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Showing posts from August, 2022

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