In Thomson's experiment,if the value of $q/m$ is the same for all positive ions striking the photographic plate,then the trace would be

$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

$A$ proton of velocity $(3\hat i + 2\hat j) \, ms^{-1}$ enters a magnetic field of $(2\hat j + 3\hat k) \, T$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times 10^8 \, C/kg$)

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:

If an electron is moving in the direction of the magnetic field $\overrightarrow{B}$ with a velocity $\overrightarrow{v}$,then the force on the electron is:

$A$ particle of mass $m$ and charge $q$ is thrown from the origin at $t = 0$ with velocity $\vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ units in a region with a uniform magnetic field $\vec{B} = 2\hat{i}$ units. After time $t = \frac{\pi m}{qB}$,an electric field $\vec{E}$ is switched on such that the particle moves in a straight line with constant speed. $\vec{E}$ may be: