1. Given: A proton, a deuteron, and an $\alpha$-particle move with the same momentum in a uniform magnetic field.

2. Magnetic force on a charged particle: $F = qvB\sin\theta$ Also, since momentum $p = mv$, we can write $F = qB\sin\theta \cdot v$

   For same $p$ and same field, it is more convenient to use: $v = \frac{p}{m}$ so $F = qB\sin\theta \cdot \frac{p}{m}$

   Since $p$, $B$, and $\sin\theta$ are same for all three particles, $F \propto \frac{q}{m}$

3. Charge and mass of each particle:

   • Proton: charge $= +e$, mass $= m$
   • Deuteron: charge $= +e$, mass $= 2m$
   • $\alpha$-particle: charge $= +2e$, mass $= 4m$

4. Force ratio: $F_p : F_d : F_\alpha = \frac{e}{m} : \frac{e}{2m} : \frac{2e}{4m}$ $= 1 : \frac{1}{2} : \frac{1}{2}$ Multiplying by 2: $F_p : F_d : F_\alpha = 2 : 1 : 1$

5. Speed ratio: Since momentum is same, $v = \frac{p}{m} \implies v \propto \frac{1}{m}$

   Therefore, $v_p : v_d : v_\alpha = \frac{1}{m} : \frac{1}{2m} : \frac{1}{4m}$ $= 1 : \frac{1}{2} : \frac{1}{4}$ Multiplying by 4: $v_p : v_d : v_\alpha = 4 : 2 : 1$

6. Final result:

   • Magnetic force ratio $= 2:1:1$
   • Speed ratio $= 4:2:1$

So the correct option is A.