Current loop rotating to give magnetic moment

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The discussion centers on calculating the magnetic moment of a rotating non-conducting ring with a radius of 0.816 cm and a total charge of 6.76 µC, rotating at an angular speed of 1.73 rad/s. The user initially attempted to find the current using the formula qω/2π and then multiplied it by the area of the ring, but this approach did not yield the correct result. They later tried integrating over the ring's radius using charge density, but still faced discrepancies in their final answer. The user is seeking clarification on their method and why their calculations did not align with the expected result of approximately 3.89 E-10. The conversation highlights the complexities involved in deriving the magnetic moment from a rotating charged ring.
nautola
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Homework Statement


A uniform non-conducting ring of radius
0.816 cm and total charge 6.76 µC rotates
with a constant angular speed of 1.73 rad/s
around an axis perpendicular to the plane of
the ring that passes through its center.
What is the magnitude of the magnetic
moment of the rotating ring?

r = .00816 m
q = 6.76E-6 C
ω = 1.73 rad/s



Homework Equations


μ = IA
I = dq/dt
A = \pir^2



The Attempt at a Solution


I took the current and said it was equal to qω/2∏, since that gives charge/time. Then I multiplied by area. When that didn't work I decided to take the same approach but integrating from 0 to ∏/2 with r replaced with (rcosθ), to be the radius of any point on the loop, making a circle as a function of angle. Then I said charge was equal to λr dθ, since the charge is uniform. I took that function and integrated it: λω(r^2)/2∫(cosθ)^2 dθ, 0,∏/2. Then I multiplied by 4, for each of the quarters of the loop.
The idea is that each infinitesimally small point on the loop has a charge and will behave like a charge orbiting and integrating over all of the possible radii gives the combined magnetic moment. However, this didn't work and I lost points. I don't understand why.
 
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qω/2∏ * area should be correct. Did you get 3.89 E-10 ?
 
I had almost that...but not quite within 1%. Which is what our online homework requires. So my approach was correct, just not some value.
Thanks.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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