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sohan singh singh Grade: 12
        


 

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

7 years ago

Answers : (3)

Vijay Luxmi Askiitiansexpert
357 Points
										

Dear Sohan,


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7 years ago
Gokul Joshi AskiitiansExpert-IITK
42 Points
										

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.



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7 years ago
SAGAR SINGH - IIT DELHI
879 Points
										

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)


 







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Sagar Singh


B.Tech IIT Delhi






7 years ago
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