At $t = 0$,a positively charged particle of mass $m$ is projected from the origin with velocity $u_0$ at an angle $37^o$ from the $x$-axis as shown in the figure. $A$ constant magnetic field $\vec{B_0} = B_0 \hat{j}$ is present in space. After a time interval $t_0$,the velocity of the particle may be:

  • A
    $u_0 \left[ \frac{\sqrt{39}}{2} \hat{i} + \frac{3}{5} \hat{j} - \frac{\hat{k}}{\sqrt{3}} \right]$
  • B
    $u_0 \left[ \frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{6}} \right]$
  • C
    $u_0 \left[ \frac{\sqrt{39}}{10} \hat{i} + \frac{3}{5} \hat{j} + \frac{1}{4} \hat{k} \right]$
  • D
    $u_0 \left[ \frac{\sqrt{39}}{10} \hat{i} + \frac{3}{5} \hat{j} + \frac{1}{2} \hat{k} \right]$

Explore More

Similar Questions

An electron has mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is moving with a velocity of $10^6 \ m/s$. It enters a region where a magnetic field exists. If it describes a circle of radius $0.10 \ m$,the intensity of the magnetic field must be:

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)

An electron and a proton having same momenta enter perpendicular into a magnetic field then

$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)

An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo