Aurora Borealis

Imagine a magnetic charge as shown in the figure of strength m m , its magnetic field vector can be expressed as k r r 2 \dfrac {kr}{|r|^{2}} where r r is the radial position vector of any point from the magnetic charge.

An electric charge has been released in its vicinity at a distance of r o r_{o} with a radial velocity component v o v_{o} and some arbitary angular velocity w o w_{o} about the cone's axis and it now moves in the magnetic field of the magnetic charge.

It moves on the surface of a cone of half angle α \alpha with its vertex at the magnetic charge (Refer to figure above).

If its radial speed v v at any time can be related to other quantities as

( f ) v o ( e ) L g ( a ) m ( c ) sin ( b ) ( α ) [ 1 r o ( d ) 1 r ( d ) ] = ( f ) v ( e ) \displaystyle {(f) v_{o}}^{(e)} - \frac { L^{g}}{(a) m^{(c)} \sin^{(b)}(\alpha )} \left [ \frac {1}{{r_{o}}^{(d)}} - \frac {1}{r^{(d)}} \right ] = {(f) v^{(e)}}

For positive integers a , b , c , d , e , f , g a,b,c,d,e,f,g , evaluate 1 2 a b c d e f g \frac {1}{2} abcdefg

Details and Assumptions

L L is the Initial angular momentum about cones axis

m m is the mass of electrical charge

r r is the radial distance

v v is the radial velocity magnitude

v o v_{o} is the Initial radial velocity

you may want to try this problem first

D e t a i l s Details

  • A magnetic charge is simply a magnetic analogue of electrical charge, it produces a radial magnetic field and though its existence is uncertain. Assume it exists for this problem.

  • The force upon an electric charge in a magnetic field is given by q ( v × B ) q(v \times B)

  • The explicit independence of k k from the relation is not an error.

  • (Fun Fact) The beautiful northern lights or Aurora Borealis is the result of this phenomenon where charged particles funnel in like a cone into the north pole and keep moving in the surface of the cone, Though yes ours is a highly idealised and simplfied situation.


The answer is 16.

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1 solution

I am sorry @Mvs Saketh

The values of c , d , e c,d,e can be easily obtained as 2 2 by dimensional analysis. After that, it just takes one or two tries to get the answer as 16 16 .

If it wasn't chem day after, I would have surely solved this one.

You really excel at spotting short cuts dont you , but thanks, now i will change the question a bit to make it harder :) sure, try it later,

Mvs Saketh - 6 years, 3 months ago

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Remove the power of L. That is a dead give away.

Raghav Vaidyanathan - 6 years, 3 months ago

How This Charge is moving ? Does This is moving Helical Circular Path ?

Karan Shekhawat - 6 years, 3 months ago

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It is winding up helically with changing radius on the surface of cone but not uniformly, infact imagine it as climbing up non uniformly on a helical path and then crossing the magnetic charge and trace a similar cone on the other side and then again mirror(come) back in and repeat the process

Mvs Saketh - 6 years, 3 months ago

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@Mvs Saketh Thanks ! It is Beautiful Problem. Is it also Created by you ? Also Can You Please tell , does energy enylysis is better approach or Force anylysis ? Please Give some Hints . I want to analyse It , So please Gives only some Hints. Thanks Saketh Sir

Karan Shekhawat - 6 years, 3 months ago

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@Karan Shekhawat Dont sir me yet bro,, i was inspired to make it from the question of sir david which i have tagged, well here are the hints,

Angular momentum about cone's axis remains constant, since magnetic force doesnt provide any torque about it as you shall see if you analyse

Energy remains constant but that is not of so much use for the radial motion,

And yes force analsys is a must, but you must be very careful to not miss out any forces, as far as the radial motion is concerned, you only need the component ofcentripetal force along radial direction , you will find that the magnetic force has no component along radial direction as expected

Mvs Saketh - 6 years, 3 months ago

@Mvs Saketh are you sure b is +ve integer ? I'am getting b=-2,

I'am getting final answer as v r 2 = v o 2 L 2 sin 2 α m 2 ( 1 r o 2 1 r 2 ) \displaystyle{{ { v }_{ r } }^{ 2 }={ { v }_{ o } }^{ 2 }-\cfrac { { L }^{ 2 }\sin ^{ 2 }{ \alpha } }{ { m }^{ 2 } } \left( \cfrac { 1 }{ { { r }_{ o } }^{ 2 } } -\cfrac { 1 }{ { { r } }^{ 2 } } \right) }

Deepanshu Gupta - 6 years, 2 months ago

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