
A thin non conducting horizontal disc of mass m having total charge q distributed uniformly over its surface can rotate freely about its axis. initially, when the disk is stationary a magnetic field b directed perpendicular to the plane is switched on at t=0. Find the angular velocity ω acquired by disc as a function of time, if B=kt, where t is the time.
Answer
219.9k+ views
Hint: Angular velocity is defined as the rate of change of angular displacement of an object and the axis about which the object is rotating. It is a vector quantity. The rotational equivalent of linear force is called Torque. In simple terms, Torque is the tendency of a force to turn or twist. Non-conductor is a substance that does not conduct heat or electricity.
Complete solution:
Mass of disc = m
Magnetic field = B
Time = t
B = kt
Angular velocity, \[\omega \]=?
We know that the Electric field-induced due to time-dependent magnetic field is given by
$\smallint E.dr = - \pi {r^2}\dfrac{{dB}}{{dt}}$
\[ \Rightarrow {\rm E} \times 2\pi r = - \pi {r^2}\dfrac{d}{{dt}}\left( {kt} \right)\]
\[ \Rightarrow {\rm E} \times 2\pi r = - \pi {r^2}k\]
\[ \Rightarrow {\rm E} = \dfrac{{rk}}{2}\]
Now the torque is given by the equation
$\tau = \smallint \dfrac{q}{{\pi {r^2}}} \times 2\pi rdr \times \dfrac{{rk}}{2} \times r$
\[ \Rightarrow \tau = \smallint \dfrac{{qk}}{{{r^2}}}{r^3}dr\]
\[ \Rightarrow \tau = \dfrac{{qk}}{{{r^2}}}\dfrac{{{r^4}}}{4}\]
\[ \Rightarrow \dfrac{{m{r^2}}}{2}\alpha = \dfrac{{qk{r^2}}}{4}\]
\[ \Rightarrow \alpha = \dfrac{{qk}}{{2m}}\]
We know that angular velocity is given by, $\omega = \alpha \times t$
$ \Rightarrow \omega = \dfrac{{qk}}{{2m}}t$
Hence the angular velocity as a function of time is $\omega = \dfrac{{qk}}{{2m}}t.$
Note: 1) Torque can be considered as a special case of Moment.
2) The angular velocity direction of the vector perpendicular to the plane of rotation is given by the right-hand rule.
3) Magnetic field defines the magnetic influence on moving magnetized materials and electric currents. When a charge is moving in a magnetic field, it experiences a force that is perpendicular to its velocity and also to the magnetic field.
4) The magnetic field and the electric field are interrelated to each other and both of them are the components of the electromagnetic force.
5) Electric charge is the property of matter which causes it to experience a force when placed in an electromagnetic field. Protons and Electrons are the two types of charges.
Complete solution:
Mass of disc = m
Magnetic field = B
Time = t
B = kt
Angular velocity, \[\omega \]=?
We know that the Electric field-induced due to time-dependent magnetic field is given by
$\smallint E.dr = - \pi {r^2}\dfrac{{dB}}{{dt}}$
\[ \Rightarrow {\rm E} \times 2\pi r = - \pi {r^2}\dfrac{d}{{dt}}\left( {kt} \right)\]
\[ \Rightarrow {\rm E} \times 2\pi r = - \pi {r^2}k\]
\[ \Rightarrow {\rm E} = \dfrac{{rk}}{2}\]
Now the torque is given by the equation
$\tau = \smallint \dfrac{q}{{\pi {r^2}}} \times 2\pi rdr \times \dfrac{{rk}}{2} \times r$
\[ \Rightarrow \tau = \smallint \dfrac{{qk}}{{{r^2}}}{r^3}dr\]
\[ \Rightarrow \tau = \dfrac{{qk}}{{{r^2}}}\dfrac{{{r^4}}}{4}\]
\[ \Rightarrow \dfrac{{m{r^2}}}{2}\alpha = \dfrac{{qk{r^2}}}{4}\]
\[ \Rightarrow \alpha = \dfrac{{qk}}{{2m}}\]
We know that angular velocity is given by, $\omega = \alpha \times t$
$ \Rightarrow \omega = \dfrac{{qk}}{{2m}}t$
Hence the angular velocity as a function of time is $\omega = \dfrac{{qk}}{{2m}}t.$
Note: 1) Torque can be considered as a special case of Moment.
2) The angular velocity direction of the vector perpendicular to the plane of rotation is given by the right-hand rule.
3) Magnetic field defines the magnetic influence on moving magnetized materials and electric currents. When a charge is moving in a magnetic field, it experiences a force that is perpendicular to its velocity and also to the magnetic field.
4) The magnetic field and the electric field are interrelated to each other and both of them are the components of the electromagnetic force.
5) Electric charge is the property of matter which causes it to experience a force when placed in an electromagnetic field. Protons and Electrons are the two types of charges.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

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

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

