A charge $+ Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards

A proton (mass $m$ ) accelerated by a potential difference $V$  flies through a uniform transverse magnetic field $B.$ The field occupies a region of space by width $'d'$. If $\alpha $ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin\,\alpha $ will be

A charge having $q/m$ equal to $10^8\, C/kg$ and with velocity $3 \times 10^5\, m/s$ enters into a uniform magnetic field $0.3\, tesla$ at an angle $30^o$ with direction of field. The radius of curvature will be ……$cm$

An electron moving with a speed $u$ along the positive $x-$axis at $y = 0$ enters a region of uniform magnetic field $\overrightarrow B = – {B_0}\hat k$ which exists to the right of $y$-axis. The electron exits from the region after some time with the speed $v$ at co-ordinate $y$, then

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