# Euler-Lagrange equations of a current-loop pendulum in a magnetic field

I am reading "Nonlinear Electromechanics", by Dmitry Skubov and Kamil S. Khodzhaev, Springer 2008. Here is the relevant and freely available chapter.

Essentially, a loop of area $$S$$, mass $$m$$, moment of inertia $$I$$, self inductance $$L$$, carrying current $$i$$ and rigidly connected to a pendulum bar of length $$l$$ is placed in a magnetic field $$B_0\sin\nu t$$. The Lagrangian of the system is:

$$L=T-V=\frac{1}{2}I\dot{\theta}^2-\frac{1}{2}Li^2-B_0Si\sin{\nu t}\sin\theta-mgl(1-\cos\theta)$$ where $$\theta$$ is the pendulum's deviation from the vertical. Euler-Lagrange equation is $$\frac{\partial L}{\partial q}-\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})=0.$$ If we take the generalized variables to be $$\theta$$ and $$i$$ then we should get: $$-B_0Si\sin{\nu t}\cos\theta-mgl\sin\theta-I\ddot{\theta}=0$$ $$Li+B_0S\sin{\nu t}\sin\theta=0$$ However, the equations in the book (2.2.2) are different from these two. The first differs by signs, while the second is completely different. What is wrong?

=== EDIT: Added the equations in question === $$I\ddot{\theta}-B_0Si\sin{\nu t}\cos\theta+mgl\sin\theta=0$$ $$L\dot i+B_0S\sin{\nu t}\cos\theta \dot \theta+B_0S\nu\cos{\nu t}\sin\theta+Ri=0$$

Recall that magnetic energies add to the Lagrangian while electric energies subtract from the Lagrangian--this is easily proven looking at the Lorentz force in the Lagrangian formalism. That is to say, the Lagrangian should be defined as $$\mathcal L=T-V+W_b-W_e$$ where $W_n$ are the magnetic ($b$) and electric ($e$) energies. Thus, your Lagrangian is $$\mathcal L=\frac12I\dot\theta^2-mgl\left(1-\cos\theta\right)+\frac12Li^2+B_0S\sin\nu t\sin\theta i$$ which takes care of your sign error in the first part.

The missing terms in your second equation likely comes from your dissipation function (deduced from the existence of the $\dot\theta$ and $R$ terms) that will probably take the form $$\Psi=\alpha i^2+\beta\dot\theta^2$$ It's not (yet) clear to me the origins of $\alpha$ and $\beta$ here (which are easily determined via direct differentiation), but hopefully you'll see that it's not simply the Lagrangian from classical mechanics applied to a new situation, there are some extra features.

The system of equations required here are called the Lagrange-Maxwell equations (defined in Section 1.2 of your book): \begin{align} \frac{d}{dt}\frac{\partial W}{\partial i}+\frac{\partial\Phi}{\partial i}+\frac{\partial V}{\partial g}&=E\\ \frac{d}{dt}\frac{\partial T}{\partial\dot\theta}-\frac{\partial\left(T+W\right)}{\partial \theta}+\frac{\partial\left(\Pi+V\right)}{\partial \theta}&=Q \end{align} where

• $W=\frac12Li^2$ is the self-inductance term
• $V=\frac12\sum_{j=1}^N\int \epsilon E^2dV$ is the energy of the electric field
• $\Phi=\frac12Ri^2$ is the thermal dissipation in the resistor
• $T$ is the kinetic energy
• $\Pi$ the potential energy
• $E$ the emf
• $Q$ the generalized forces

which is slightly different from the Lagrangian mechanics you've come to know.

In your case, $V=E=Q=0$, leaving \begin{align} \frac{d}{dt}\frac{\partial W}{\partial i}+\frac{\partial\Phi}{\partial i}&=0\\ \frac{d}{dt}\frac{\partial T}{\partial\dot\theta}-\frac{\partial\left(T+W\right)}{\partial \theta}+\frac{\partial\Pi}{\partial \theta}&=0 \end{align} Inserting the forms of $W$, $\Phi$, $T$, and $\Pi$ you have returns the form of 2.2.2 from your book (it does for me, at least).