Calculating Flux Change and Meaning of Rotating Ring Diameter?

dikmikkel
Messages
160
Reaction score
0

Homework Statement


A ring can rotate about a horizontal axis(x), and a diameter placed on the x-axis. A uniform field is perpendicular to the ring -B0*y. The diameter of the ring is D. it spins with constant angular velocity ω around the x-axis. At at time t = 0 the ring is entirely in the xy plane

How do i find the flux change per time? And what does it mean that it can rotate around a diameter.
I've attached a figure.


Homework Equations


Flux = \int \vec{B}\cdot d\vec{a}


The Attempt at a Solution


At a time t = 0 the flux through the loop is 0 and i tried to write a solution using that B is constant:
Flux = B\int sin(\theta (t))d\vec{a}
Flux/dt = B\int cos(\theta(t) ) \omega d\vec{a}
 

Attachments

  • problem.png
    problem.png
    14.2 KB · Views: 509
Physics news on Phys.org
Nwm.
I found out.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top