Two particles $x$ and $y$ have equal charges and possess equal kinetic energy. They enter a uniform magnetic field and describe circular paths of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is

When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix

An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

Bohr model is applied to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a transverse magnetic field $B$. The energy of the charged particle in the $n^{\text{th}}$ level will be $[h = \text{Planck's constant}]$

$A$ particle of charge per unit mass $\alpha$ is released from the origin with a velocity $\vec{v} = v_0 \hat{i}$ in a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. If the particle passes through $(0, y, 0)$,then $y$ is equal to