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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) \l

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 infinitesima

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

Force between two long and parallel current-carrying conductor

Derivation of Force between two long and parallel current-carrying conductors: Let us consider: The two long straight, parallel conductors = $PQ$ and $RS$ The length of the conductor= $l$ The distance between the parallel conductor = $r$ The current flowing in conductor $PQ$ = $i_{1}$ The current flowing in conductor $RS$ = $i_{2}$ The magnetic field due to conductor $PQ$ = $B_{1}$ The magnetic field due to conductor $RS$ = $B_{2}$ The magnetic force on conductor $PQ$= $F_{1}$ The magnetic force on conductor $RS$= $F_{2}$ Force Between Parallel Current Carrying Conductor Now Consider the magnetic force on conductor $RS$ is i.e. $F_{2}=i_{2}B_{1}l sin\theta$ Where $\theta$ is the angle between the magnetic field and length element of conductor i.e. $\theta=90^{\circ}$ so above equation can be written as, $F_{2}=i_{2}B_{1}l sin 90^{\circ}$ $F_{2}=i_{2}B_{1}

Force on current carrying conductor in uniform magnetic field

Derivation of force on current-carrying conductor in uniform magnetic field: Let us consider: The length of the conductor - $l$ The cross-section area of the current carrying conductor - $A$ The current flow in a conductor- $i$ The drift or average velocity of the free electrons - $v_{d}$ The current-carrying conductor is placed in a magnetic field - $B$ The total number of free electrons in the current carrying conductor - $N$ Force on current carrying conductor in the uniform magnetic field Now the magnetic force on one free electron in a conductor - $F'= ev_{d}B sin\theta \qquad(1)$ The net force on the conductor is due to all the free electrons present in the conductor $F=N\: F' \qquad(2)$ Let $N$ is the number of free electrons per unit volume of conductor. So the total number of free electrons in the $Al$ volume of the conductor will be $N=nAl \qquad(3)$ Now substitute the value of $N$ and $F'$ from above equation $(1)$ and equat

Principle, Construction and Working of Current Carrying Solenoid

Current Carrying Solenoid: The Solenoid is an artificial magnet which is used for different purposes. Principle of Solenoid: The principle of the solenoid is based on the "Ampere Circuital Law" and its magnetic field is raised due to the current carrying a circular loop. Construction of Solenoid: The current carrying solenoid is consist of insulated cylindrical material and conducting wire. The conducting wire like copper is wrapped closely around the insulated cylindrical material ( like cardboard, clay, or plastic). The end faces of the conducting wire are connected to the battery. Long Current Carrying Solenoid Working: When the electric current flow in the solenoid then a field (i.e. Magnetic field) is produced around and within the current carrying solenoid. This magnetic field is produced in solenoid due to circular loops of the solenoid and the direction of the magnetic field is depend upon the direction of the electric current flow i

Comparison of Step Index and Graded Index Fibres

Comparison of Step Index Fibres and Graded Index Fibres(GRIN)→ S.No. Step Index Fibre Graded Index Fibre 1. In a step-index fibre, the refractive index of the core a constant value. In graded-index fibre, the refractive index in the core decreases continuously in a nearly parabolic manner from a maximum value at the centre of the core to a constant value at the core-cladding interface. 2. For a step-index fibre, the variation of refractive index is mathematically expressed as, $\begin{cases} & \mu(r)=\mu_{1} \qquad 0 < r < a \quad for (core)\\ & \mu(r)=\mu_{2} \qquad r >a \quad for(Cladding)\\ \end{cases} \\ Where \: \mu_{1} > \mu{2} $ Parabolic refractive index variation in GRIN fibre is mathematically expressed as, $ \begin{cases} & \mu^{2}(r)=\mu^{2}_{1} \left[ 1- \left(\frac{r}{\alpha} \right)^{2} \

Comparison of Single Mode and Multimode Index Fibres

Comparison of Single-Mode Index Fibres and Multimode Index Fibres→ S.No. Single Mode Index Fibre Multimode Index Fibre 1. In single mode index fibre, the diameter of the core is very small and is of the same order as the wavelength of light to be propagated. It is in the range $5\mu m - 10 \mu m$. The Cladding diameter is about $125 \mu m$. In multimode index fibre, the diameter of the core is large. It is in the range $30\mu m - 100 \mu m$. The Cladding diameter is in the range $125 \mu m - 500 \mu m$. 2. The difference in refractive indices of the core and cladding material is very small. The difference in the refractive indices of the core and the cladding materials is large. 3. In single-mode fibre, only a single mode is propagated. In multi-mode fibre, a large number of modes can be propagated. 4. Single mod fibre does require a much more sophisticated li

