A particle of mass m and charge q moving with | vedclass

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

Question

(A, C) The particle enters region-$c$ if the radius of its circular path $R$ is greater than the width $L$ of region-$b$.Since $R = \frac{mv}{qB}$, th

Vedclass · Accepted Answer

Time spent in region-$b$ is independent of velocity $v$

Vedclass · Answer

(A, C) The particle enters region-$c$ if the radius of its circular path $R$ is greater than the width $L$ of region-$b$.Since $R = \frac{mv}{qB}$, the condition for entering region-$c$ is $\frac{mv}{qB} > L$, which implies $v > \frac{qBL}{m}$. Thus, option $A$ is correct.Since the magnetic field $B$ is uniform and the velocity vector $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$, the magnetic force provides the necessary centripetal force, making the path of the particle circular in region-$b$. Thus, option $C$ is correct.The time spent in region-$b$ is given by $t = \frac{	heta}{\omega}$, where $	heta$ is the angle subtended by the arc at the center and $\omega = \frac{qB}{m}$ is the angular velocity. Since the angle $	heta$ depends on the radius $R$ (and thus on velocity $v$), the time spent in region-$b$ depends on the velocity $v$. Therefore, option $D$ is incorrect.

Similar Questions

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $6 \times 10^{-4} \ T$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \ kg$ and charge of electron is $1.6 \times 10^{-19} \ C$). (in $cm$)

$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then, it travels:

Vedclass Test Series

Exam Paper Generator

Online Exam Module