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The problem statement: "A conducting ring of radius 'r' is held stationary in a magnetic field $\vec{B}$ which, initially has a magnitude of $B_0$ and is inclined at $45^\text{o}$ to the area vector $\vec{A}$. At $t=0$, the supports holding the ring stationary are removed, and the $\vec{B}$ field starts to increase at a constant rate $\frac{dB}{dt}$. The direction of field is always along the initial direction. Find the linear momentum of the ring's centre of mass, and the angular momentum of the ring about its centre of mass

(i) as a function of $\theta$, a parameter dependent on time, defined as the angle between the vectors $\vec{A}$ and $\vec{B}$,

(ii) as a function of time, an independent parameter "

My attempt:

The torque about the centre of mass is given $\vec{\Gamma}$ = $I\vec{A}\times\vec{B}$ where I is the instantaneous current in the loop.

I work this current to be $$I = \frac{1}{R}\frac{d(\vec{A}\cdot \vec{B})}{dt}$$ where $R$ is the resistance of the loop (which the question surprisingly doesn't define).

The consolidated expression of the torque as a function of theta and time (taking magnitudes as well) gives, $$\lvert{\Gamma}\rvert = \frac{\left\lvert \vec{A} \right\rvert^2{\left\lvert \vec{B} \right\rvert^2}}{R}\sin\theta \cos\theta\frac{d\theta}{dt}$$

Here is where I seem to get stuck, since the induced magnetic field in the ring, will produce a field that seems very hard to work with.

So I looked to restructure the problem through flux, instead of the field distribution. Defining instantaneous flux through the loop as $\phi = LI$, where $I$ is the instantaneous current and $L$ is the self-inductance of the loop, $$L = \frac{\mu_o\pi r}{2}$$

This gives the flux $$\phi = \frac{1}{R} \frac{\mu_o\pi r}{2} \frac{d\phi}{dt}$$

I'm not sure I can solve this differential equation directly, since the flux itself is likely a function in $t$.

Again the problem seems to dissolve into finding the flux through a loop for an electromagnetically induced field, where the field itself is not distributed uniformly, and I hypothesize this field is proportional to $\sqrt{x^2+y^2}$, $x$ and $y$ being natural coordinates of any point in the plane of the loop.

I thought of parametrizing this into polar coordinates in the plane of the loop, then integrating the instantaneous field over the surface area, but the field itself is dependent on the instantaneous flux. All of the variables seem very dependent on each other, and it seems impossible to find a non-recurrent relation to $t$ or $\theta$.

Here, I was pretty skeptical of my initial approach; so assuming I obtain a closed form expression of the flux as a function of time (or $\theta$), it would then be necessary to differentiate this to use in an expression for torque at any time, and the integral of this would give the change in angular momentum. The procedure feels pretty apparent, however hard to apply.

Could an energy-based approach be used to find the angular momentum, since the initial and final conditions are way easier to solve for? I decided against this in the presence of non-conservative fields, (though this probably won't make a difference in mechanical energy, but equating the two seems to be the core of this approach, and that doesn't feel physically right.)

As for the linear momentum of the centre of mass, is that just going to be equal to zero? I fail to see, qualitatively, how any unbalanced forces would act on the loop as $d\vec{l} \times \vec{B}$ around a loop should come out to be $0$.

I also read the answers through Motion in a time-dependent uniform magnetic field The question there seems to deal with a very general description of motion of a charged particle in a time-dependent magnetic field, and the answers invoke math that I don't know how to interpret. Due to this lack of knowledge of mathematical methods, it is perhaps better for me to know a qualitative analysis of the situation; though I'm very curious to know the path to the solution as well.

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  • $\begingroup$ Just a minor comment, that differential equation is of the form $\phi = \lambda \dot{\phi}$ which is straightforward to solve as $\phi = \phi(0)e^{\lambda t}$. $\endgroup$
    – Triatticus
    Aug 16, 2022 at 17:46
  • $\begingroup$ But as I said, shouldn't the flux be a function itself dependent on time? Doesn't this integration assumes that the flux is an independent variable? $\endgroup$
    – DVnyT
    Aug 17, 2022 at 9:01
  • $\begingroup$ A differential equation is an equation between a function of a dependant variable and at least one derivative of the function with respect to that dependant variable. If you haven't taken a class on differential equations yet this might be a point of confusion here $\endgroup$
    – Triatticus
    Aug 17, 2022 at 14:39

1 Answer 1

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I think your calculation of the induced emf is wrong. e(t) has two components, an external one due to the variation of the flux of the external magnetic field and an internal one generated by the variation of the magnetic field created by the current I(t). The equations you have to solve are:

$$ \tau = \frac{dL}{dt} $$ $$ L= \frac{ m A}{2 \pi } \frac{ d\theta(t) }{d t }$$

$$ F=\frac{dP}{dt}$$ $$ \overrightarrow{F}=- \overrightarrow{ \nabla }\big(- \overrightarrow{ \mu }. \overrightarrow{B} \big)= \overrightarrow{0}$$

$$\tau = I(t).A.B(t).sin( \theta(t) ) $$ $$ B(t)= \big(\frac{dB}{dt}\big)_{0} .t+ B_{0}$$

$$ e(t)=R.I(t)$$ $$ e(t)=- \frac{d(B(t).A.cos( \theta (t))}{dt} -L_{m}. \frac{dI(t)}{dt}$$

Where $L_{m}$ is the inductance of the loop.The formula you have used for the inductance of a circular loop uses the magnetic field at the center of the loop and assumes it is constant throughout the loop. Since it is a crude approximation and since $ 4 \pi \mu _{0} \simeq 10^{-6} $, you can try to solve your problem by iteration.

The good news: It is easy to get the linear momentum P. $$P=0$$ The bad news: You get a coupled system of differential equations for the angle $\theta(t)$ and I(t). $$\begin{cases}I(t).t.sin( \theta(t) )= \frac{m}{2 \pi.B'_{0} }\frac{ d^{ 2 }\theta(t) }{d t^{2} } \\L_{m}. \frac{dI(t)}{dt}+R.I(t)=- \frac{d(B(t).A.cos( \theta (t))}{dt} \end{cases}$$

If one neglects the self induced current, my equations differ from yours due to the sign of the emf.But you can forget about getting an analytical solution without making approximations.

$$ e=-\frac{d \phi }{dt} $$ It seems you have forgotten the negative sign.

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  • $\begingroup$ But aren't both the current and the angle time dependent themselves? Not to mention that the induced current would oppose the change in magnetic field through it. How would you reduce the entire right hand side to an expression of time. I don't think they're asking for a simple relationship, where the instantaneous current through the loop can be left as is. $\endgroup$
    – DVnyT
    Aug 17, 2022 at 8:58
  • $\begingroup$ \cdot will give $\cdot$ $\endgroup$
    – Kuhlambo
    Aug 17, 2022 at 12:38

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