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Consider a hollow cylinder. The inner radius r1=1 and the outer radius r2=2. Assume the cylinder has a current of I ampere flowing through it. a. Derive an expression for the magnetic field in the hollow part of the cylinder i.e. for r b. Derive an expression for the magnetic field between r1 and r2, i.e. for r1 c. Derive an expression for the magnetic field outside the cylinder, i.e. for r>r2.

Consider a hollow cylinder. The inner radius r1=1 and the
outer radius r2=2. Assume the cylinder has a current of I ampere flowing through it.
a. Derive an expression for the magnetic field in the hollow part of the cylinder i.e.
for rb. Derive an expression for the magnetic field between r1 and r2, i.e. for r1c. Derive an expression for the magnetic field outside the cylinder, i.e. for r>r2.

Grade:Upto college level

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
one year ago
emember when we were looking at electric fields inside and outside charged spherical shells? We used Gauss' Law to show that the field inside the shell was zero, and outside the shell the electric field was the same as the field from a point charge with a charge equal to the charge on the shell and placed at the center of the shell.

The analagous situation for Ampere's Law is a long cylindrical shell carrying a uniformly-distributed current I. Inside the hollow part of the cylinder the magnetic field is zero (an amperian loop encloses no current) and outside the cylinder the magnetic field is the same as that from a long straight wire placed on the axis of the cylinder:

B =

μoI
2πr

The most interesting case is the field inside the solid part of the shell. Let's say the shell has an inner radius a and an outer radius b. What is the field in the region a < r < b?

First we need the current density, J, the current per unit area. The area of the shell is:

A = πb2- πa2

Therefore J =

I
A

=

I
πb2- πa2

Apply Ampere's Law to an amperian loop of radius r in the solid part of the cylindrical shell.

∫B•dl=∫B dl = B∫dl = μoIenc

The left-hand side of the equation is easy to calculate.

B∫dl = B 2πr

The right-hand side is a little trickier. How much current is enclosed by the loop?

The area enclosed is:

Aenc= πr2- πa2

The enclosed current is then Ienc= J Aenc

Ienc = J (πr2- πa2) =

I (r2- a2)
b2- a2

Putting all the pieces of Ampere's Law together gives:

B 2πr =

μoI (r2- a2)
b2- a2

B =

μoI (r2- a2)
2πr (b2- a2)

in the region a < r < b

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