$A$ particle with $10^{-11} \, C$ of charge and $10^{-7} \, kg$ mass is moving with a velocity of $10^8 \, m/s$ along the $y$-axis. $A$ uniform static magnetic field $B = 0.5 \, T$ is acting along the $x$-direction. The force on the particle is:

$A$ particle moving with velocity $v$ having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d = \frac{3mv}{5qB}$ at an angle of $53^{\circ}$ to the boundary of the magnetic field. Find the angle $\theta$ shown in the diagram.

$A$ proton with a kinetic energy of $2.0 \, eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} \, T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $.......... \, cm$. (Take,mass of proton $= 1.6 \times 10^{-27} \, kg$ and charge on proton $= 1.6 \times 10^{-19} \, C$)

Particles having positive charges occasionally come with high velocity from the sky towards the earth. On account of the magnetic field of the earth,they would be deflected towards the

$A$ 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 and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is