Guest

Consider a thin cylinder’s shell with radius r and mass m (inertia mr^2). Two sides of the surface have different characteristics. Outer surface of the shell is a conductor with resistivity ρ and thickness δ(δ http://yfrog.com/3t005pg] HERE l IS THE LENGTH OF THE CYLINDER


 

Consider a thin cylinder’s shell with radius r and mass m (inertia mr^2). Two sides of the surface have different characteristics. Outer surface of the shell is a conductor with resistivity ρ and thickness δ(δ<http://yfrog.com/3t005pg] HERE l IS THE LENGTH OF THE CYLINDER

Grade:12

3 Answers

Vijay Luxmi Askiitiansexpert
357 Points
13 years ago

Dear Sohan,

Please Post your query only once.

 

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly.

We are all IITians and here to help you in your IIT JEE preparation.

All the best.

 

Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Regards

Vijay Luxmi

Gokul Joshi AskiitiansExpert-IITK
42 Points
13 years ago

Dear Sohan,

It seems you have not posted a complete question. 

Please feel free to post as many doubts on our discussion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly.


We are all IITians and here to help you in your IIT JEE preparation.All the best.

Regards,
Askiitians Experts,
Gokul  Joshi




SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear sohan,

The surface charge density = σ = charge per m^2
Charge = σ * area of the inner surface
The area of the inner surface = 2 * π * r * l
l = length of cylinder

Charge = σ * 2 * π * r * l Coulombs

If we let the length of the cylinder = 1 meter,
The amount of charge on a 1 meter length of the cylinder = σ * 2 * π * r

1 amp of current = 1 coulomb per second
When the angular velocity = ω, the cylinder is rotating ω radians per second.
1 radian per second means a point on the inner surface is moving at a linear velocity of 1 radius per second. So the charged particles on the inner surface is moving at a linear velocity of (ω * r) meters per second.

This means the charge on a 1 meter length of the cylinder is moving at a velocity of (ω * r) meters per second.

The charge on a 1 meter length of the cylinder = σ * 2 * π * r
(σ * 2 * π * r) Coulombs of charge is moving around in a circle at a linear velocity of (ω * r) meters per second.

1 amp of current = 1 coulomb per second
Amps = coulomb/meter * meters/sec
Amps = (σ * 2 * π * r) * (ω * r)

So, the current flowing on the inner surface of the cylinder = (σ * 2 * π * r) * (ω * r) amps

This amount of current flowing in a circle on the inner surface of the cylinder produces a magnetic field. This magnetic field induces a current in the outer surface of the shell.

The strength of the β field is inversely proportional to the square of the distance between the source of the β field and the inner surface of the shell. The distance = r + δ. Since δ<<r , we can say that the distance = r

So, the β field at the outer surface of the shell = μo * (σ * 2 * π * r) * (ω * r) ÷ r^2
r * r ÷ r^2 = 1

So, the β field at the outer surface of the shell = μo * (σ * 2 * π) * (ω)

This β field will induce a voltage in the outer surface of the shell, which is a conductor with resistivity ρ and thickness δ

The induced voltage = β outer * (length) * (velocity)
(length) * (velocity) = area per second.

Induced voltage = β outer * area per second.

Area per second is the distance that the area of the inner surface moves each second.
The area of the inner surface = 2 * π * r * length

Linear velocity of cylinder = (ω * r) meters per second.
This means the outer surface of the cylinder is moving (ω * r) meters per second in a circular path.

This means that a surface with area of (2 * π * r) m^2 is moving (ω * r) m/s

Area per second = (2 * π * r) * (ω * r)
We let the length of the cylinder = 1 meter

Area per second = (2 * π * r) * (ω * r)
β outer = μo * (σ * 2 * π) * (ω)

Induced voltage = μo * (σ * 2 * π) * (ω) * (2 * π * r) * (ω * r)

V = I * R
Resistance = (resistivity * length) ÷ area
Length = 1
Resistance = ρ ÷ (2 * π * r)

Current = V ÷ R

Current = μo * (σ * 2 * π) * (ω) * (2 * π * r) * (ω * r) ÷ ρ ÷ (2 * π * r)

angular acceleration = τ ÷ [mr^2 + (1/ ρ * π * δ * l * μo^2 * σ^2 * r^2 * ω)

 

We are all IITians and here to help you in your IIT JEE preparation.

All the best.

 If you like this answer please approve it....

win exciting gifts by answering the questions on Discussion Forum

 

Sagar Singh

B.Tech IIT Delhi

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free