B Magnetic pendulum and electric energy....

AI Thread Summary
A pendulum oscillating inside a coil connected to a galvanometer can generate electricity due to Faraday's law, but this does not violate the conservation of energy. The initial potential energy of the pendulum is converted into kinetic energy, which dissipates through friction and air resistance, along with energy lost as heat due to induced current in the coil's resistance. Lenz's law plays a crucial role, as it creates a counteracting force that slows the pendulum down, leading to damped harmonic motion. The energy generated is not free; it comes from the energy initially supplied to the pendulum. Thus, while electricity is generated, it adheres to the principles of energy conservation.
KedarMhaswade
Messages
35
Reaction score
6
TL;DR Summary
A simple pendulum oscillating inside a coil obviously does not generate electricity, or does it?
While reading about electromagnetism from the OpenStax books with my son (and doing some experiments), he asked this question.

Suppose I hang a pendulum and make it oscillate inside a coil connected to a Galvanometer as shown in the schematic diagram:

1643125804531.png


Hopefully the image is clear enough. His argument is that the analysis of the pendulum from Newtonian mechanics suggests that the initial potential energy that I provide to it (##mgh##) is converted into its kinetic energy (##\frac{1}{2} mv^2##). Then the energy dissipates through losses as pendulum eventually comes to a standstill. Thus, the energy I provide in raising the magnet initially is finally lost in friction or to the surrounding air.

But if the magnet is made to oscillate inside a coil, according to Faraday's law, electricity is generated in the coil as the Galvanometer needle shows the deflection. The pendulum does not know that it is made to oscillate inside a coil and it ends up inducing electromotive force in it (as a mere side-effect).

Have I not generated electrical energy violating the law of conservation of energy?

How will you refute this claim?

Does Lenz's law come into the picture and slow down the magnet (since ##F = -\frac{d\phi}{dt}##) which would not have happened had the magnet pendulum not oscillated inside the coil?
 
Physics news on Phys.org
KedarMhaswade said:
Summary:: A simple pendulum oscillating inside a coil obviously does not generate electricity, or does it?

Does Lenz's law come into the picture and slow down the magnet (since F=−dϕdt) which would not have happened had the magnet pendulum not oscillated inside the coil?
Absolutely. The induced current in the coil generates heat because the coil has non-zero resistance. That's where the energy goes. The magnet will execute damped harmonic motion and come to a stop regardless of air resistance. You may wish to read about eddy current braking here.
 
  • Like
  • Informative
Likes vanhees71, russ_watters, PeroK and 1 other person
KedarMhaswade said:
The pendulum does not know that it is made to oscillate inside a coil
Putting something in bold doesn't make it true. There is a force exerted on the pendulum with the coil that does not exist without the coil. So in the sense that the pendulum "knows" anything it does know that it is made to oscillate inside a coil.

KedarMhaswade said:
Have I not generated electrical energy violating the law of conservation of energy?

How will you refute this claim?
By pointing out the recognized fact that it violates the conservation of energy so it must be wrong.
 
Dale said:
Putting something in bold doesn't make it true.
True ;-). Wouldn't the button say "True" otherwise?
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top