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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

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6 years ago

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6 years ago
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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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6 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)

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6 years ago

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