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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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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
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
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