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(C) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2\pi m}{qB}$.For a proton,$T_p = \frac{2\pi m_p}{eB}$.For

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$45$

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(C) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2\pi m}{qB}$.For a proton,$T_p = \frac{2\pi m_p}{eB}$.For an $\alpha$-particle,$m_{\alpha} = 4m_p$ and $q_{\alpha} = 2e$. Thus,$T_{\alpha} = \frac{2\pi (4m_p)}{(2e)B} = 2 \left( \frac{2\pi m_p}{eB} ight) = 2T_p$.When the proton's velocity changes direction by $90^{\circ}$,it has completed one-fourth of its circular path,meaning the time elapsed is $t = \frac{T_p}{4}$.In this same time $t$,the $\alpha$-particle covers an angle $	heta_{\alpha} = \omega_{\alpha} t = \left( \frac{2\pi}{T_{\alpha}} ight) \left( \frac{T_p}{4} ight) = \left( \frac{2\pi}{2T_p} ight) \left( \frac{T_p}{4} ight) = \frac{\pi}{4} = 45^{\circ}$.Since the particles were projected in opposite directions,let the initial velocity of the proton be $\vec{v}_p = v_0 \hat{i}$ and the $\alpha$-particle be $\vec{v}_{\alpha} = -v_0 \hat{i}$.After time $t$,the proton's velocity is $\vec{v}_p' = v_0 \hat{j}$ (rotated $90^{\circ}$ counter-clockwise).The $\alpha$-particle's velocity rotates $45^{\circ}$ clockwise from its initial direction (opposite to the proton's rotation due to opposite charge sign and opposite initial direction),so $\vec{v}_{\alpha}' = v_0 (\cos 45^{\circ} (-\hat{i}) + \sin 45^{\circ} (-\hat{j})) = -v_0 \frac{1}{\sqrt{2}} \hat{i} - v_0 \frac{1}{\sqrt{2}} \hat{j}$.The angle between $\vec{v}_p'$ and $\vec{v}_{\alpha}'$ is found using the dot product: $\cos 	heta = \frac{\vec{v}_p' \cdot \vec{v}_{\alpha}'}{|v_p'||v_{\alpha}'|} = \frac{(0)(-\frac{1}{\sqrt{2}}) + (1)(-\frac{1}{\sqrt{2}})}{(1)(1)} = -\frac{1}{\sqrt{2}}$.Thus,$	heta = 135^{\circ}$. However,re-evaluating the geometry: the proton turns $90^{\circ}$ and the $\alpha$-particle turns $45^{\circ}$ in the opposite sense relative to the initial axis. The angle between them is $180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$.

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