Question

An electron and a proton enter a region of uniform magnetic field in a direction perpendicular to the field with the same kinetic energy. They revolve in circular paths of radii \[{r_e}\] and \[{r_p}\] respectively. Then
A. \[{r_e} = {r_p}\]
B. \[{r_e} < {r_p}\]
C. \[{r_e} > {r_p}\]
D. \[{r_e} \le {r_p}\]

Answer 1

Hint: In the given question, we need to determine the relation between \[{r_e}\] and \[{r_p}\]. For this, we need to use the formula of radius of circular path of a charged particle and the kinetic energy. So, we need to express the formula of radius of circular path in terms of kinetic energy. After that we have to take the ratio of \[{r_e}\] and \[{r_p}\] to get the desired result.





Formula used:
The following formulae are used to solve the given question.
The radius of the circular path of charged particles is \[r = \dfrac{{mv}}{{qB}}\].
Here, \[r\] indicates the radius of curvature of the path of a charged particle with mass \[m\] as well as charge \[q\] traveling at a speed \[v\] perpendicular to a magnetic field of intensity \[B\].
Also, the kinetic energy of a particle is \[K.E. = \dfrac{1}{2}m{v^2}\]
Here, \[m\] is the mass and \[v\] is the velocity.
Also, the pressure is the product of mass and velocity.
Mathematically, it is represented as \[p = m \times v\]
Here, \[p\] is the momentum.





Complete answer:
We know that the radius of the circular path of a charged particle is \[r = \dfrac{{mv}}{{qB}}\]…. \[\left( 1 \right)\]
Also, the kinetic energy of a particle is \[K.E. = \dfrac{1}{2}m{v^2}\]
But \[p = m \times v\]
This gives, \[K.E. = \dfrac{{{p^2}}}{{2m}}\]
So, we get \[{p^2} = 2m\left( {K.E.} \right)\]
By taking square root on both sides, we get
\[p = \sqrt {2m\left( {K.E.} \right)} \]
Now, the equation \[\left( 1 \right)\] becomes \[r = \dfrac{p}{{qB}}\]
Thus, \[r = \dfrac{{\sqrt {2m\left( {K.E.} \right)} }}{{qB}}\]
But \[B\] and \[K.E.\] are constant.
Thus, we can say that \[r\alpha \dfrac{{\sqrt m }}{q}\] ….. \[\left( 2 \right)\]
Hence, we get \[\dfrac{{{r_e}}}{{{r_P}}} = \sqrt {\dfrac{{{m_e}}}{{{m_P}}}} \left( {\dfrac{{{q_P}}}{{{q_e}}}} \right)\]
But \[{q_e} = {q_P}\]
This gives, \[\dfrac{{{r_e}}}{{{r_P}}} = \sqrt {\dfrac{{{m_e}}}{{{m_P}}}} \]
But \[{m_e} < {m_P}\]
Hence, we can say that \[{r_e} < {r_p}\].
Therefore, the correct option is (B).




Note:Many students make mistakes in the simplification part. That means, it is necessary to apply the proper formulae to get the desired result. Also, we need to analyze the relation between the radius of the circular path of electrons and protons.