A particle of mass m and carrying a charge q | vedclass

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

Question

(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it undergoes uniform circular motion.The magnetic force prov

Vedclass · Accepted Answer

Depends on magnetic field and $(q/m)$ of the particle

Vedclass · Answer

(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it undergoes uniform circular motion.The magnetic force provides the necessary centripetal force: $qvB = \frac{mv^2}{r}$.From this,the radius of the orbit is $r = \frac{mv}{qB}$.The time period $T$ is the time taken to complete one full circle: $T = \frac{2\pi r}{v}$.Substituting the value of $r$: $T = \frac{2\pi (mv/qB)}{v} = \frac{2\pi m}{qB}$.Thus,the time period $T$ is independent of velocity $v$ and radius $r$,and depends only on the mass $m$,charge $q$,and magnetic field $B$. Therefore,it depends on the magnetic field and the charge-to-mass ratio $(q/m)$.

$A$ particle of mass $m$ and carrying a charge $q$ enters with a velocity $v$ perpendicular to a uniform magnetic field $B$. The time period of rotation of the particle:

Similar Questions

In the Thomson experiment for finding $e/m$ for electrons,the beam of electrons is replaced by a beam of muons (particles with the same charge as electrons but with a mass $208$ times that of electrons). The condition for no deflection is satisfied if:

An electron is projected with uniform velocity along the axis inside a current-carrying long solenoid. Then:

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module