$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field:

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2} \, T$. If the value of $e/m$ is $1.76 \times 10^{11} \, C/kg$,the frequency of revolution of the electron is

An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $A$ lies in the same plane at a distance $2R$ from the centre. Point $A$ records the magnetic field produced by the $\alpha$ particle. If the minimum time interval between two successive times at which $A$ records zero magnetic field is $t$,find the angular speed $\omega$ in terms of $t$.

$A$ particle having charge $q$ enters a region of uniform magnetic field $\vec{B}$ (directed inwards) and is deflected a distance $x$ after travelling a distance $y$. The magnitude of the momentum of the particle is: