
The radius of circular path of an electron when subjected to a perpendicular magnetic field is
A . $\dfrac{Be}{mv}$
B . $\dfrac{me}{Bv}$
C . $\dfrac{mB}{ev}$
D . $\dfrac{mv}{Be}$
Answer
163.8k+ views
Hint:In order to answer this question, we have to use the concept that a charged particle moving in a magnetic field experiences a magnetic force that occasionally causes it to deflect. The charge, speed, and magnetic field strength of the particle all affect how strong the magnetic force is.
Formula used:
The magnetic force:
F=qvB; here, q denotes the charge, v the particle's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
Complete answer:
The centripetal force counterbalances the magnetic pull that acts on a moving particle perpendicular to its velocity. The equilibrium formula states the following:
F(magnetic force)=${F_c}$(centripetal force) - (i)
The centripetal force here operating on the electron is known as ${F_c}$. We are aware that a electron is subject to the force, F, of a magnetic field represented by,
F=evB – (ii)
Here, e denotes the charge of the electron, v the electron's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$ - (iii)
Here m is the mass of the electron, v is the velocity of the electron, and r is the radius.
We will now substitute equation (iii) and (ii) in equation (i) to find the value of r.
$evB=\dfrac{m{{v}^{2}}}{r}$
$r=\dfrac{mv}{eB}$
The correct answer is D.
Note:A force is applied to a particle when the magnetic field and particle are moving perpendicular to one another. If the magnetic field and the particle are moving in the same direction, there is no force exerting on the particle.
Formula used:
The magnetic force:
F=qvB; here, q denotes the charge, v the particle's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
Complete answer:
The centripetal force counterbalances the magnetic pull that acts on a moving particle perpendicular to its velocity. The equilibrium formula states the following:
F(magnetic force)=${F_c}$(centripetal force) - (i)
The centripetal force here operating on the electron is known as ${F_c}$. We are aware that a electron is subject to the force, F, of a magnetic field represented by,
F=evB – (ii)
Here, e denotes the charge of the electron, v the electron's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$ - (iii)
Here m is the mass of the electron, v is the velocity of the electron, and r is the radius.
We will now substitute equation (iii) and (ii) in equation (i) to find the value of r.
$evB=\dfrac{m{{v}^{2}}}{r}$
$r=\dfrac{mv}{eB}$
The correct answer is D.
Note:A force is applied to a particle when the magnetic field and particle are moving perpendicular to one another. If the magnetic field and the particle are moving in the same direction, there is no force exerting on the particle.
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

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
