If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field,then the orbit will be

When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $V_1$,it experiences a force $F$. If the potential difference is changed to $V_2$,the force experienced by the electron in the same magnetic field is $2F$. Then,the ratio of potential differences $\frac{V_2}{V_1}$ is:

$A$ rectangular region $ABCD$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. $A$ narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $BC$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$,the magnetic field has to be

Two infinitely long straight wires $A$ and $B$,each carrying current $I$,are placed on the $x$ and $y$-axes,respectively. The current in wires $A$ and $B$ flows along $-\hat{i}$ and $\hat{j}$ directions,respectively. The force on a charged particle having charge $q$,moving from position $r = d(\hat{i} + \hat{j})$ with velocity $v = v\hat{i}$ is:

$A$ uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is

$A$ proton and an $\alpha$-particle enter a magnetic field perpendicularly with the same velocity. If the proton takes $25 \mu \text{sec}$ to complete $5$ revolutions,what is the time period of the $\alpha$-particle in $\mu \text{sec}$?