Rotating Disk of charge

[amirks]

Consider the case of a disk of radius a, carrying a surface charge density σ

that varies linearly with the distance from the centre of the disk. If the disk rotates about

its symmetry axis with an angular velocity of ω, calculate the magnetic field B a

distance z above the midpoint of the disk. What is the electric field at the same point?

 

so what I did was I said digma= cr c=constant and v=omega r

so surface current I=sigma v = c omega r^2

 

now using amperian loop we have int (B.dl) = mu I(enc)

i used an amparian loop with one vertical side corssing the centre with width of r and height of z above the disk (and below, and we have two vertical parts)

so int (B.dl) = B4z

and I(enc) = c omega r^2 r

not sure if any of these are right, and also I thought of doing this by thinking of the disk as many narrow loops of charge and then find the current at each loop and then the field and ....

 

plz helllp!!

[mooeypoo]

Okay, it's very late here, but let's see if I can at least get you going on the right direction.

 

The total charge in the disk, would be Q:

[math]Q=\sigma (\pi R^2)[/math]

 

What you need to do is divide this into very thin strips, "dr".

 

Find the current relative to "dr". That will enable you to find the magnetic field later.

 

Remember this: http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law

And take the 'r' (the vector where you measure the magnetic field) to be an arbitrary point above the rotating disk. This will probably be the most annoying calculation (I remember I always got confused with all the 'parts' of that vector, but just watch out for it.

I would consider taking a coordinate system that might help more, maybe cylindrical.

 

For the Biot Savart law, you need the current. You can find the current by what I gave you in the beginning.

 

Try this, slowly, don't panic, take it step by step. Those questions are just confusing, but they're not horribly hard. I have a huge exam tomorrow so I don't really have time to sit and solve this question, but if you are still stuck tomorrow afternoon, I'll try to lead you more.

 

Try tackling it, though.

 

Good luck!

 

~moo

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Merged post follows:

Consecutive posts merged

BTW, I just noticed (and it's different a bit from what you said in the chatroom) -- NOTICE that the question asks for the magnetic field *at the symmetry*.. that is, above the axis of rotation, so *not* at an arbitrary point!

 

That makes your caculation MUCH easier. Much. Think of the symmetry and the current density (at the 'dr' small rings) and use that for the calculation of the B magnetic field.