Charged fractal in a magnetic field

Electricity and Magnetism Level 5

We generate our fractal as follows:

A stick of total length $L$ is taken. A certain amount of charge is uniformly distributed along it's length.

Now, one third of the stick is removed from the middle. We are now left with two sticks which are one third the size of the original stick. Remove one third from the middle, from these two sticks...and from the smaller sticks left, ad infinitum.

What we are left with looks like the figure. Now, after these infinite iterations, let the fractal that we get have a mass $M$ and a total charge $Q$ residing on it.

The fractal is spun with an angular velocity $\omega$ about the center of the original stick.

A magnetic field of intensity $B$ exists parallel to the axis of rotation.

Let the magnitude of rotational kinetic energy associated with the fractal be $A$ and the magnitude of it's magnetic potential energy be $B$ .

Then find $A+B$ in joules.

Details and Assumptions

$M=12 \text{ kg}, Q=6 C, B = 1 \text{ T}, \omega = 9 \text{ rad/s}, L = 7 \text {m}$
Assume the standard convention for magnetic potential energy (meaning, for what configuration the P.E is taken to be 0).

Looking for a challenge in optics? Try this

The answer is 3142.125.

1 solution

Вук Радовић
Feb 3, 2015

The only tricky part here is finding a moment inertia of our fractal. So lets say that $I_O=kML^2$ where O is center of mass of whole system.

Then moment of inertia of left and right side separately around point O is $I=I_A+m(L/3)^2$ where A is center of mass of left side.

Then $I_A=k\frac{M}{2}\frac{L}{3}^2$ .

And know we know that $I_O=2I$ . Solving this equation you will find that

$k=\frac{1}{8}$ .

Using gyromagnetic ratio formula you know that

$p_m=\frac{Q}{2L}I_O\omega$ .

$E_k=I_O\frac{\omega}{2}$

$W=p_mB$

Good work :) ...The only thing left to do is mention why the gyromagnetic ratio is valid here(easy enough).

Shashwat Shukla - 6 years, 4 months ago

Thanks ! Yeah, it's easier part.

Вук Радовић - 6 years, 4 months ago

can you please explain ..... gyromagnetic ratio ?? @Shashwat Shukla

Karan Shekhawat - 6 years, 2 months ago

Very sorry for the late reply, I was busy with JEE Mains and I hadn't seen your comment.

Well, it states that $\frac{M}{L}=\frac{q}{2m}$

where M is the magnetic moment of the body, L is it's angular momentum, q is the charge on the body and m is it's mass.

Now, the thing is, this concept is applicable only for $spherically \quad symetric$ bodies like cylinders and spheres but not cubes and cuboids.

Just as important is the fact that it holds only for bodies with a $uniform$ mass and charge distribution.

It's easy to see why:

1)Prove this result for a ring.

2)Hence prove that this result holds for any body that can be broken up into rings. A disc is composed of rings and 3d objects like spheres are made of discs.

That's about it, I guess :)

@Karan Shekhawat

Shashwat Shukla - 6 years, 2 months ago

@Shashwat Shukla – Thank you very much for explaining ! Also don't you think in above solution It should be $E=Iw^2/2$ instead of $E=Iw/2$ Byk 's solution !

Karan Shekhawat - 6 years, 2 months ago

@Karan Shekhawat – You're welcome :)

And yes, it should be $E=Iw^2/2$ .

Shashwat Shukla - 6 years, 2 months ago

I believe this should be level 4, maybe. Only 3% people could solve it.

Aalap Shah - 6 years, 2 months ago

Yes, I can't understand it either.

Shashwat Shukla - 6 years, 2 months ago

Charged fractal in a magnetic field

Looking for a challenge in optics? Try this

The answer is 3142.125.

1 solution

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