MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM

DETAILS

MRP

DISCOUNT

FINAL PRICE

Total Price: Rs.

There are no items in this cart.

Continue Shopping

Continue Shopping

Menu

` `

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

6 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 winexciting giftsby answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.Regards

Vijay Luxmi

6 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

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)

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....

winexciting giftsby answering the questions on Discussion Forum

Sagar Singh

B.Tech IIT Delhi

6 years ago

Think You Can Provide A **Better Answer ?**

Answer & Earn Cool Goodies

Have any Question? Ask Experts

Post Question

copyright © 2006-2017 askIITians.com

info@askiitians.com