$A$ particle of mass $m$,charge $Q$,and kinetic energy $K$ enters a transverse uniform magnetic field of induction $B$. After $3$ $seconds$,the kinetic energy of the particle will be .......$K$.

$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

$A$ proton moving with a velocity of $2.5 \times 10^7 \, m/s$ enters a magnetic field of intensity $2.5 \, T$ making an angle of $30^o$ with the magnetic field. The force on the proton is:

$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

An electron moves at a right angle to a magnetic field of $1.5 \times 10^{-2} \, T$ with a speed of $6 \times 10^7 \, m/s$. If the specific charge of the electron is $1.7 \times 10^{11} \, C/kg$,the radius of the circular path will be.....$cm$

$A$ proton and an $\alpha$-particle enter a magnetic field perpendicularly with the same velocity. If the proton takes $25 \mu \text{sec}$ to complete $5$ revolutions,what is the time period of the $\alpha$-particle in $\mu \text{sec}$?