A proton and an $\alpha$-particle are simultaneously projected in opposite direction into a region of uniform magnetic field of $2 \mathrm{mT}$ perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by $90^{\circ}$. Then at this time, the angle between the velocity vectors of proton and $\alpha$-particle is

Key Idea In a circular motion, a body changes its direction by $90^{\circ}$ in one-fourth of its time period. Given, magnetic field, $B=2 \mathrm{mT}$ Let $T_p$ be the time-period of revolution of proton in
$$ m_\alpha \simeq 4 m p \text { and } q_\alpha=2 q^p $$ and $T_\alpha$ be the time-period of revolution of $\alpha$-particle in the magnetic field,
Now, the ratio of time-period of proton and $\alpha$-particle, by dividing Eq. (i) and (ii), $$ \begin{aligned} & \frac{T_p}{T_\alpha}=\frac{1}{2} \\ & \Rightarrow \quad T_\alpha=2 T_p \\ & \end{aligned} $$ Hence, the time-period of $\alpha$-particle is double of the proton, i.e. if proton covers $90^{\circ}$ of angle from its starting, then $\alpha$-particle will cover $45^{\circ}$ of the angle. $\therefore$ Hence, the correct option is (c).