$A$ particle with charge-to-mass ratio $\frac{q}{m} = \alpha$ is shot with a speed $v$ towards a wall at a distance $d$,moving perpendicular to the wall. The minimum value of the magnetic field $\vec{B}$ that exists in this region,perpendicular to the velocity of the particle,such that the particle does not hit the wall is:

  • A
    $\frac{v}{\alpha d}$
  • B
    $\frac{2v}{\alpha d}$
  • C
    $\frac{v}{2\alpha d}$
  • D
    $\frac{v}{4\alpha d}$

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Similar Questions

$A$ particle of charge $q$ and mass $m$ moving with a velocity $v$ along the $x$-axis enters the region $x > 0$ with a uniform magnetic field $B$ along the $\hat{k}$ direction. The particle will penetrate this region in the $x$-direction up to a distance $d$ equal to:

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

An electron,moving in a uniform magnetic field of induction of intensity $\vec{B}$,has its radius directly proportional to

$A$ mixed beam of $He^+$ and $O^{2+}$ ions (mass of $He^+ = 4 \ amu$ and that of $O^{2+} = 16 \ amu$) passes through a region of a constant perpendicular magnetic field. If the kinetic energy of all the ions is the same,then:

An electron (mass $m$) is accelerated through a potential difference of $V$ and then it enters a magnetic field of induction $B$ normal to the field lines. The radius of the circular path is ($e$ = electronic charge).

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