An $\alpha$-particle of $1 \text{ MeV}$ kinetic energy is moving on a circular path in a uniform magnetic field. The kinetic energy of a proton,to be moved in the same magnetic field on a circular path of double radius,is $..... \text{ MeV}$.

$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

An electron of mass $m$ and charge $e$ is projected normally from the surface of a sphere of radius $a$ with speed $v_0$ into a uniform magnetic field $B$ perpendicular to the plane of the paper. The centre of the sphere is at a distance $b$ from a wall. If the electron strikes the wall symmetrically with respect to the $x$-axis,what should be the magnetic field $B$ such that the charged particle just touches the wall?

$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$, then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)

An electron gun is placed inside a long solenoid of radius $R$ on its axis. The solenoid has $n$ turns per unit length and carries a current $I$. The electron gun shoots an electron along the radius of the solenoid with speed $v$. If the electron does not hit the surface of the solenoid,the maximum possible value of $v$ is (all symbols have their standard meaning):

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)