If a charged particle enters a uniform magnetic field normally with a certain velocity,then the time period of revolution of the particle

An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along

$A$ magnetic field $\overrightarrow{B} = B_0 \hat{j}$ exists in the region $a < x < 2a$ and $\overrightarrow{B} = -B_0 \hat{j}$ in the region $2a < x < 3a$,where $B_0$ is a positive constant. $A$ positive point charge moving with a velocity $\overrightarrow{v} = v_0 \hat{i}$,where $v_0$ is a positive constant,enters the magnetic field at $x = a$. The trajectory of the charge in this region can be like,

$A$ particle with charge $q$,moving with a momentum $p$,enters a uniform magnetic field normally. The magnetic field has magnitude $B$ and is confined to a region of width $d$,where $d < \frac{p}{Bq}$. The particle is deflected by an angle $\theta$ in crossing the field.

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

An electron having mass $9.1 \times 10^{-31} \ kg$,charge $1.6 \times 10^{-19} \ C$ and moving with velocity of $10^6 \ ms^{-1}$ enters a region where a magnetic field exists. If it describes a circle of radius $0.2 \ m$,then the intensity of the magnetic field must be . . . . . . $\times 10^{-5} \ T$.