Difference between Fraunhofer and Fresnel diffraction

Difference between Fraunhofer Diffraction and Fresnel Diffraction→ S.No. Fresnel Diffraction Fraunhofer Diffraction 1. The distance between source to slit and slit to screen is finite. The distance between source to slit and slit to screen is infinite. 2. The shape of the incident wavefront on the slit is spherical or cylindrical. The shape of the incident wavefront on the slit is plane. 3. The shape of the incident wavefront on the screen is spherical or cylindrical. The shape of the incident wavefront on the screen is a plane. 4. There is a path difference created between the rays before entering the slit. This path difference depends on the distance between the source and slit. There is not any path difference between the rays before entering the slit. 5. Path difference between the rays forming the diffraction pattern depends on the

Principle construction and working of potentiometer

Potentiometer- An ideal voltmeter that does not change the original potential difference, needs to have infinite resistance. But a voltmeter cannot be designed to have infinite resistance. The potentiometer is one such instrument that does not draw any current from the circuit and still measures the potential difference. so it behaves as an ideal voltmeter. "A potentiometer is an instrument. This is used to measure the potential difference between two points of an electric circuit and emf of a cell." Principle- The principle of the potentiometer depends upon the potential gradient along the wire i.e. "When a constant current flows in a wire then the potential drops per unit length of the wire". Construction- A potentiometer consists of a long wire $AB$ of uniform cross-section, usually, this wire is $4 m$ to $10 m$ long and it is made of the material having high resistivity and low-temperature coefficient such as manganin or constant. Usually,

Metre Bridge OR Slide Wire Bridge

What is Metre Bridge? It is the simplest practical application of the Wheatstone's bridge that is used to measure an unknown resistance. Principle: Its working is based on the principle of Wheatstone's Bridge. When the Wheatstone's bridge is balanced $\frac{P}{Q}=\frac{R}{S}$ Construction: It consists of usually one-meter long manganin wire of uniform cross-section, stretched along a meter scale fixed over a wooden board and with its two ends soldered to two L-shaped thick copper strips $A$ and $C$. Between these two copper strips, another copper strip is fixed so as to provide two gaps $mn$ and $m_{1}n_{1}$. A resistance box (R.B.) is connected in the gap $mn$ and the unknown resistance $S$ is connected in the gap $m_{1}n_{1}$. A cell of emf $E$, Key $(K)$, and rheostat are connected across $AC$. A movable jockey and a galvanometer are connected across the $BD$, as shown in the figure. Metre Bridge Or Slide Wire Bridge Working: In

Wheatstone's Bridge

It is an arrangement of four resistance used to determine one of this resistance quickly and accurately in terms of the remaining three resistance. Objective: To find the unknown resistance with the help of the remaining three resistance. Principle of Wheatstone Bridge:  The principle of Wheatstone bridge is based on the principle of Kirchhoff's Law. Construction: A Wheatstone bridge consists of four resistance $P$,$Q$,$R$, and $S$. This resistance is connected to form quadrilateral $ABCD$. A battery of EMF $E$ is connected between point $A$ and $C$ and a sensitive galvanometer is connected between point $B$ and $D$ Which is shown in the figure below. Diagram of Wheatstone's Bridge Working: To find the unknown resistance $S$, The resistance $R$ is to be adjusted like there is no deflection in the galvanometer. which means that there is not any flow of current in the arm $BD$. This condition is called "Balanced Wheatstone bridge" i.e

Kirchhoff's laws for an electric circuits

Kirchhoff's laws:  Kirchhoff had given two laws for electric circuits i.e. Kirchhoff's Current Law or Junction Law Kirchhoff's Voltage Law or Loop Law Kirchhoff's Current Law or Junction Law: Kirchhoff's current law state that The algebraic sum of all the currents at the junction in any electric circuit is always zero. $\sum_{1}^{n}{i_{n}}=0$ Sign Connection:   While applying the KCL, the current moving toward the junction is taken as positive while the current moving away from the junction is taken as negative. The flow of Current in a junction So from figure,the current $i_{1}$,$i_{2}$,$i_{5}$ is going toward the junction and the current $i_{3}$,$i_{4}$, So $\sum{i}= i_{1}+i_{2}+(-i_{3})+(-i_{4})+i_{5}$ According to KCL $\sum{i}= 0$, Now the above equation can be written as $i_{1}+i_{2}+