The motion of a charged particle can be used to distinguish between a magnetic field and electric field in a certain region by firing the charge

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3}\,T$. The radius of path of the circle will approximately be :

If a charged particle enters perpendicularly in the uniform magnetic field then

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

Statement $-1$ : Path of the charge particle may be straight line in uniform magnetic field.
Statement $-2$ : Path of the charge particle is decided by the angle between its velocity and the magnetic force working on it

A proton with a kinetic energy of $2.0\,eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3}\,T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is $..........cm$ (Take, mass of proton $=1.6 \times 10^{-27}\,kg$ and Charge on proton $=1.6 \times 10^{-19}\,kg)$