An electron and a proton are moving on straight parallel paths with the same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement$(s)$ is/are true?
$(A)$ They will never come out of the magnetic field region.
$(B)$ They will come out travelling along parallel paths.
$(C)$ They will come out at the same time.
$(D)$ They will come out at different times.

  • A
    $(B)$ and $(C)$
  • B
    $(B)$ and $(D)$
  • C
    $(A)$ and $(B)$
  • D
    $(A)$ and $(D)$

Explore More

Similar Questions

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

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 $cm$)

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 $= 9.1 \times 10^{-31} \, kg$)

$A$ magnetic field $\overrightarrow{B} = B_0 \hat{j}$ exists in the region $a < x < 2a$ and $\overrightarrow{B} = -B_0 \hat{j}$ in the region $2a < x < 3a$,where $B_0$ is a positive constant. $A$ positive point charge moving with a velocity $\overrightarrow{v} = v_0 \hat{i}$,where $v_0$ is a positive constant,enters the magnetic field at $x = a$. The trajectory of the charge in this region can be like,

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

Difficult
View Solution

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