A uniform magnetic field vec{B} = B_0 hat{j} | vedclass

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

Question

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.The radius of this circular path

Vedclass · Accepted Answer

$\frac{B_0 q d}{m}$

Vedclass · Answer

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.The radius of this circular path is given by $R = \frac{mv}{qB_0}$.The particle is projected from $(d, 0, 0)$ towards the negative $x$-axis. The $y-z$ plane is the plane where $x = 0$.For the particle not to hit the $y-z$ plane,the radius of the circular path must be less than or equal to the distance $d$ from the $y-z$ plane.Thus,$R \leq d$.Substituting the expression for $R$: $\frac{mv}{qB_0} \leq d$.Solving for $v$: $v \leq \frac{B_0 q d}{m}$.Therefore,the maximum value of $v$ is $v_{	ext{max}} = \frac{B_0 q d}{m}$.

$A$ uniform magnetic field $\vec{B} = B_0 \hat{j}$ exists in space. $A$ particle of mass $m$ and charge $q$ is projected towards the negative $x$-axis with speed $v$ from the point $(d, 0, 0)$. The maximum value of $v$ for which the particle does not hit the $y-z$ plane is

Similar Questions

$A$ 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:

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?

$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

An electron with energy $0.1 \, keV$ moves at a right angle to the Earth's magnetic field of $1 \times 10^{-4} \, Wb/m^2$. The frequency of revolution of the electron will be. (Take mass of electron $= 9.0 \times 10^{-31} \, kg$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module