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Step-by-Step Solution
Step 1: Identify the Relevant Physical Law
The magnetic force on a charged particle moving in a uniform magnetic field with velocity $v$ is given by:
$F = q \, v \, B \, \sin \theta,$
where $q$ is the charge of the particle, $v$ is its speed, $B$ is the magnetic field, and $\theta$ is the angle between the velocity and magnetic field. In this question, assume the motion is perpendicular to the field for maximum force, so $\sin \theta = 1.$ Thus,
$F = q \, v \, B.$
Step 2: Express Speed in Terms of Momentum
Each of the three particles (proton, deuteron, and alpha particle) is moving with the same momentum $p.$ Recall that:
$p = m \, v \quad \Rightarrow \quad v = \dfrac{p}{m}.$
Hence, the magnetic force can also be written as:
$F = q \, \left(\dfrac{p}{m}\right) B = \left(\dfrac{q}{m}\right) p \, B.$
Since $p$ and $B$ are common for all three, the force ratio depends on $q/m.$
Step 3: Determine Charge-to-Mass Ratios
Proton: Charge $q = e$, Mass $m_p = m.$
Deuteron (nucleus of deuterium): Charge $q = e$, Mass $m_d = 2m.$
Alpha particle ($\alpha$): Charge $q = 2e$, Mass $m_\alpha = 4m.$
Step 4: Calculate the Ratio of Magnetic Forces
Using $F \propto \dfrac{q}{m}$ for each particle:
Proton: $ \dfrac{e}{m} \quad,\quad$
Deuteron: $ \dfrac{e}{2m} \quad,\quad$
Alpha: $ \dfrac{2e}{4m} = \dfrac{e}{2m}.
The ratio is:
$
F_\text{proton} : F_\text{deuteron} : F_\alpha
= \dfrac{e}{m} : \dfrac{e}{2m} : \dfrac{e}{2m}
= 2 : 1 : 1.
$
Step 5: Calculate the Ratio of Speeds
Since $v = \dfrac{p}{m}$, for each particle with the same momentum $p$:
Proton: $v_p = \dfrac{p}{m}, \quad$
Deuteron: $v_d = \dfrac{p}{2m}, \quad$
Alpha: $v_\alpha = \dfrac{p}{4m}.
The ratio is:
$
v_p : v_d : v_\alpha
= \dfrac{p}{m} : \dfrac{p}{2m} : \dfrac{p}{4m}
= 4 : 2 : 1.
$
Final Answer
The ratio of the magnetic forces on the proton, deuteron, and alpha particle is $2 : 1 : 1,$ and their speeds are in the ratio $4 : 2 : 1.$
