$A$ proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha$-particle (mass = $4m$ and charge = $+2e$) so that it can revolve in the path of the same radius?

In a mass spectrometer used for measuring the masses of ions,the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant,the ratio $\left(\frac{\text{charge on the ion}}{\text{mass of the ion}}\right)$ will be proportional to

The distance between two plates is $1 \ cm$ and the potential difference is $1000 \ V$. $A$ magnetic field $B = 1 \ T$ is applied. If an electron passes through without any deflection,what will be its velocity?

$A$ 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:

$A$ uniform magnetic field $\vec{B} = B_0 \hat{j}$ exists in space. $A$ particle of mass $m$ and charge $q$ is projected towards the negative $x$-axis with speed $v$ from the point $(d, 0, 0)$. The maximum value of $v$ for which the particle does not hit the $y-z$ plane is

$A$ metallic block carrying current $I$ is subjected to a uniform magnetic induction $\overrightarrow{B}$ as shown in the figure. The moving charges experience a force $\overrightarrow{F}$ given by ........... which results in the lowering of the potential of the face ........ Assume the speed of the carriers to be $v$.