If a charged particle moves through a magnetic field with a velocity that has both perpendicular and parallel components,what is the path followed by the particle?

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

$A$ particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec{E} = E_0 \hat{i}$ and a magnetic field $\vec{B} = B_0 \hat{k}$. Its velocity at $(x_0, y_0, 0)$ is $(4 \hat{i} + 3 \hat{j})$. The value of $x_0$ is:

In a region,steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. $A$ charged particle is released from rest in this region. The path of the particle will be a

$A$ nucleus at rest undergoes $\alpha -$ decay according to the equation ${}_{92}^{226}X \to Y + \alpha$. At $t=0$,the emitted $\alpha -$ particle enters a region of space where a uniform magnetic field $\vec{B} = B_0 \hat{i}$ and an electric field $\vec{E} = E_0 \hat{i}$ exist. The $\alpha -$ particle enters the region with velocity $\vec{v} = v_0 \hat{j}$ at $x = 0$. After a time $t = \left( \sqrt{3} \times 10^7 \frac{m_{\alpha}}{q_{\alpha} E_0} \right)$,the particle is found to have twice its initial speed. The initial speed $v_0$ of the $\alpha -$ particle is:

Two electrons,$e_1$ and $e_2$ of mass $m$ and charge $q$ are injected into the perpendicular direction of the magnetic field $B$ such that the kinetic energy of $e_1$ is double than that of $e_2$. The relation of their frequencies of rotation,$f_1$ and $f_2$ is