$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be

$A$ charged particle with charge $q$ enters a region of constant,uniform,and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{v}$ perpendicular to both $\vec{E}$ and $\vec{B}$,and comes out without any change in the magnitude or direction of $\vec{v}$. Then:

$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

$A$ monoenergetic $(18 \;keV)$ electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \;G$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \;cm$ $(m_{e} = 9.11 \times 10^{-31} \;kg)$.

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

$A$ particle of charge per unit mass $\alpha$ is released from the origin with a velocity $\vec{v} = v_0 \hat{i}$ in a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. If the particle passes through $(0, y, 0)$,then $y$ is equal to