$A$ magnetic field is applied on an electron moving with a velocity of $10^7 \,m/s$ at an angle of $30^{\circ}$ to the magnetic field. The time period of revolution of the electron in a circular path of radius $2 \,m$ is . . . . . .

$A$ charged particle of charge $q$ and mass $m$ gets deflected through an angle $\theta$ upon passing through a square region of side $a$,which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side,what is the speed of the particle?

An $\alpha$-particle of $1 \, MeV$ energy moves on a circular path in a uniform magnetic field. Then,the kinetic energy of a proton in the same magnetic field for a circular path of double radius is ...... $MeV$.

$A$ 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 $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

An electron $(e^-)$ is moving in the North direction and the magnetic field at this place is in the upward direction. Then,the electron $(e^-)$ will be deviated towards:

$A$ proton and a deuteron are both accelerated through the same potential difference and enter a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$,assuming the neutron and proton masses are nearly equal,the radius of the proton's path will be