$A$ small block of mass $20 \,g$ and charge $4 \,mC$ is released on a long smooth inclined plane of inclination angle $45^{\circ}$. $A$ uniform horizontal magnetic field of $1 \,T$ is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is (in $\,s$)

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:

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

$A$ particle having a charge of $10\,\mu C$ and $1\,\mu g$ mass moves along a circular path of $10\, cm$ radius in the presence of a uniform magnetic field of $0.1\, T$. When the charge is at point $P$,a uniform electric field is applied in the region such that the charge moves tangentially with a constant speed. The value of the electric field is......$V/m$

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The angle of incidence is $45^{\circ}$. The radius of the circular portion of the path is:

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]