Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

Two particles $X$ and $Y$ having equal charges are being accelerated through the same potential difference. Thereafter,they enter normally into a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $X$ and $Y$ is:

$A$ particle of charge $1.0 \times 10^{-16} \ C$ moves through a uniform magnetic field $B=B_0(\hat{i}+4 \hat{j}) \ T$. The particle velocity at some instant is $v=(2 \hat{i}+4 \hat{j}) \ ms^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \ N$. The magnitude of $B_0$ is (in $T$)

$A$ strong magnetic field is applied on a stationary electron. Then,the electron

$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: 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 $v = 10^7 \, m/s$)