$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .

An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma$. Electric intensity in the space between the plates is $E$. $A$ uniform magnetic field $B$ also exists in that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\vec{E}$ and $\vec{B}$ without any change in direction. The time taken by the electron to travel a distance $\ell$ in the space is

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?

The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the Earth's magnetic field points down. The electron is deflected towards:

The region between $y = 0$ and $y = d$ contains a magnetic field $\vec B = B\hat k$. $A$ particle of mass $m$ and charge $q$ enters the region with a velocity $\vec v = v\hat i$. If $d = \frac{mv}{2qB}$,the acceleration of the charged particle at the point of its emergence at the other side is