An electron revolves in a circle of radius 0. | vedclass

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

Question

(C) Given: radius $r = 0.4 	ext{ Å} = 0.4 	imes 10^{-10} 	ext{ m}$, speed $v = 10^6 	ext{ m/s}$, charge $q = 1.6 	imes 10^{-19} 	ext{ C}$.The mo

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given: radius $r = 0.4 	ext{ Å} = 0.4 	imes 10^{-10} 	ext{ m}$, speed $v = 10^6 	ext{ m/s}$, charge $q = 1.6 	imes 10^{-19} 	ext{ C}$.The moving electron constitutes an electric current $i$ given by $i = \frac{q}{T}$, where $T$ is the time period.Since $T = \frac{2\pi r}{v}$, we have $i = \frac{qv}{2\pi r}$.Substituting the values:$i = \frac{1.6 	imes 10^{-19} 	imes 10^6}{2\pi 	imes 0.4 	imes 10^{-10}} = \frac{1.6 	imes 10^{-13}}{0.8\pi 	imes 10^{-10}} = \frac{2 	imes 10^{-3}}{\pi} 	ext{ A}$.The magnetic field $B$ at the center of a circular current loop is $B = \frac{\mu_0 i}{2r}$.Substituting the values:$B = \frac{4\pi 	imes 10^{-7}}{2 	imes 0.4 	imes 10^{-10}} 	imes \frac{2 	imes 10^{-3}}{\pi} = \frac{4\pi 	imes 10^{-7} 	imes 2 	imes 10^{-3}}{0.8\pi 	imes 10^{-10}} = \frac{8\pi 	imes 10^{-10}}{0.8\pi 	imes 10^{-10}} = 10 	ext{ T}$.

Similar Questions

Find the magnetic field at $O$.

$A$ straight conductor of length $32 \,cm$ carries a current of $30 \,A$. Magnetic induction at a point in air at a perpendicular distance of $12 \,cm$ from the mid-point of the conductor is (in $G$)

The magnetic field intensity at the centre of a circular wire of radius $0.1 \,m$ carrying a current of $0.2 \,A$ is:

At what distance from a long straight wire carrying a current of $12 \, A$ will the magnetic field be equal to $3 \times 10^{-5} \, Wb/m^2$?

The magnitude of magnetic induction at the mid-point $O$ due to the current arrangement as shown in the figure will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module