The figure shows three situations when an electron with velocity $\vec v$ travels through a uniform magnetic field $\vec B$. In each case,what is the direction of the magnetic force on the electron?

$A$ particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$. There is a region in which the magnetic field is absent,as shown. The particle after entering the region collides elastically with a rigid wall. The time after which the velocity of the particle becomes anti-parallel to its initial velocity is

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. The $e/m$ ratio for it will be:

$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

An electron (mass = $9.0 \times 10^{-31} \ kg$ and charge = $1.6 \times 10^{-19} \ C$) is moving in a circular orbit in a magnetic field of $1.0 \times 10^{-4} \ Wb/m^2$. Its period of revolution is:

$A$ 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 $= 1.6 \times 10^{-19} \, C$)