$A$ beam of electrons,initially at rest,is accelerated by a potential $V$. This beam experiences a force $F$ in a uniform magnetic field. The accelerating potential is increased to $V^{\prime}$ and the force experienced by the electrons in the same magnetic field becomes $2F$. The ratio $\frac{V}{V^{\prime}}$ is:

Two ions having masses in the ratio $1: 1$ and charges $1: 2$ are projected into a uniform magnetic field perpendicular to the field with speeds in the ratio $2: 3$. The ratio of the radii of circular paths along which the two particles move is :

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

$A$ 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 $\alpha$-particle and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is