Magnetic Dipole Moment of a Revolving Electron

What is value of bohr magneton?

In Bohr's hydrogen like atom magnetic dipole moment in ${\mathrm{n}}^{\mathrm{th}}$ orbit is expressed as.

When charged particle moves in circle then vectorial representation of magnetic dipole moment is ( if $\overrightarrow{\mathrm{L}}$ is angular momentum)

An electron revolves around the nucleus of hydrogen like atom then magnetic moment of electron is

The relation between the magnetic moment $\mu$ of a revolving electron around the nucleus with principal quantum number $n$ is

When the space within a current-carrying toroid is filled with aluminium, the susceptibility of aluminium is $2.1\times {10}^{-5}$. The percentage increase in the magnetic field $B$ is $2.1\times {10}^{-K}$. Find the value of K.

Choose the option that contains the correct value of Bohr Magneton.

Idenitfy the correct option in relation with the gyromagnetic ratio of an electron in sodium atom.

The numerical value of gyromagnetic ratio of an electron is given by:

What is the magnetic moment associated with the orbital motion of the electron if an electron of charge $e$ moves in a circular orbit of radius $r$ around the nucleus at a frequency $\nu$?

What is the correct value of Bohr magneton?

A particle of charge q and mass m moves in a circular orbit of radius r with angular speed $\omega$ . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$ . The associated magnetic moment $m$ is given by

Magnetic moment due to the motion of the electron in ${\mathrm{n}}^{\mathrm{t}\mathrm{h}}$ energy state of hydrogen atom is proportional to

In a hydrogen atom, the electron is making $6.6\times {10}^{15} \mathrm{r}\mathrm{e}\mathrm{v} {\mathrm{s}}^{-1}$ around the nucleus in an orbit of radius $0.528 Å$. The magnetic moment ($\mathrm{A} {\mathrm{m}}^2$) will be

In a hydrogen atom, the electron is making $6.6\times {10}^{15} \mathrm{re}\mathrm{v} {\mathrm{s}}^{-1}$around the nucleus in an orbit of radius $0.528 \mathrm{Å}.$The magnetic moment $\left(\mathrm{A}{\mathrm{m}}^2\right)$ will be

The magnetic moment of an electron orbiting in a circular orbit of radius r with a speed v is equal to:

An electron moving around the nucleus with an angular momentum $l$ has a magnetic moment

An electron moving around the nucleus with an angular momentum $L$ has a magnetic moment of

An electron is moving in an orbit of the radius $R$ with a time period $T$ as shown in the figure. The magnetic moment produced may be given by |$e$| represents the magnitude of the electron charge.

     
$\left|\text{A}\right|=\pi {\text{R}}^2$