} m \: B \: sin\theta d\theta$

$W= m \: B \: \left[ -cos\theta \right]_{\theta_{1}}^{\theta_{2}} $

$W= m \: B \: \left( cos\theta_{1} - cos\theta_{2} \right) $

This work is stored in the form of potential energy $U$ of the current loop :

$U= m \: B \: \left( cos\theta_{1} - cos\theta_{2} \right) $

If $\theta_{1}=90^{\circ}$ and $\theta_{2}= \theta$

$U= m \: B \: \left( cos90^{\circ} - cos\theta \right) $

$U= - m \: B \: cos\theta $

$U= - \overrightarrow{m} . \overrightarrow{B}$

Thus, a current loop has minimum potential energy when $\overrightarrow{m}$ and $\overrightarrow{B}$ are parallel and maximum potential energy when $\overrightarrow{m}$ and $\overrightarrow{B}$ are antiparallel.

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Magnetic Dipole Moment of Current carrying loop

Current carrying Loop or Coil or Solenoid:

The current carrying loop (or Coil or solenoid) behaves like a bar magnet. A bar magnet with the north and south poles at its ends is a magnetic dipole, so a current -loop is also a magnetic dipole.

Equation of Magnetic Dipole Moment of Current carrying Loop:

When a current loop is suspended in a magnetic field, it experiences the torque which tends to rotate the current loop to a position in which the axis of the loop is parallel to the field. So the magnitude of the torque acting on the current loop in the uniform magnetic field $\overrightarrow{B}$ is given by:

$\tau=iAB sin\theta \qquad(1)$

Where $A$ - Area of the current loop

We also know that when the electric dipole is placed in the electric field, it also experiences the torque which tends to rotate the electric dipole in the electric field. So the magnitude of the torque on the electric dipole in the uniform electric field $\overrightarrow{E}$ is given by:

$\tau=pE sin\theta \qquad(2)$

Where $p$ - The magnitude of the electric dipole moment

Now compare the equation $(1)$ and equation $(2)$ and we can conclude that the current loop also has a magnetic dipole moment just like an electric dipole have an electric dipole moment. The magnetic dipole moment is associated with the current in the loop and the area of the current loop. It is represented by $\overrightarrow {m}$. So the magnitude of the magnetic dipole moment of current carrying loop is:

$m=iA$

The vector form of the magnetic dipole moment current carrying loop is

$\overrightarrow{m} = i\overrightarrow{A}$

The magnetic dipole moment of current carrying coil: If the current-carrying loop has $N$ number of turns (i.e current carrying coil) then the magnetic dipole moment of current carrying coil:

$m=NiA$

The vector form of the magnetic dipole moment of the current carrying coil is

$\overrightarrow{m} =N i\overrightarrow{A}$

The magnetic dipole moment of Circular Loop: Let us consider the circular loop of radius $a$ in which current $i$ is flowing the magnitude of the magnetic dipole moment of the circular loop:

$m=i A$

Here the area $A$ of the circular loop is $\pi a^{2}$ then the magnitude of the magnetic dipole moment of the circular loop is:

$m=i \pi a^{2} \qquad(3)$

The magnetic field at the center of the current carrying a circular loop in terms of current is:

$B=\frac{\mu_{\circ}i}{2a}$

Now substitute the value of $i$ from equation $(3)$ in the above equation then the magnetic field at the center of the current carrying circular loop in terms of magnetic dipole moment is:

$B=\frac{\mu_{\circ}m}{2\pi a^{3}}$

$B=\frac{\mu_{\circ}}{4\pi} \frac{2m}{a^{3}}$

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