Infinitely long cylindrical shell (magnetic field at centre)

[Alex_Neof]

Homework Statement

An infinitely long hollow cylinder of radius (a) carries a constant current (i). Use Biot - Savart's law and show that the magnetic field at the centre of the cylinder is zero.

Also show, using symmetry arguments and Ampere's law, the magnetic field is zero in any point inside the cylinder.

Homework Equations

Biot - Savart's law.

The Attempt at a Solution

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I know through symmetry and the right hand rule you get a magnetic field of zero at the centre, but how would one show this mathematically. I think evaluating a double integral would do this.

Firstly I evaluated an integral (with limits from 0 to pi) for the magnetic field at a perpendicular distance (a) from one straight infinitely long wire and obtained
B = (μ0)*(di) / 2*pi*a.

Where di is a fraction of the total current i for that one wire.

Now where I am finding difficulty is finding a way to evaluate the second integral for a circle for loads of wires which would produce an infinitely long hollow cylinder (cylindrical shell).

How would I produce the second integral ?

Thank you.
Update: second integral containing (μ0)*(di) / 2*pi*a. The limits would be from 0 to 2*pi

 

[Goddar]

Hi. This is pretty straightforward, there are no tricks involved with wires and such.
In the first part, start from Biot-Savart law:
- express the current density J in cylindrical coordinates in terms of the current I (with its vector-direction) and delta/theta-functions;
- plug this into Biot-Savart's, knowing that you're evaluating at the origin (r = 0);
- i'll let you unfold all this but here's the important step: convert the resulting vector direction in Cartesian coordinates (because unless it's z it's not constant in cylindrical coordinates and you can't take it out of the integral), then perform the integral.
I can't be more specific since you're not saying in what direction the current is flowing (although I'm assuming along the cylinder), but you should manage from here.

For the second part, Ampere's law with the right Amperian loop will get you through it.