$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $times$)

  • A
    $4$
  • B
    $12$
  • C
    $8$
  • D
    $16$

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$A$ proton, a deuteron, and an $\alpha$-particle enter a region of a uniform magnetic field perpendicular to their velocities with the same kinetic energy. If $r_p, r_d,$ and $r_\alpha$ are the radii of the circular paths of these particles, respectively, then:

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$A$ particle of specific charge $q / m = \pi \text{ C kg}^{-1}$ is projected from the origin towards the positive $X$-axis with a velocity $10 \text{ ms}^{-1}$ in a uniform magnetic field $\vec{B} = -2 \hat{k} \text{ T}$. The velocity $\vec{v}$ of the particle after time $t = \frac{1}{12} \text{ s}$ will be (in $\text{ ms}^{-1}$):

As shown in the figure,the uniform magnetic field between the two identical plates is $B$. There is a hole in the bottom plate. If a particle of charge $q$,mass $m$,and energy $E$ enters this magnetic field through this hole,then the particle will not collide with the upper plate provided:

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?

What is the use of a mass spectrometer?

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