Hello everyone,

 

I thought about this problem just for a while and couldn't exactly figure out what would happen. I don't have the necessary equipment, so can't perform the experiment.

 

Suppose a wire attached to a metallic ball hangs, such that it is free to move to and fro [see figure]. A conducting base is provided below the ball, such that the surface of the base just touches the ball, providing negligible friction. There are two concave magnets providing a uniform magnetic field to the ball. The ball and the wire and the base all are connected to a circuit with AC supply. So, as soon as the key is pressed, the current flows. Accordingly, a force will cause the pendulum ( or rather the ball) to move in a direction perpendicular to the magnetic field. With the alternation of the direction of current, the direction of motion will change too.

My question is what will happen to the system. Will the ball show oscillatory motion with the change in the direction of current? Or will it collapse due to high impedance? Also, please write the equation of motion for the above system.

 

 

 

Please see that I haven't made the image turn. I hope it's still as clear and understandable.

Edited April 16, 20178 yr by Sriman Dutta

Why concave magnets?

 

I guess the result would be a driven harmonic oscillator.

https://en.m.wikipedia.org/wiki/Harmonic_oscillator

And the equation?

 

Those would be needlessly complex with the inhomoheneous magnetic field between the concave magnets.

 

In general, you can just use those found on Wikipedia. More specific can be found under "sinusoidal driving force" and "simple pendulum".

Edited April 17, 20178 yr by Bender

I figured out this:

[math] m\frac{d^2\theta}{dt^2}X l = -mgsin\theta + \int_L I_0 sin \omega t dl X B [/math]

Edited April 17, 20178 yr by Sriman Dutta

The sine in sin(theta) is usually left out, which is a good approximation for small angles.

The B should be in the integral because the field is not homogeneous. For simplicity, you could replace the integral with F_0 sin(omega t)

Yes. The sine theta part is left out often yo make it linear.

But why should B be in integral? We are taking the summation of each current that flows through each infinitesimal length element of the wire.

No, we sum the force on each length of the wire, and B is different over each length of the wire.

In the given setup, it would be quite complex to calculate B.

Edited April 17, 20178 yr by Bender

Oh I see!

The force on any wire is given by

[math]F = I \int_L dl X B [/math]

Edited April 18, 20178 yr by Sriman Dutta