$A$ 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 accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

$A$ nucleus at rest undergoes $\alpha -$ decay according to the equation ${}_{92}^{226}X \to Y + \alpha$. At $t=0$,the emitted $\alpha -$ particle enters a region of space where a uniform magnetic field $\vec{B} = B_0 \hat{i}$ and an electric field $\vec{E} = E_0 \hat{i}$ exist. The $\alpha -$ particle enters the region with velocity $\vec{v} = v_0 \hat{j}$ at $x = 0$. After a time $t = \left( \sqrt{3} \times 10^7 \frac{m_{\alpha}}{q_{\alpha} E_0} \right)$,the particle is found to have twice its initial speed. The initial speed $v_0$ of the $\alpha -$ particle is:

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ 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 $= 1.6 \times 10^{-19} \ C$)

An electron and a proton of equal linear momentum enter in a direction perpendicular to a uniform magnetic field. If the radii of their circular paths are $r_e$ and $r_p$ respectively,then $\frac{r_e}{r_p}$ is equal to - (mass of electron $= m_e$,mass of proton $= m_p$)

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.