anonymous
  • anonymous
hi we have a disk with uniform charge density "sigma" that rotate around of axis that transmitted from it's origin,find magnetic moment of disk?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ok, Total charge on the disk is: \[\ Q=\sigma \pi R^2 \]where R is the radius of the disk If you divide the disk into small rings(width dr) you can write the current in that ring as:\[\ dI=\frac{dQ}{T}\]where T- time period Since \[\ dI =2 \sigma \pi rdr \] and \[\ T=\frac{2\pi}{\omega}\] you can rewrite dI as:\[\ dI =2 \sigma \pi rdr \frac{\omega}{2 \pi} \]\[\ dI = \sigma r \omega dr \] Magnetic moment is defined as: \[\ m=IA\] where I-current, A-area \[\ dm=dI \pi r^2\]\[\ dm= \sigma r \omega dr \pi r^2\]\[\ \int dm= \int \sigma \omega \pi r^3 dr\]\[\ m=\frac{\pi}{4}\sigma \omega R^4\] That would be it.
anonymous
  • anonymous
Also, put a hat on m and omega, since it is a vector...
anonymous
  • anonymous
thnx

Looking for something else?

Not the answer you are looking for? Search for more explanations.