B How to use magnets in levitation?

  • B
  • Thread starter Thread starter woodturner
  • Start date Start date
AI Thread Summary
It is possible to float a wooden lid over a box using magnets without an electric power source, though it presents challenges. To achieve this, the rim of the box should be lined with north-up facing magnets, while the lid's edge should feature north-down facing magnets, creating a shallow magnetic bowl effect. This configuration helps maintain the lid's horizontal position with the assistance of gravity. However, the floating effect can be easily disrupted by misalignment or side forces. For added stability, making the lid circular and spinning it may enhance the levitation effect.
woodturner
Messages
1
Reaction score
0
How did you find PF?: google search

I wish to find out if its possible to float a wooden lid over a box using magnets. without using electric power source,if so how?
 
Physics news on Phys.org
It is possible - but tricky.

I would try lining the rim of the box with north-up facing magnets.
Then I would line the edge of the lid with north-down facing magnets.
The lid magnets should outline a slightly smaller area than the rim magnets - so when the lid is on, it is sitting in a shallow magnetic bowl. That way, gravity will assist in keeping the lid horizontally positioned.

The problem will be that the repulsive force keeping the lid floating can be readily disrupted by side forces created when everything isn't aligned well.
 
If I recall correctly you have to make the lid circular and spin it.
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Thread 'Does Poisson's equation hold due to vector potential cancellation?'
Imagine that two charged particles, with charge ##+q##, start at the origin and then move apart symmetrically on the ##+y## and ##-y## axes due to their electrostatic repulsion. The ##y##-component of the retarded Liénard-Wiechert vector potential at a point along the ##x##-axis due to the two charges is $$ \begin{eqnarray*} A_y&=&\frac{q\,[\dot{y}]_{\mathrm{ret}}}{4\pi\varepsilon_0 c^2[(x^2+y^2)^{1/2}+y\dot{y}/c]_{\mathrm{ret}}}\tag{1}\\...
Back
Top