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

Mixed $H{e^ + }$ and ${O^{2 + }}$ ions (mass of $H{e^ + } = 4\,\,amu$ and that of ${O^{2 + }} = 16\,\,amu)$ beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then

A beam of protons with speed $4 \times 10^{5}\, ms ^{-1}$ enters a uniform magnetic field of $0.3\, T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to….$cm$

(Mass of the proton $=1.67 \times 10^{-27}\, kg$, charge of the proton $=1.69 \times 10^{-19}\,C$)

An electron has mass $9 \times {10^{ – 31}}\,kg$ and charge $1.6 \times {10^{ – 19}}C$ is moving with a velocity of ${10^6}\,m/s$, enters a region where magnetic field exists. If it describes a circle of radius $0.10\, m$, the intensity of magnetic field must be

An electron moving with a velocity ${\vec V_1} = 2\,\hat i\,\, m/s$ at a point in a magnetic field experiences a force ${\vec F_1} =  – 2\hat j\,N$ .  If the electron is moving with a velocity ${\vec V_2} = 2\,\hat j \,\,m/s$ at the same point, it experiences a force ${\vec F_2} =  + 2\,\hat i\,N$ .  The force the electron would experience if it were moving with a velocity ${\vec V_3} = 2\hat k$  $m/s$ at the same point is