JEE Exam > JEE Questions > A uniform disc of radius R having charge Q di...
Start Learning for Free
A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) is
Correct answer is '1'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A uniform disc of radius R having charge Q distributed uniformly all o...
Solution:
Given data:
- The radius of the disc, R = 1m
- The charge on the disc, Q = (1/4k) C
- The magnetic field, B = kxt^2 T
- The time for which the magnetic field is switched on, t = 15s
Let's calculate the magnetic moment of the disc first.
Magnetic Moment of the Disc:
- The magnetic moment of the disc is given by the product of the current flowing through it and the area of the disc.
- The current flowing through the disc is due to the induced electric field in it, which is given by E = -dΦ/dt, where Φ is the magnetic flux through the disc.
- The magnetic flux through the disc is given by Φ = BA, where A is the area of the disc and B is the magnetic field.
- So, the induced electric field in the disc is given by E = -d(BA)/dt = -AB(dB/dt) = -Akx^2.
- The current flowing through the disc is given by I = σE, where σ is the surface charge density of the disc.
- The surface charge density of the disc is given by σ = Q/4πR^2.
- So, the current flowing through the disc is given by I = (Q/4πR^2)(-Akx^2).
- The magnetic moment of the disc is given by M = IA = (Q/4πR^2)(-Akx^2)(πR^2) = -AkQx^2.
Let's calculate the torque acting on the disc now.
Torque on the Disc:
- The torque acting on the disc is given by the cross product of the magnetic moment of the disc and the magnetic field.
- So, the torque acting on the disc is given by τ = M × B = (-AkQx^2) × kxt^2.
- The direction of the torque is perpendicular to both the magnetic moment and the magnetic field, which is in the z-direction.
- So, the magnitude of the torque acting on the disc is given by |τ| = Ak^2Qx^4t^2.
- Substituting the given values, we get |τ| = (1/4)(1/16)(1^4)(15^2) = 1 N-m.
Therefore, the torque acting on the disc after 15 sec is 1 N-m.
Given data:
- The radius of the disc, R = 1m
- The charge on the disc, Q = (1/4k) C
- The magnetic field, B = kxt^2 T
- The time for which the magnetic field is switched on, t = 15s
Let's calculate the magnetic moment of the disc first.
Magnetic Moment of the Disc:
- The magnetic moment of the disc is given by the product of the current flowing through it and the area of the disc.
- The current flowing through the disc is due to the induced electric field in it, which is given by E = -dΦ/dt, where Φ is the magnetic flux through the disc.
- The magnetic flux through the disc is given by Φ = BA, where A is the area of the disc and B is the magnetic field.
- So, the induced electric field in the disc is given by E = -d(BA)/dt = -AB(dB/dt) = -Akx^2.
- The current flowing through the disc is given by I = σE, where σ is the surface charge density of the disc.
- The surface charge density of the disc is given by σ = Q/4πR^2.
- So, the current flowing through the disc is given by I = (Q/4πR^2)(-Akx^2).
- The magnetic moment of the disc is given by M = IA = (Q/4πR^2)(-Akx^2)(πR^2) = -AkQx^2.
Let's calculate the torque acting on the disc now.
Torque on the Disc:
- The torque acting on the disc is given by the cross product of the magnetic moment of the disc and the magnetic field.
- So, the torque acting on the disc is given by τ = M × B = (-AkQx^2) × kxt^2.
- The direction of the torque is perpendicular to both the magnetic moment and the magnetic field, which is in the z-direction.
- So, the magnitude of the torque acting on the disc is given by |τ| = Ak^2Qx^4t^2.
- Substituting the given values, we get |τ| = (1/4)(1/16)(1^4)(15^2) = 1 N-m.
Therefore, the torque acting on the disc after 15 sec is 1 N-m.
Free Test
FREE
| Start Free Test |
Community Answer
A uniform disc of radius R having charge Q distributed uniformly all o...
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?
Question Description
A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?.
A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?.
Solutions for A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer?, a detailed solution for A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? has been provided alongside types of A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A uniform disc of radius R having charge Q distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, B = kxt2, where k is a constant, x is the distance (in metre) from the centre of the disc and t is the time (in second), is switched on perpendicular to the plane of the disc. Find the torque (in N-m) acting on the disc after 15 sec. (Take 4kQ = 1 S.I.unit and R = 1m) isCorrect answer is '1'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup