
An electron moves in a circular orbit with uniform speed $v$.It produces a magnetic field $B$at the centre of the circle. The radius of the circle is proportional to
A.$\dfrac{B}{v}$
B.$\dfrac{v}{R}$
C.$\sqrt{\dfrac{v}{B}}$
D.$\sqrt{\dfrac{B}{v}}$
Answer
163.8k+ views
Hint: When an electron moves in a circular orbit with uniform speed producing a magnetic field at the centre of the circle. This is because a circular current is generated at the centre. Therefore by substituting the value of the equivalent current produced in a circular orbit in Biot-Savart law we can determine the radius of the circle.
Complete answer:
The electron moves in a circular orbit with uniform speed $v$and the magnetic field$B$ is produced at the centre of the circle. The time period of the electron that is moving in an angular orbit can be expressed as:
Time period,$T=\dfrac{2\pi r}{v}$ ……….(i)
Here $r$is the radius of the circle and $v$ the speed of the charged particle.
[since$2\pi r=$ Circumference of the circular path.]
Now the equivalent current,$i$ which is produced in a circle due to the flow of electrons is given by
$i=\dfrac{ch\arg e}{time}=\dfrac{e}{T}$ ………(ii)
Now putting the value $T$in equation (ii),
$i=\dfrac{e}{\dfrac{2\pi r}{v}}$
Or,$i=\dfrac{ev}{2\pi r}$ ……….(iii)
According to Biot-Savart law, the magnetic field $B$at a particular point induced by the flow of current can be expressed as,

$B=\dfrac{{{\mu }_{o}}i}{2r}$ ………(iv)
Here ${{\mu }_{o}}$denotes the permeability constant and $i$is equivalent current produced in a circular path.
Hence by substituting the value of the equivalent current,$i$in equation (iv) we get,
$B=\dfrac{{{\mu }_{o}}}{2r}\times \dfrac{ev}{2\pi r}$
Or,${{r}^{2}}=\dfrac{{{\mu }_{o}}ev}{4\pi B}$
Or,$r=\sqrt{\dfrac{{{\mu }_{o}}ev}{4\pi B}}$
Therefore $r$ $\alpha $ $\sqrt{\dfrac{v}{B}}$
The radius of the circle,$r$, is directly proportional to $\sqrt{\dfrac{v}{B}}$.
Thus, option (C) is correct.
Note: Magnetic permeability is used to determine materials resistance to a magnetic field. The higher the permeability, the higher will be the conductivity of magnetic lines of force. It is found that the materials like iron alloys, nickel alloys, and cobalt alloys have relatively high permeability.
Complete answer:
The electron moves in a circular orbit with uniform speed $v$and the magnetic field$B$ is produced at the centre of the circle. The time period of the electron that is moving in an angular orbit can be expressed as:
Time period,$T=\dfrac{2\pi r}{v}$ ……….(i)
Here $r$is the radius of the circle and $v$ the speed of the charged particle.
[since$2\pi r=$ Circumference of the circular path.]
Now the equivalent current,$i$ which is produced in a circle due to the flow of electrons is given by
$i=\dfrac{ch\arg e}{time}=\dfrac{e}{T}$ ………(ii)
Now putting the value $T$in equation (ii),
$i=\dfrac{e}{\dfrac{2\pi r}{v}}$
Or,$i=\dfrac{ev}{2\pi r}$ ……….(iii)
According to Biot-Savart law, the magnetic field $B$at a particular point induced by the flow of current can be expressed as,

$B=\dfrac{{{\mu }_{o}}i}{2r}$ ………(iv)
Here ${{\mu }_{o}}$denotes the permeability constant and $i$is equivalent current produced in a circular path.
Hence by substituting the value of the equivalent current,$i$in equation (iv) we get,
$B=\dfrac{{{\mu }_{o}}}{2r}\times \dfrac{ev}{2\pi r}$
Or,${{r}^{2}}=\dfrac{{{\mu }_{o}}ev}{4\pi B}$
Or,$r=\sqrt{\dfrac{{{\mu }_{o}}ev}{4\pi B}}$
Therefore $r$ $\alpha $ $\sqrt{\dfrac{v}{B}}$
The radius of the circle,$r$, is directly proportional to $\sqrt{\dfrac{v}{B}}$.
Thus, option (C) is correct.
Note: Magnetic permeability is used to determine materials resistance to a magnetic field. The higher the permeability, the higher will be the conductivity of magnetic lines of force. It is found that the materials like iron alloys, nickel alloys, and cobalt alloys have relatively high permeability.
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
