$A$ proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The magnetic field is directed into the plane of the paper. The particle enters at point $C$ with an angle of $45^{\circ}$ with the boundary. Find the time spent by the particle in the magnetic field region in $ns$.

An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off,and the same magnetic field is maintained,the electrons move

In a crossed field, the magnetic field induction is $2.0 \,T$ and electric field intensity is $20 \times 10^3 \,V/m$. At which velocity will the electron travel in a straight line without being deflected by the electric and magnetic fields?

$A$ particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly into a region of uniform magnetic field $B$ and executes a circular arc of radius $R$. Then,$\frac{q}{m}$ equals: