
An electron, a proton, a deuteron, and an alpha particle, each having speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles is respectively ${{R}_{e,}}{{R}_{p}},{{R}_{d}}$and ${{R}_{\alpha }}$. It follows that
A. ${{R}_{e}}={{R}_{p}}$
B.${{R}_{p}}={{R}_{d}}$
C.${{R}_{d}}={{R}_{\alpha }}$
D.${{R}_{p}}={{R}_{\alpha }}$
Answer
164.1k+ views
Hint: When a charged particle moves with definite velocity and enters a uniform magnetic field $B$, then it experiences a magnetic force perpendicular to the direction of motion and it travels a circular path. Then by equating magnetic force with centripetal force, we can derive the equation of the radius of a circular path.
Formula used:
The radius,$R$ of the circular path in a magnetic field,$B$can be expressed in the following way:
$R=\dfrac{mv}{qB}$
Here $m\And v$are the mass and velocity of the particle with charge $q$.
Complete answer:
When a particle carrying charge$q$, moving with velocity $\vec{v}$enters into a magnetic field $\vec{B}$, it experiences a magnetic force, $\vec{F}=q(\vec{B}\times \vec{v})$

Or,$F=q(Bv\sin {{90}^{o}})$ [Since $\vec{B}$is perpendicular to $\vec{v}$]
Or,$F=qBv$ ……..(i)
As a particle moves in a circular path, then magnetic force becomes a centripetal force $\dfrac{m{{v}^{2}}}{R}$.
Hence by equating magnetic force with centripetal force,
$qBv=\dfrac{m{{v}^{2}}}{R}$
Or,$R=\dfrac{mv}{qB}$
Here we have four charged particles: an electron($e$), a proton ($p$), a deuteron ($d$), and an alpha particle $(\alpha )$. They all have equal speed,$v$and move in a region of the constant magnetic field,$B$.
Therefore the radius of the circular path mainly depends on $\dfrac{mass(m)}{ch\arg e(q)}$ratio.
Or,$R$ $\alpha $ $\dfrac{m}{q}$
Let us check $\dfrac{m}{q}$ratio of each charged particle in the following table,
Let the Mass of a proton be $m$and charge $q$.
As we know the mass of an electron, deuterium and an alpha particle are $\dfrac{1}{1836}$, $2$ and $4$ times the mass of the proton.
Therefore $\dfrac{m}{q}$ ratio for deuterium and an alpha particle are equal, hence their radius of the circular orbit would be equal i.e,${{R}_{d}}={{R}_{\alpha }}$.
Thus, option (C) is correct.
Note:Neutron does not feel any magnetic force while other charged particles experience that force. A neutron is a neutral particle, having no charge. But for charged particle trajectory curvature is proportional to the mass by charge ratio for a definite velocity.
Formula used:
The radius,$R$ of the circular path in a magnetic field,$B$can be expressed in the following way:
$R=\dfrac{mv}{qB}$
Here $m\And v$are the mass and velocity of the particle with charge $q$.
Complete answer:
When a particle carrying charge$q$, moving with velocity $\vec{v}$enters into a magnetic field $\vec{B}$, it experiences a magnetic force, $\vec{F}=q(\vec{B}\times \vec{v})$

Or,$F=q(Bv\sin {{90}^{o}})$ [Since $\vec{B}$is perpendicular to $\vec{v}$]
Or,$F=qBv$ ……..(i)
As a particle moves in a circular path, then magnetic force becomes a centripetal force $\dfrac{m{{v}^{2}}}{R}$.
Hence by equating magnetic force with centripetal force,
$qBv=\dfrac{m{{v}^{2}}}{R}$
Or,$R=\dfrac{mv}{qB}$
Here we have four charged particles: an electron($e$), a proton ($p$), a deuteron ($d$), and an alpha particle $(\alpha )$. They all have equal speed,$v$and move in a region of the constant magnetic field,$B$.
Therefore the radius of the circular path mainly depends on $\dfrac{mass(m)}{ch\arg e(q)}$ratio.
Or,$R$ $\alpha $ $\dfrac{m}{q}$
Let us check $\dfrac{m}{q}$ratio of each charged particle in the following table,
Let the Mass of a proton be $m$and charge $q$.
Proton | Electron | deuteron | deuteron | |
Mass($m$) | $m$ | $\dfrac{m}{1836}$ | $2m$ | $4m$ |
Charge($q$) | $q$ | $q$ | $q$ | $2q$ |
$\dfrac{mass}{ch\arg e}\left( \dfrac{m}{q} \right)$ | $\dfrac{m}{q}$ | $\dfrac{m}{q\times 1836}$ | $\dfrac{2m}{q}$ | $\dfrac{4m}{2q}=\dfrac{2m}{q}$ |
As we know the mass of an electron, deuterium and an alpha particle are $\dfrac{1}{1836}$, $2$ and $4$ times the mass of the proton.
Therefore $\dfrac{m}{q}$ ratio for deuterium and an alpha particle are equal, hence their radius of the circular orbit would be equal i.e,${{R}_{d}}={{R}_{\alpha }}$.
Thus, option (C) is correct.
Note:Neutron does not feel any magnetic force while other charged particles experience that force. A neutron is a neutral particle, having no charge. But for charged particle trajectory curvature is proportional to the mass by charge ratio for a definite velocity.
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 Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Charging and Discharging of Capacitor
