An electron is revolving around a proton, pro | vedclass

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

Question

(A) The magnetic field $B$ at the center of a circular orbit due to a revolving electron is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi i}{r}$.S

Vedclass · Accepted Answer

$10^{17} \, rad/s$

Vedclass · Answer

(A) The magnetic field $B$ at the center of a circular orbit due to a revolving electron is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi i}{r}$.Since the current $i = \frac{q}{T} = \frac{e}{2\pi/\omega} = \frac{e\omega}{2\pi}$, we substitute this into the formula:$B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi (e\omega / 2\pi)}{r} = 10^{-7} \cdot \frac{e\omega}{r}$.Given $B = 16 \, Wb/m^2$, $e = 1.6 	imes 10^{-19} \, C$, and $r = 1 \, Å = 10^{-10} \, m$:$16 = 10^{-7} \cdot \frac{1.6 	imes 10^{-19} \cdot \omega}{10^{-10}}$.$16 = 10^{-7} \cdot 1.6 	imes 10^{-9} \cdot \omega$.$16 = 1.6 	imes 10^{-16} \cdot \omega$.$\omega = \frac{16}{1.6 	imes 10^{-16}} = 10 	imes 10^{16} = 10^{17} \, rad/s$.

An electron is revolving around a proton, producing a magnetic field of $16 \, Wb/m^2$ in a circular orbit of radius $1 \, Å$. Its angular velocity will be:

Similar Questions

Consider the circular loop carrying current $i$ as shown in the figure. The magnetic field at the central point $O$ is

The magnetic induction at the centre $O$ is:

If the radius of a coil is halved and the number of turns is doubled,then the magnetic field at the centre of the coil,for the same current,will

$A$ dielectric circular disc of radius $R$ carries a uniform surface charge density $\sigma$. If it rotates about its axis with angular velocity $\omega$,the magnetic field at the center of the disc is:

In the following figure,a wire bent in the form of a regular polygon of $n$ sides is inscribed in a circle of radius $a$. The net magnetic field at the centre will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module