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 :

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

$A$ particle having a charge of $10.0\,\mu C$ and mass $1\,\mu g$ moves in a circle of radius $10\,cm$ under the influence of a magnetic field of induction $0.1\,T$. When the particle is at a point $P$, a uniform electric field is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is......$V/m$

$A$ particle of mass $M$ and charge $q$ moves with a constant velocity $V$ in the positive $x$-direction. $A$ constant magnetic field $B$ exists in the negative $z$-direction between $x = a$ and $x = b$. Find the minimum value of $V$ such that the particle enters the region $x > b$.

$A$ 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

At $t = 0$,a positively charged particle of mass $m$ is projected from the origin with velocity $u_0$ at an angle $37^o$ from the $x$-axis as shown in the figure. $A$ constant magnetic field $\vec{B_0} = B_0 \hat{j}$ is present in space. After a time interval $t_0$,the velocity of the particle may be: