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Yes there is and your age is completely irrelevant - study mathematics. For there is no theoretical physics without mathematics.What follows is designed to neither impress nor scare but to set up a realistic expectation of things to come before we give a concrete advice on what to do*. As a side effect, you get to benefit from the information provided by a person who can look in the rear view mirror.Colleges’ relevant curricula and accents or emphasis differ across space and time but to give you one, statistically insignificant, sample, in the mid 1980-ies I took the courses on Mathematical Analysis, Linear Algebra, Complex Analysis, Fourier Analysis, Functional Analysis, Calculus of Variations, Integral Equations, Differential Equations, Analytic Geometry, Differential Geometry, Theory of Probability, Combinatorial Analysis, Algebra, Group Theory, Statistics, Vector and Tensor Calculus, Equations of Mathematical Physics and so on.I am not sure whether I should include the computer sciency or computational courses here or not so just in case look up the (big red) book titled “Numerical Recipes in C” by Press, Teukolsky, Vetterling and Flannery for clues.When in college* do not take the traditional calculus - take the analysis, proofs-based, courses where proof is king. In my time a fundamental mathematical result could not be used unless it was previously proven in a public setting - during a lecture.To give you a taste, in my first, fall, semester we had a total of [math]96[/math] hours of just lectures on real analysis, plus the daily seminars where we discussed theory and solved problems.That means that my analysis professor read at least one lecture per day and at least once a week my day began with what in a student slang we called (in Russian) пара, meaning a pair, that is - a [math]45[/math]-minute proofs-based lecture, short break, another [math]45[/math]-minute lecture.Why mathematical proofs for theoretical physics majors. Because mathematical proofs are a combination of the rigid discipline of rigor and the fluidity or looseness of creativity. You will need both. Not necessarily in that order.Our mathematical curriculum ran at least one semester ahead of the physics curriculum (most of the time).All my math and physics exams were verbal - we were given a (set of) problem(s) and some time to solve it/them. When called, you would take a seat at the professor’s desk by the chalkboard for a quality face-to-face time and have a discussion.None of the teachers cared about the answer - they were interested in your reasoning, blow by blow. We want to see the only thing that matters - how you think. Not how uncle Bob thinks. Not how aunt Sally thinks. You:in a steady state a massless, absolutely rigid rod of length [math]l[/math][math][/math] is rotating in a horizontal plane at a constant angular velocity [math]\omega_0[/math] with a small bead tied to it [math]l_0[/math] units of distance away from the axis of rotation. At some point in time the thread holding the bead in place is cut and the bead begins to slide without friction off the rodFind: 1) the linear velocity of the bead relative to the rod when the bead jumps off the rod; 2) the separation time - how long will it take the bead to begin its free flightProfessor: Расказывайте (do tell).You: we will use the Lagrangian mechanics to solve this problem. Choose the plane of rotation to be such a position at which the system has zero potential energy ([math]U = 0[/math]):Since there is no friction in the system, its total kinetic energy [math]T[/math] is comprised of the kinetic energy of the bead’s rotation and the kinetic energy of the bead’s motion with respect to the rod (because the rod itself is massless and possesses no angular momentum of its own):[math]T = \dfrac{J\omega_0^2}{2} + \dfrac{mv^2}{2} \tag*{}[/math]where [math]J[/math] is the bead’s moment of inertia and [math]v[/math] is the bead’s linear velocity relative to the rod.If [math]x[/math] is the distance of the bead from the axis of rotation after the thread has been cut then its moment of inertia [math]J[/math] is:[math]J = mx^2 \tag*{}[/math]Since there is no friction in the system, the angular momentum is conserved which means that it must be the case that[math]\dfrac{ml_0^2\omega_0^2}{2} = \dfrac{mx^2\omega^2}{2} \tag*{}[/math]For the above equality to hold it follows that the quantity [math]x^2\omega[/math] must remain constant over time. In words: the square of the bead’s distance from the axis of rotation times the bead’s instantaneous angular velocity is constant.Which means that, in general, as [math]x[/math] grows [math]\omega[/math] must diminish and, left to its own devices, the system must eventually slow down.We, thus, have two options: 1) the bead’s angular velocity does remain constant or 2) the bead’s angular velocity diminishes.For the bead’s angular velocity to stay constant an external, properly calibrated, input must be provided to compensate for the loss (of the said angular velocity).Professor: Рассмотрите первый вариант (consider option one).You: in that case the Lagrangian of the system [math]L[/math][math][/math] is:[math]L[/math][math] = \dfrac{mx^2\omega_0^2}{2} + \dfrac{m\dot x^2}{2} \tag*{}[/math]where [math]x[/math] is the sole generalized coordinate that fully describes the (2-space) motion of the bead which, further, means that, since we have one equation per generalized coordinate, the Euler-Lagrangian equations in our case are:[math]\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot x} = \dfrac{\partial L}{\partial x} \tag*{}[/math][math]\dfrac{\partial L}{\partial \dot x} = m \dot x \tag*{}[/math][math]\dfrac{d(m \dot x)}{dt} = m \ddot x \tag*{}[/math][math]\dfrac{\partial L}{\partial x} = mx\omega_0^2 \tag*{}[/math][math]m \ddot x = mx\omega_0^2 \tag*{}[/math][math]\ddot x = \omega_0^2 x \tag{1}[/math]since the bead’s mass remains constant at all times.Professor: Как будете решать уравнение? (how will you be solving (1)?)You: by separating the variables - spell out the definition of the second derivative over time:[math]d\dot x = \omega_0^2 x\;dt \tag*{}[/math]Multiply both sides by [math]\dot x[/math]:[math]\dot x\;d\dot x = \omega_0^2 x\;dt\;\dot x = \omega_0^2 x\;dt\;\dfrac{dx}{dt} = \omega_0^2 xdx \tag*{}[/math]Integrate both sides collecting the constant of integration on one side:[math]\dot x^2 = \omega_0^2 x^2 + C \tag*{}[/math]Since initially the bead was at rest with respect to the rod:[math]x(t=0) = l_0, \; \dot x(t=0) = 0 \tag*{}[/math][math]C = - \omega_0^2l_0^2 \tag*{}[/math]Hence:[math]\dot x^2 = \omega_0^2 x^2 - \omega_0^2l_0^2 \tag*{}[/math]and[math]\dot x = \omega_0\sqrt{x^2 - l_0^2} \tag*{}[/math]Professor: Почему потеряли второй корень? (why did you lose the second root?)You: since the linear velocity grows over time, the derivative has to remain positive.Prepare the final separation of variables:[math]\dfrac{dx}{dt} = \omega_0\sqrt{x^2 - l_0^2} \tag*{}[/math][math]\dfrac{dx}{\sqrt{x^2 - l_0^2}} = \omega_0dt \tag*{}[/math]Integrate both sides: the left - from [math]l_0[/math] to [math]x[/math], the right - from [math]0[/math] to [math]t[/math]:[math]\displaystyle \int\limits_{l_0}^x\dfrac{dx}{\sqrt{x^2 - l_0^2}} = \omega_0t \tag*{}[/math]Professor: Как посчитаете интеграл? (how will you compute the integral?)You: via a hyperbolic substitution[math]x = l_0\cosh z, \; dx = l_0\sinh z\;dz \tag*{}[/math][math]\sqrt{x^2 - l_0^2} = l_0\sinh z \tag*{}[/math]Hence:[math]\displaystyle \int dz = z + C \tag*{}[/math]and, recalling the inverse function of [math]\cosh z[/math]:[math]z = \log\left(\dfrac{x}{l_0} \pm \sqrt{\dfrac{x^2}{l_0^2} - 1}\right) \tag*{}[/math]of which we take only the positive branch, since [math]x[/math] varies between [math]l_0>0[/math] and [math]+\infty[/math], we have:[math]\displaystyle \int dz = \log\left(\dfrac{x}{l_0} + \sqrt{\dfrac{x^2}{l_0^2} - 1}\right) + C = \tag*{}[/math][math]\log\left(x + \sqrt{x^2 - l_0^2}\right) + C' \tag*{}[/math]because we have absorbed the constant [math]\log l_0[/math] into [math]C’[/math]. Thus:[math]\log\left(x + \sqrt{x^2 - l_0^2}\right)\Bigg|_{l_0}^x = \tag*{}[/math][math]\log\left(x + \sqrt{x^2 - l_0^2}\right) - \log l_0 = \omega_0 t \tag*{}[/math]from where:[math]\dfrac{x}{l_0} + \sqrt{\dfrac{x^2}{l_0^2} - 1} = e^{\omega_0t} \tag{2}[/math]Getting rid of the radical, we have:[math]\dfrac{x^2}{l_0^2} - 1 = e^{2\omega_0t} - 2\;\dfrac{x}{l_0}\;e^{\omega_0t} + \dfrac{x^2}{l_0^2} \tag*{}[/math]from where:[math]2\;\dfrac{x}{l_0}\;e^{\omega_0t} = e^{2\omega_0t} + 1 \tag*{}[/math]and, hence:[math]x(t) = l_0\;\dfrac{e^{\omega_0t} + e^{-\omega_0t}}{2} \tag{3}[/math]Differentiating (3) over time once, we find the [math]\dot x(t)[/math]:[math]\dot x(t) = \omega_0l_0\;\dfrac{e^{\omega_0t} - e^{-\omega_0t}}{2} \tag{4}[/math]Differentiating (4) over time once, we find the [math]\ddot x(t)[/math]:[math]\ddot x(t) = \omega_0^2l_0\;\dfrac{e^{\omega_0t} + e^{-\omega_0t}}{2} \tag{5}[/math]To find the bead’s separation time [math]t_s[/math] put [math]x = [/math][math]l[/math][math][/math] in (2) and solve it for [math]t_s[/math]:[math]\dfrac{l}{l_0} + \sqrt{\dfrac{l^2}{l_0^2} - 1} = e^{\omega_0t_s} \tag*{}[/math]from where:[math]t_s = \dfrac{1}{\omega_0}\;\log\left(\dfrac{l}{l_0} + \sqrt{\dfrac{l^2}{l_0^2} - 1}\right) \tag{6}[/math]We can find the separation velocity [math]v_s[/math] either the hard or the easy way.The hard way would be inserting [math]t_s[/math] from (6) into (4) and solving the result for [math]v_s[/math]:[math]\dot x(t_s) = v_s = \omega_0l_0\;\dfrac{e^{\omega_0t_s} - e^{-\omega_0t_s}}{2} \tag*{}[/math](and, multiplying something by a conjugate, you are encouraged to do just that to prove to yourself that you will get the same result)As a budding theoretical physicist, however, you are expected to be wily and do it the easy way and that is why, among other reasons, we advertise the blow by blow process so much - we go back to our intermediate relationship which we have established during the process of separation of variables:[math]\dot x = \omega_0\sqrt{x^2 - l_0^2} \tag*{}[/math]and put [math]x = [/math][math]l[/math][math][/math] into it:[math]v_s = \omega_0\sqrt{l^2 - l_0^2} \tag{7}[/math]Professor: Можете ли Вы дать геометрическую интерпретацию данного движения? (can you give a geometric interpretation of this motion?)You (probably drawing on the chalkboard): the bead’s motion is the superposition of two independent motions - rotation and sliding. Thus, the bead either rolides or slitates (or feel free to coin a better term) in such a way that if in the current instance its distance from the axis of rotation is a side of a right triangle then in the next instance that distance plays the role of the hypotenuse and so on:which results in the bead tracing a logarithmic spiral in the fixed plane. The vector sum of the above radial and perpendicular (to the rod) velocities is, of course, tangential to the spiral.*As you can see, you are expected to be in a firm command of the mathematical apparatus known by you at the moment and you will spend the rest of your life expanding that apparatus and, who knows, may be inventing a new one.Start studying mathematical proofs from the ground up, little by little, do not get ahead of yourself. Intellectual maturity is more important then an artificially fast reached milestone. This is not speed dating. Elementary number theory, elementary combinatorics, Euclidean geometry are some sample starting points.Right now attend a mathematics and a physics circle or its equivalent - a specialized school that offers these two subjects only and covers both in depth. In the lack thereof - find a mentor.Later on enroll into a college that offers the major you are interested in - so far there is no substitute to getting your ass systematically kicked into shape, academically speaking.

TL;DR answer: the inductor is one of the four fundamental circuit elements.I’ll answer under a theoretical perspective, just forgetting how in the real world the devices (inductors, capacitors, etc…) are “created”. The practical side has already been tackled by previous answers.There are four fundamental electrical variables:The voltage [math]v(t)[/math]The current [math]i(t)[/math]The charge [math]q(t)[/math]And the flux [math]\phi(t)[/math]The resistor relates voltage and current[math]v(t)=R i(t)[/math]The capacitor relates charge and current[math]i(t)=\frac{dq(t)}{dt}[/math]The inductor relates flux and voltage[math]v(t)=\frac{d\phi(t)}{dt}[/math]and finally, there is a (almost) missing circuit element called memristor that relates charge and flux.[math]\phi(t)=M q(t)[/math]These relations are organized in a beautiful picture (taken from http://www.maisfisica.com/tag/memristor-o-que-e-um-memristor-semicondutor-memristencia/).Leon Chua proposed the memristor as the “missing circuit element” in a theoretical paper in 1971 (L. O. Chua, “Memristor-the missing circuit element,” IEEE Trans. Circuit Theory, vol. 18, pp. 507 – 519, Sep. 1971. Can be downloaded in http://www.cpmt.org/scv/meetings/chua.pdf) At the time almost no (real) evidence of a physical memristor was known (if my memory is not tricking me, I read once that at the time the memristor could be associated to some electro-chemical processes observed in some ionic solutions). An artificial memristor was also fabricated then with the help of opamps. It was only in 2008 that the fabrication of a real memristor was released in public (giving a boost to HP shares at the time) and nowadays is used in the fabrication of memories. To see details on memristors see the references in an abstract of one of Chua’s talks (http://sti.epfl.ch/files/content/sites/sti/files/shared/sel/pdf/Abstract_Prof_Chua.pdf).For linear inductors and capacitors we have the auxiliary equations[math]q(t) = C v(t) \qquad \phi(t) = [/math][math]L[/math][math] i(t)[/math]and so we arrive at the commonly displayed equations for those elements:[math]i(t)=C \frac{dv(t)}{dt}[/math] (in the capacitor)[math]v(t)=L \frac{di(t)}{dt}[/math] (in the inductor)So, finally, let’s answer the question. I hope it is clear now that the inductor is one of the four fundamental circuit elements which relates the flux and the current, two of the four fundamental circuit variables. If the inductor was missing, then one piece of the “Physical Universe” wouldn’t be explained (it is a bit exagerated, but…). Moreover, there are many practical applications of inductors in circuits where this device cannot be replaced by others, as other answers already mentioned.

If you carefully examine the data from the questions being asked on Quora on a daily basis you will eventually come to the realization that almost everything about quantum physics is misunderstood .It’s not that surprising though. Quantum physics diverges from your classical intuition and unless you spend a lot of time towards developing a quantum mechanical intuition, you will never understand it and even if you do spend a lot of time doing that, you will find yourself feeling perplexed a number of times.Eventually, at some point, some things will feel almost natural to you, therefore you won’t waste a lot of time thinking about how quantum mechanics fits in your classical perception of the world. I think the moment you realize that quantum physics is not a small scale approximation of classical physics but rather it’s the other way around, then you are on the right track towards understanding quantum physics.For the general public it’s even harder (compared to physicists or mathematicians) because, well, most people do not even know classical physics. If you don’t understand how the tangible world that you actually experience through your senses works, it is nearly impossible to understand the quantum behavior of nature. You cannot build a house by starting with the roof - you have to dig a hole and set the foundations. The same analogy holds for physics.In addition to that, consider the fact that quantum mechanics utilizes several mathematical tools, many of which are quite advanced. Most people have (at best) a high school level knowledge of mathematics and while this might be more than enough for other cases, for quantum physics it is not. This means that people who know neither physics nor mathematics, are at a great disadvantage.I mean, let’s say you want to calculate the path followed by a particle going from point [math]a[/math] to point [math]b[/math]. One of the methodologies that can be used is path integrals. Feynman showed that the amplitude for going from [math]a[/math] to [math]b[/math] is given by the kernel [math]K(b,a)[/math] which is:[math]K(b,a)=\displaystyle\int^b_a e^{(i/\hbar)S[a,b]}\mathcal Dx(t)\tag* {}[/math]where [math]S[a,b][/math] is the classical action given by:[math] S[a,b] = \displaystyle\int^{t_b}_{t_a}L(\dot x,x,t)dt\tag* {}[/math]and [math]\mathcal Dx(t)[/math] denotes integration over all possible paths.This is crazy!It basically means that you “just” write down all the possible (which is an infinite number of) paths the particle can follow, each one with a certain probability corresponding to it and if you add them all up you will figure out what actually happens! This is crazy both intuitively and computationally. After all you will have to calculate something that looks like this:[math]K(b,a)=\displaystyle\int^b_a e^{(i/\hbar)\int^{t_b}_{t_a}L(\dot x,x,t)dt}\mathcal Dx(t)\tag* {}[/math]However, despite the counter-intuitive nature of the concept leading to the general expression above, as it turns out this is precisely true!As a kicker, add the fact that people who lack the necessary ingredients to understand quantum physics usually resort to reading pop-sci versions of quantum physics. This might be a good choice if all you want is a superficial “understanding” of quantum theory but if you actually have questions that have to be answered, I’m afraid there’s no shortcut and popularized quantum physics only causes further confusion after that superficial level.If you have questions that you feel that you have to know the answer, unfortunately, you will have to put in the effort. Nevertheless, I promise you, it is going to be worthwhile!