Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}\,tesla$ with speed of $6 \times 10^7\,m/s$. If the specific charge of the electron is $1.7 \times 10^{11}\,C/kg$. The radius of circular path will be......$cm$

If a proton, deutron and $\alpha - $ particle on being accelerated by the same potential difference enters perpendicular to the magnetic field, then the ratio of their kinetic energies is

An $\alpha - $ particle travels in a circular path of radius $0.45\, m$ in a magnetic field $B = 1.2\,Wb/{m^2}$ with a speed of $2.6 \times {10^7}\,m/\sec $. The period of revolution of the $\alpha - $ particle is

The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to....... $m$

(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$

An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B$. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of

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