As shown in the figure,the uniform magnetic field between the two identical plates is $B$. There is a hole in the bottom plate. If a particle of charge $q$,mass $m$,and energy $E$ enters this magnetic field through this hole,then the particle will not collide with the upper plate provided:

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:

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.

For a positively charged particle moving in a $x-y$ plane initially along the $x$-axis,there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond $P$. The curved path is shown in the $x-y$ plane and is found to be non-circular. Which one of the following combinations is possible?

Consider a long thin conducting wire carrying a uniform current $I$. $A$ particle having mass $M$ and charge $q$ is released at a distance $a$ from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of $x$ is [$\mu_0$ is vacuum permeability]

An $\alpha$-particle travels in a circular path of radius $0.45\,m$ in a magnetic field $B = 1.2\,Wb/m^2$ with a speed of $2.6 \times 10^7\,m/s$. The period of revolution of the $\alpha$-particle is: