$A$ proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to the direction of the magnetic field. If both the particles move around the magnetic field in circles of equal radii,the ratio of the momentum of the proton to the alpha particle $\left( \frac{P_p}{P_\alpha} \right)$ is

$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters vertically downward into a horizontal magnetic field of induction $\vec{B}$. The acceleration of the particle is:

An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

Two long parallel conductors $S_{1}$ and $S_{2}$ are separated by a distance $10 \, cm$ and carry currents of $4 \, A$ and $2 \, A$ respectively. The conductors are placed along the $x$-axis in the $X-Y$ plane. There is a point $P$ located between the conductors (as shown in the figure). $A$ charged particle of $3 \pi \, C$ is passing through the point $P$ with velocity $\overrightarrow{v} = (2 \hat{i} + 3 \hat{j}) \, m/s$; where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$ axes respectively. The force acting on the charged particle is $4 \pi \times 10^{-5} (-x \hat{i} + 2 \hat{j}) \, N$. The value of $x$ is:

$A$ proton and an $\alpha$-particle enter a uniform magnetic field perpendicularly with the same speed. If a proton takes $25 \ \mu s$ to make $5$ revolutions,then the periodic time for the $\alpha$-particle would be ........ $\mu s$.

$A$ deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at a right angle to the field. If $r_{d}$ and $r_{p}$ are the radii of their circular paths respectively,then the ratio $\frac{r_{d}}{r_{p}}$ will be $\sqrt{x} : 1$ where $x$ is ..........