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 same radius.......$MeV$

A proton beam is going from north to south and an electron beam is going from south to north. Neglecting the earth's magnetic field, the electron beam will deflected (Zero gravity)

Which of the following particle will describe the smallest circle when projected with the same  velocity perpendicular to the magnetic field ?

Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $\mathrm{X}$ and $\mathrm{Y}$ is :

A charged particle carrying charge $1\,\mu C$  is moving with velocity $(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })\, ms ^{-1} .$ If an external magnetic field of $(5 \hat{ i }+3 \hat{ j }-6 \hat{ k }) \times 10^{-3}\, T$ exists in the region where the particle is moving then the force on the particle is $\overline{ F } \times 10^{-9} N$. The vector $\overrightarrow{ F }$ is :

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6: 5$ and their respective masses ratio is $9: 4$. Then, the ratio of their charges will be.