$A$ charge moves in a circular path perpendicular to a magnetic field. The time period of revolution is independent of

$A$ 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 ........ $\mu s$.

In Thomson's experiment,if the value of $q/m$ is the same for all positive ions striking the photographic plate,then the trace would be

$A$ proton and an $\alpha$-particle,having kinetic energies $K_{p}$ and $K_{\alpha}$ respectively,enter a magnetic field at right angles. The ratio of the radii of the trajectories of the proton to that of the $\alpha$-particle is $2:1$. The ratio of $K_{p}:K_{\alpha}$ is:

An electron $(e^-)$ is moving parallel to a long current-carrying wire as shown in the figure. Calculate the magnetic force acting on the electron. (Given: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)

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: