$H^+$,$He^+$,and $O^{++}$ are projected in a uniform transverse magnetic field with equal accelerating potential. If their masses are $1 \, a.m.u.$,$4 \, a.m.u.$,and $16 \, a.m.u.$ respectively,find the ratio of their radii.

$A$ proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities,the ratio of the radii $r_p : r_{\alpha}$ of the circular paths described by them will be

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ enters a uniform magnetic field of $0.3 \ T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $= 1.67 \times 10^{-27} \ kg$, charge of the proton $= 1.6 \times 10^{-19} \ C$)

$A$ velocity selector is to be constructed to select ions with a velocity of $6 \,km \,s^{-1}$. If the electric field used is $400 \,V \,m^{-1}$, then the magnetic field to be used is

$A$ charged particle moving in a magnetic field $B$ has velocity components both along $B$ and perpendicular to $B$. The path of the charged particle will be:

An electron experiences a force $\vec{F} = (4.0\,\hat{i} + 3.0\,\hat{j}) \times 10^{-13}\,N$ in a uniform magnetic field when its velocity is $\vec{v}_1 = 2.5\,\hat{k} \times 10^7\,m/s$. When the velocity is redirected to $\vec{v}_2 = (1.5\,\hat{i} - 2.0\,\hat{j}) \times 10^7\,m/s$,the magnetic force on the electron is zero. The magnetic field $\vec{B}$ is: