SidDash

SHM Phasors This post is only an extension to the previous article on AC Phasors. The reader is advised to read the article on AC Phasors here  first, to gain an elementary idea on phasors. In this article, we consider phasors used in describing Simple Harmonic Motion. Simple Harmonic Motion: The motion governed by a force proportional to the displacement of the particle from the mean position and directed towards the mean position is called simple harmonic motion. i.e. for an ideal particle of mass `m`, we have, ` F = -kx` i.e.   ` m \frac{\text{d}^{2}x}{\text{dt}^{2}} = -kx ` Suppose we have the mass particle `m` attached to a spring that applies force according to the above equation. We stretch it from the mean position to some distance say `a` and then release it.  The differential equation of this motion is then, ` \frac{\text{d}^{2}x}{\text{dt}^{2}} = - \frac{k}{m}x` Since both `k` and `m` are positive quantities, we can let `\frac{k}{m} = w^{2}`, for some `w...

Magnetic Moment of A Rotating Charged Rod Magnetic Moment: In general for any closed loop magnetic moment is given by formula `\vec{μ}= ni \vec{A}` , where ` \vec{A}` represents the area vector of the loop determined as usual by right hand thumb rule. Now, the primary problem for us is like this: Prob. A rod with charge density ` ρ = ρ_{0}\frac{x}{L} ` is rotated about its one end along an axis perpendicular to its length. Then find the magnetic moment of the setup. For a Revolving Charged Particle: Before going into the problem, it is important to realise that we have been given a charged rod, and not a simple wire loop wherein a current flows. However as the rod shall rotate, the charges will tranverse through space thus creating an apparent current. In any circular path, when a charged particle say q revolves it appears to create and average current,  `<i> = qf = \frac{q}{T} = \frac{qω}{2π}` Thus in that case,  ` μ = (\frac{qw}{2π})(π r^{2}) = \frac{qωr^{2}}{2}` F...