Question

A neutron, a proton, an electron and an $\alpha$ particle enter a region of constant magnetic field with equal velocity. The $\overrightarrow{B}$ is along the perpendicularly inward to the plane of the paper. The path followed by each particle is shown in the figure.


The path followed by $\alpha$ particle is represented by:

• A. $\text{B}$
• B. $\text{A}$
• C. $\text{D}$
• D. $\text{C}$

Answer 1

The correct option is A $\text{B}$

The charge on the neutron, $q_n=0$

From relation,

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

$F_n=0$

Hence, it will go undeviated, represented by path $\text{C}$.

And we know that electron has negative charge so it will follow path $\text{D}$ opposite to $\text{A \& B}$.

Now rest of the charged particle will move on a circular path, whose radius is;

$r=\frac{mv}{qB}$

For same value of $V$ and $B$ as mention in question,

$r\propto \frac{m}{q}....\left(1\right)$

$\because m_{\alpha }=4m_p;q_{\alpha }=2q_p$

$\frac{r_{\alpha }}{r_p}=\frac{m_{\alpha }}{q_{\alpha }}\frac{q_p}{m_p}=\frac{4m_p}{2q_p}\frac{q_p}{m_p}=2$

$\therefore r_{\alpha }>r_p$

Both of these positive charge will move/deviate in the same sense.
Hence, path $\text{A}$ and $\text{B}$ represent moment of proton and $\alpha$ particle.

$\because r_B>r_A$

Therefore, $\text{B}$ represents path of $\alpha$particle.