$A$ charged particle moving in a magnetic field experiences a resultant force:

$A$ particle of mass $m$ and charge $q$ is incident on the $XZ$ plane with velocity $v$ in a direction making an angle $\theta$ with a uniform magnetic field applied along the $X$-axis. The nature of motion performed by the particle is . . . . . . .

$A$ particle of mass $0.6\, g$ and having a charge of $25\, nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4\, m/s$ in a uniform magnetic field. If the particle continues to move with uniform velocity,the value of the magnetic induction is $(g = 10\, m/s^2)$.

$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,

$A$ block of mass $m$ and charge $q$ is released on a long smooth inclined plane. $A$ magnetic field $B$ is constant,uniform,horizontal,and parallel to the surface as shown. Find the time from the start when the block loses contact with the surface.

When a charged particle enters a uniform magnetic field,its kinetic energy