The magnetic field induction produced at the | vedclass

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

Question

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

Vedclass · Accepted Answer

$n^{-5}$

Vedclass · Answer

(B) The magnetic field $B$ at the center of a circular orbit is given by $B = \frac{\mu_0 I}{2r}$,where $I$ is the equivalent current and $r$ is the radius of the orbit.From Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$. Since $v = \omega r$,we have $mr^2\omega = \frac{nh}{2\pi}$.For a hydrogen atom,the radius $r \propto n^2$. Substituting this into the angular momentum equation: $m(n^2)^2\omega \propto n$,which implies $\omega \propto \frac{n}{n^4} = n^{-3}$.The equivalent current $I$ is given by $I = \frac{e}{T} = \frac{e\omega}{2\pi}$,so $I \propto \omega \propto n^{-3}$.Now,substituting $I \propto n^{-3}$ and $r \propto n^2$ into the magnetic field formula: $B \propto \frac{I}{r} \propto \frac{n^{-3}}{n^2} = n^{-5}$.Therefore,the magnetic field induction is proportional to $n^{-5}$.

The magnetic field induction produced at the centre of orbit due to an electron revolving in $n^{\text{th}}$ orbit of a hydrogen atom is proportional to

Similar Questions

Two long straight parallel wires are separated by a distance $2d$. Each wire carries a current $I$ in the same direction. The magnetic field at a point $P$ midway between them is

The magnetic field on the axis of a circular loop of radius $R = 100\,cm$ carrying current $I = \sqrt{2}\,A$,at a point $x = 1\,m$ away from the centre of the loop is given by:

$A$ circular coil carrying current has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{27}$th of its value at the centre of the coil is:

$A$ solenoid of $1000$ turns per metre has a core with relative permeability $500$. Insulated windings of the solenoid carry an electric current of $5 \, A$. The magnetic flux density produced by the solenoid is (permeability of free space $= 4 \pi \times 10^{-7} \, H/m$)

$A$ charged particle $A$ with charge $q = 2 \, C$ has a velocity $v = 100 \, m/s$. When it passes through point $A$ with the velocity in the direction shown in the figure,the strength of the magnetic field at point $B$ due to this moving charge is ....... $\mu T$ $(r = 2 \, m)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module