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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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7 years ago

357 Points
```										Dear Sohan,

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.

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7 years ago
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)

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

Sagar Singh
B.Tech IIT Delhi

```
7 years ago
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