Question

Two $\alpha $ - particles have the ratio of their velocities as $3:2$ on entering a magnetic field. If they move in different circular paths, then the ratio of the radii of their paths is:
$A)\text{ }2:3$
$B)\text{ 3}:2$
$C)\text{ 9}:4$
$D)\text{ 4}:9$

Answer 1

Hint: This problem can be solved by directly using the formula for the radii of the circular paths for a charged particle moving in a magnetic field in terms of its speed. By using the ratio of the two velocities given in the question, we can get the required ratio of the radii of the two circular paths.

Formula used:
$R=\dfrac{mv}{qB}$


Complete step by step solution:
We will use the formula for the radius of the circular path of a charged particle moving in a magnetic field, for the two particles and get the required ratio.
Let the velocity of the first $\alpha $ - particle be ${{v}_{1}}$ and that of the second one be ${{v}_{2}}$.
Now, according to the question,
${{v}_{1}}:{{v}_{2}}=3:2$
$\therefore \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{3}{2}$ --(1)
Now, let the radii of the circular paths of the first and second $\alpha $ - particles be ${{R}_{1}}$ and ${{R}_{2}}$ respectively.
We have to find ${{R}_{1}}:{{R}_{2}}$.
Now an $\alpha $ - particle is nothing but the nucleus of a helium atom. Let the charges on the first and second $\alpha $- particles be $q$ for both.
Let the mass of both the $\alpha $- particles be $m$.
Now, the radius $R$ of the circular path of a charge $q$ of mass $m$ moving in a magnetic field of magnitude $B$ with a speed $v$ is given by
$R=\dfrac{mv}{qB}$ --(2)
Now, using (2), we get,
${{R}_{1}}=\dfrac{m{{v}_{1}}}{qB}$ --(3)
${{R}_{2}}=\dfrac{m{{v}_{2}}}{qB}$ --(4)
Dividing (3) by (4), we get,
$\therefore \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\dfrac{m{{v}_{1}}}{qB}}{\dfrac{m{{v}_{2}}}{qB}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}$
Using (1), we get,
$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{3}{2}=3:2$
$\therefore {{R}_{1}}:{{R}_{2}}=3:2$

Hence, the required ratio of the radii of their circular paths is $3:2$.

Therefore, the correct option is $B)\text{ 3}:2$.


Note:We could have also solved this problem by considering the fact that the radii of the circular path are directly proportional to the speed of the charged particle in the magnetic field, which is the result we reached upon any way in the penultimate step of the calculation process. Following this method would take out many unnecessary variables like the charge, mass and magnitude of the magnetic field out of the equation. However, this process would only be valid here since both the particles are identical and move in the same magnetic field. Therefore, it is better to write the full formula for both the particles and then get the ratio and cancel whatever variables can be cancelled during the calculation.