$A$ charged particle moves through a magnetic field perpendicular to its direction. Then

An electron with energy $0.1 \, keV$ moves at a right angle to the Earth's magnetic field of $1 \times 10^{-4} \, Wb/m^2$. The frequency of revolution of the electron will be. (Take mass of electron $= 9.0 \times 10^{-31} \, kg$)

Uniform magnetic fields of different strengths ($B_1$ and $B_2$),both normal to the plane of the paper,exist as shown in the figure. $A$ charged particle of mass $m$ and charge $q$,at the interface at an instant,moves into the region $2$ with velocity $v$ and returns to the interface. It continues to move into region $1$ and finally reaches the interface. What is the displacement of the particle during this movement along the interface? (Consider the velocity of the particle to be normal to the magnetic field and $B_2 > B_1$)

An electron having a charge $e$ moves with a velocity $\vec{v}$ in the positive $x$-direction. $A$ magnetic field $\vec{B}$ acts on it in the positive $y$-direction. The force on the electron acts in (where the outward direction is taken as the positive $z$-axis):

$A$ charged particle is moving in a uniform magnetic field in a circular path with radius $R$. When the energy of the particle is doubled,then the new radius will be

Two ions have equal masses but one is singly ionized and the second is doubly ionized. They are projected from the same place in a uniform transverse magnetic field with the same velocity. Then:
$(a)$ Both ions will move along circles of equal radii.
$(b)$ The radius of the circle described by the singly ionized charge is double the radius of the circle described by the doubly ionized charge.
$(c)$ Both circles do not touch each other.
$(d)$ Both circles touch each other.