An electron moves in a Bohr orbit. The magnet | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The magnetic field $B$ at the centre of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$,where $I$ is the current and $r$ is the radiu

Vedclass · Accepted Answer

$n^{-5}$

Vedclass · Answer

(D) The magnetic field $B$ at the centre of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$,where $I$ is the current and $r$ is the radius of the orbit.For an electron revolving in an orbit,the current $I = \frac{e}{T} = \frac{ev}{2\pi r}$,where $v$ is the velocity and $T$ is the time period.Substituting $I$ into the magnetic field formula: $B = \frac{\mu_0 ev}{4\pi r^2}$.According to Bohr's model,the radius $r \propto n^2$ and the velocity $v \propto n^{-1}$.Substituting these proportionalities: $B \propto \frac{v}{r^2} \propto \frac{n^{-1}}{(n^2)^2} = \frac{n^{-1}}{n^4} = n^{-5}$.Therefore,the magnetic field at the centre is proportional to $n^{-5}$.

An electron moves in a Bohr orbit. The magnetic field at the centre is proportional to

Similar Questions

The ratio between the total acceleration of the electron in a singly ionized helium atom and a hydrogen atom (both in the ground state) is:

$A$ hydrogen atom in the ground state absorbs $ 10.2 \text{ eV} $ of energy. The orbital angular momentum of the electron is increased by:

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

An electron revolving in the $n^{\text{th}}$ Bohr orbit has a magnetic moment $\mu_n$. If $\mu_n \propto n^x$,the value of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module