An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

Two very long,straight,parallel wires carry steady currents $I$ and $-I$ respectively. The distance between the wires is $d$. At a certain instant of time,a point charge $q$ is at a point equidistant from the two wires,in the plane of the wires. Its instantaneous velocity $v$ is perpendicular to the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

$A$ 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 $\,s$)

$A$ narrow electron beam passes undeviated through an electric field $E = 3 \times 10^4 \ V/m$ and an overlapping magnetic field $B = 2 \times 10^{-3} \ Wb/m^2$. If the electric field and magnetic field are mutually perpendicular,the speed of the electrons is: