Question

A proton with speed v moves in a circle of radius Rp perpendicular to a uniform magnetic field of magnitude B. An α-particle moves with the same speed v in a circle of radius Ra perpendicular to the same field. Given that the alpha particle has four times the mass of the proton and twice its charge, what is the relation between Ra and Rp?
a. Ra = Rp/4
b. Ra = (1/2) Rp/2
c. Ra = Rp
d. Ra = 2 Rp
e. Ra = 4 Rp

Answer 1

Answer:

$R_{\alpha}=2R_p$

Explanation:

Let $m_p\ and\ m_{\alpha }$ are the mass of proton and the alpha particles. Let $R_p\ and\ R_{\alpha}$ are the radius of proton and the alpha particles. Given that the alpha particle has four times the mass of the proton and twice its charge.

$m_\alpha =4m_p$

and

$q_\alpha =2q_p$

The centripetal force is balanced by the magnetic force as :

$qvB=\dfrac{mv^2}{R}$

$R=\dfrac{mv}{qB}$

For proton,

$R_p=\dfrac{m_pv}{q_pB}$

For alpha particle,

$R_{\alpha}=\dfrac{m_{\alpha}v}{q_{\alpha} B}$

$R_{\alpha}=\dfrac{4m_pv}{2q_pB}$

$R_{\alpha}=2\dfrac{m_pv}{q_pB}$

$R_{\alpha}=2R_p$

So, the relation between $R_p\ and\ R_{\alpha}$ is :

Option (d) : $R_{\alpha}=2R_p$

Answer 2

The relation Ra and Rp will be "Ra = 2 Rp".

Magnetic field:

Let,

• Mass of proton and alpha particles = $m_p$ and $m_\alpha$
• Radius of proton and alpha particles = $R_p$ and $R_{\alpha}$

According to the question,

$m_\alpha = 4 m_p$, and

$q_\alpha = 2 q_p$

The centripetal force balanced by magnetic force:

→ $qvB = \frac{mv^2}{R}$

→     $R = \frac{mv}{qB}$

For Proton,

→ $R_p = \frac{m_pv}{q_p B}$

For Alpha,

→ $R_\alpha = \frac{m_\alpha v}{q_\alpha B}$

        $= \frac{4 m_p v}{2 q_p B}$

        $= 2 P_p$

Thus the above answer i.e., "Option d" is correct.  

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