
An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, then
A. The path of proton shall be more curved than that of electron
B. The path of proton shall be less curved than that of electron
C. Both are equally curved
D. Path of both will be straight line
Answer
163.2k+ views
Hint: When a charged particle enters into a region of perpendicular magnetic field of uniform magnetic field strength then it experiences a magnetic force in the direction perpendicular to the initial velocity. So the resulting path of the motion is circular in nature.
Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.
\[p = mv\], here p is the magnitude of the momentum of the body of mass m moving with speed v.
Complete answer:
Let the mass of the electron is \[{m_e}\]and the mass of the proton is \[{m_p}\]
The charge on the electron and the proton is same in magnitude \[{q_e} = {q_p} = e\]
It is given that the electron and proton is moving with equal momentum,
\[{m_e}{v_e} = p\]
\[{m_p}{v_p} = p\]
On dividing both the expressions of the momentum, we get
\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{{m_p}}}{{{m_e}}}\]
Using the radius formula,
The radius of the circular path of the electron is,
\[{r_e} = \dfrac{{{m_e}{v_e}}}{{Be}}\]
\[{r_e} = \dfrac{p}{{Be}}\]
The radius of the circular path of the proton is,
\[{r_p} = \dfrac{{{m_p}{v_p}}}{{Be}}\]
\[{r_p} = \dfrac{p}{{Be}}\]
On dividing both the expressions for the radius of the circular path,
\[\dfrac{{{r_e}}}{{{r_p}}} = \dfrac{{\left( {\dfrac{p}{{Be}}} \right)}}{{\left( {\dfrac{p}{{Be}}} \right)}}\]
\[{r_e} = {r_p}\]
So, the radius of the circular path for the electron is equal to the radius of the circular path for the proton.
Therefore, the correct option is (C).
Note:As the nature of the charge on proton and electron is opposite, so the direction of the deflection will be opposite to each other though the radius of the circular path will be equal.
Formula used:
\[r = \dfrac{{mv}}{{Bq}}\], here r is the radius of the circular path when a charged particle of mass m and charge q enters into a region of magnetic field strength B with speed v.
\[p = mv\], here p is the magnitude of the momentum of the body of mass m moving with speed v.
Complete answer:
Let the mass of the electron is \[{m_e}\]and the mass of the proton is \[{m_p}\]
The charge on the electron and the proton is same in magnitude \[{q_e} = {q_p} = e\]
It is given that the electron and proton is moving with equal momentum,
\[{m_e}{v_e} = p\]
\[{m_p}{v_p} = p\]
On dividing both the expressions of the momentum, we get
\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{{m_p}}}{{{m_e}}}\]
Using the radius formula,
The radius of the circular path of the electron is,
\[{r_e} = \dfrac{{{m_e}{v_e}}}{{Be}}\]
\[{r_e} = \dfrac{p}{{Be}}\]
The radius of the circular path of the proton is,
\[{r_p} = \dfrac{{{m_p}{v_p}}}{{Be}}\]
\[{r_p} = \dfrac{p}{{Be}}\]
On dividing both the expressions for the radius of the circular path,
\[\dfrac{{{r_e}}}{{{r_p}}} = \dfrac{{\left( {\dfrac{p}{{Be}}} \right)}}{{\left( {\dfrac{p}{{Be}}} \right)}}\]
\[{r_e} = {r_p}\]
So, the radius of the circular path for the electron is equal to the radius of the circular path for the proton.
Therefore, the correct option is (C).
Note:As the nature of the charge on proton and electron is opposite, so the direction of the deflection will be opposite to each other though the radius of the circular path will be equal.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
