I forgot to write that the answer at the back says something completely otherwise (its only an answer and not a solution)
\sqrt{\ \frac{\ \nu _{o}^{2}}{2} +\sqrt{\left(\frac{\ \nu _{o}^{2}}{2}\right)^{2} +\frac{q^{2} \nu _{o}^{2}}{4\pi ^{2} d^{2} mC}}}
Ion traps are very complex, but one of my Physics Olympiad textbooks (this is not homework) presents a simplified model of a resonating charged particle in an ion trap which according to me is wrong, there is either something missing or something overspecified here.
A tuned circuit...
I forgot to write that the answer at the back says something completely otherwise (its only an answer and not a solution)
$$\sqrt{\ \frac{\ \nu _{o}^{2}}{2} +\sqrt{\left(\frac{\ \nu _{o}^{2}}{2}\right)^{2} +\frac{q^{2} \nu _{o}^{2}}{4\pi ^{2} d^{2} mC}}}$$
Ion traps are very complex, but one of my Physics Olympiad textbooks presents a simplified model of a resonating charged particle in an ion trap
A tuned circuit consists of an inductor and a parallel plate capacitor (capacitance C and plate separation d). It has a resonating frequency ##\nu...
Yes, sir, I understand that. However, I no longer want to use shell theorem, I want to mathematically conclude that any density/velocity profile of the cloud normalized to given conditions will end up in the same max radius, not relying on the force equations at the edge, hence I need a way to...
@jbriggs444 Thanks for the answer. Yeah, that seems understandable. So no compulsion in assuming that.
That aside, I tried my best to solve this using a general radial density variation rho(r) which is, let's say, dependant on the maximum radius of the cloud at any moment.
The density profile...
An asteroid of mass M explodes into a spherical homogenous cloud in free space. Due to energy received by the explosion, the cloud expands and the expansion is spherically symmetric. At an instant, when the radius of the cloud is R, all of its particles on the surface are observed receding...
I think I found the solution, the textbook directly suggested the use of PE=-M.B, however we know that
Del(PE)=M.B(1-cos(theta))=PEtheta - PE0
consider two separate loops of magnetic moment M1 and M2 (square and circle respectively)
Del(PE1)=M1.B(1-cos(theta))=PEtheta - PE0
and...
Thanks for that, although I wanted a deeper explanation as to why potential energy is still defined by the same way, P=M.B is found by integrating Torque =MxB w.r.t (theta), here though theta remains the same and M is changing, I have found an integral by mapping every elemental length on the...
How would you go about calculating the work done in morphing a square current-carrying loop into a circular current-carrying loop, without change in length while maintaining the same angular orientation with an external magnetic field.
My book suggests defining P(potential energy) = M.B (dot...