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(A) Let the energy of the proton be $E_p$ and the alpha particle be $E_{\alpha}$. Given $\frac{E_p}{E_{\alpha}} = \frac{1}{4}$.Since $E = \frac{1}{2}m

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$1: 2$

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(A) Let the energy of the proton be $E_p$ and the alpha particle be $E_{\alpha}$. Given $\frac{E_p}{E_{\alpha}} = \frac{1}{4}$.Since $E = \frac{1}{2}mv^2$,we have $\frac{\frac{1}{2}m_p v_p^2}{\frac{1}{2}m_{\alpha} v_{\alpha}^2} = \frac{1}{4}$.Given $m_{\alpha} = 4m_p$,we substitute: $\frac{m_p v_p^2}{4m_p v_{\alpha}^2} = \frac{1}{4} \Rightarrow \frac{v_p^2}{v_{\alpha}^2} = 1 \Rightarrow v_p = v_{\alpha}$.The magnetic force is $F = qvB \sin(	heta)$. Since $	heta = 90^{\circ}$,$F = qvB$.The ratio of forces is $\frac{F_p}{F_{\alpha}} = \frac{q_p v_p B}{q_{\alpha} v_{\alpha} B} = \frac{q_p}{q_{\alpha}}$.Since $q_{\alpha} = 2q_p$,the ratio is $\frac{q_p}{2q_p} = \frac{1}{2}$.

$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

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