Trajectory of a Particle in a Magnetic Field

Here is a problem shared with me.

At any instant of time, let the particle's position and velocity be:

r=x i^+y j^\vec{r} = x \ \hat{i} + y \ \hat{j} v=vx i^+vy j^\vec{v} = v_x \ \hat{i} + v_y \ \hat{j}

Applying Newton's second law at this instant:

ma=m(v˙x i^+v˙y j^)=q(v×B)kvm\vec{a} =m(\dot{v}_x \ \hat{i} + \dot{v}_y \ \hat{j})= q(\vec{v}\times \vec{B}) - k \vec{v}

Simplifying gives:

mv˙x=qBvykvx (1)m\dot{v}_x = -qBv_y -kv_x \ \dots(1) mv˙y=qBvxkvy (2)m\dot{v}_y = qBv_x-kv_y \ \dots(2)

Let:

a=kma = \frac{k}{m} b=qBmb = \frac{qB}{m}

The differential equations become:

v˙x=avxbvy (3)\dot{v}_x = -av_x -bv_y \ \dots(3) v˙y=bvxavy (4)\dot{v}_y = bv_x - av_y \ \dots(4)

vx(0)=u ; vy(0)=0v_x(0)=u \ ; \ v_y(0) = 0

Differentiating (3) wrt. time:

v¨x=av˙xbv˙y\ddot{v}_x = -a\dot{v}_x -b\dot{v}_y

Plugging in v˙y\dot{v}_y above gives:     v¨x=av˙xb(bvxavy)\implies \ddot{v}_x = -a\dot{v}_x - b (bv_x - av_y )     v¨x=av˙xb2vx+abvy (5)\implies \ddot{v}_x = -a\dot{v}_x - b^2 v_x + abv_y \ \dots(5)

Again, looking at (3):

bvy=v˙xavxbv_y = - \dot{v}_x - av_x Plugging the above in (5) and simplifying gives:

v¨x+2av˙x+(a2+b2)vx=0\ddot{v}_x + 2a\dot{v}_x + (a^2+b^2)v_x=0

A similar operation can be performed and the following equation can be obtained: v¨y+2av˙y+(a2+b2)vy=0\ddot{v}_y + 2a\dot{v}_y + (a^2+b^2)v_y=0

vx(0)=u ; v˙x(0)=au ; vy(0)=0 ; v˙y(0)=buv_x(0)=u \ ; \ \dot{v}_x(0) = -au \ ; \ v_y(0)=0 \ ; \ \dot{v}_y(0) = bu

Let:

2a=p2a = p a2+b2=qa^2 + b^2 = q

This implies:

v¨x+pv˙x+qvx=0\ddot{v}_x + p\dot{v}_x + qv_x=0

The roots of the above characteristic equation are:

k1=p+p24q2=a+ibk_1 = \frac{-p+\sqrt{p^2-4q}}{2} = -a + ib k2=pp24q2=aibk_2 = \frac{-p-\sqrt{p^2-4q}}{2}=-a-ib

Where i=1i= \sqrt{-1}

Finally, the general solution reads:

vx=eat(c1sinbt+c2cosbt)v_x = \mathrm{e}^{-at} \left(c_1 \sin{bt} + c_2 \cos{bt}\right) vy=eat(c3sinbt+c4cosbt)v_y = \mathrm{e}^{-at} \left(c_3 \sin{bt} + c_4 \cos{bt}\right)

The above constants c1,2,3,4c_{1,2,3,4} can be obtained by applying initial conditions.

Know that:

x(0)=y(0)=0x(0) = y(0) =0

Integrating the velocities will give the displacements along each direction. It can be seen that as time increases, the particle eventually comes to rest. So the displacement when the particle comes to rest can be found as such:

xD=limtx(t)x_D = \lim_{t \to \infty} x(t) yD=limty(t)y_D = \lim_{t \to \infty} y(t)

The required answer is, therefore:

D=xD2+yD2D = \sqrt{x_D^2 + y_D^2}

I have not done all the calculations as doing so will take me time. I have laid out the steps to solve this problem.


Edit: I have solved for the particle's motion numerically with the following parameters:

B=q=m=u=1B = q=m = u=1

The trajectory is plotted for various kk as follows. One can see that when k=0k=0 one gets the classical circular motion result. For the other cases, the motion is a stable spiral trajectory.

Note by Karan Chatrath
1 week, 5 days ago

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@Karan Chatrath

Lil Doug - 1 week, 5 days ago

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@Karan Chatrath will the particle come to at the rest after t=t=\infty ?

Lil Doug - 1 week, 5 days ago

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Yes, the particle will come to rest as tt \to \infty since the expression for the velocity components has a decaying exponent.

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath Using two initial conditions , how can we find 44 constants?

Lil Doug - 1 week, 5 days ago

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vx(0)=u ; v˙x(0)=Au ; vy(0)=0 ; v˙y(0)=Buv_x(0)=u \ ; \ \dot{v}_x(0) = -Au \ ; \ v_y(0)=0 \ ; \ \dot{v}_y(0) = Bu

You missed the above line in the note, so I am adding it here too.

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath oh yess sorry
If you don't mind we should not write it as B, because there are two B which leads to confusion.

Lil Doug - 1 week, 5 days ago

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@Lil Doug That is a good point. I have edited the note

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath after integration of velocity , 2 more constants are waiting us.

Lil Doug - 1 week, 5 days ago

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x(0)=y(0)=0x(0) = y(0) =0

You missed this line too, so I am adding it here. Use these two conditions to solve for the constants.

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath That really interesting.
I wish I could see particle whole motion in graph.

Lil Doug - 1 week, 5 days ago

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I will update this note after some time. But if you put k=0    a=0k=0 \implies a=0 the particle moves in a circle as expected.

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath acha to fir magnetic force aur drag force barabar ho jaayega to particle ruk jayega .
Right?, aisa aisa hua to bahut interesting ho jaayega

Lil Doug - 1 week, 5 days ago

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@Karan Chatrath if you don't mind,if it takes very much time. To write all thing in latex, which maybe a waste of time, then I suggest you to use pen and page.

Lil Doug - 1 week, 5 days ago

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@Karan Chatrath After doing calculations I am getting c1=0c_{1}=0 c2=uc_{2}=u c3=uc_{3}=u c4=0c_{4}=0

Lil Doug - 1 week, 5 days ago

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I have not checked this myself. But from a dimensional standpoint, the result seems okay.

I have edited the note by adding some plots. Hope this helps.

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath after that

Lil Doug - 1 week, 5 days ago

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Based on this, as tt \to \infty:

xD=aua2+b2x_D = \frac{au}{a^2 + b^2} yD=bua2+b2y_D = \frac{bu}{a^2 + b^2}

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath and yes the answer is D=kmuq2B2+k2i+qBmuq2B2+k2j\Large \vec{D}=\frac{kmu}{q^{2}B^{2}+k^{2}} \vec{i}+ \frac{qBmu}{ q^{2}B^{2}+k^{2} } \vec{j}
Greetings we just rocked!
But your contribution is 95% :)

Lil Doug - 1 week, 5 days ago

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Well done! Glad that I could help

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath You help always. Love you sir .
But this was very difficult, what do you think.

Lil Doug - 1 week, 5 days ago

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@Lil Doug Solving by hand makes this a difficult problem. It is easy to make calculation mistakes

Karan Chatrath - 1 week, 5 days ago

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@Karan Chatrath

I was looking at phase portraits earlier; thanks for this one!

Krishna Karthik - 1 week, 5 days ago

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The plots that I have shared are not phase portraits.

Karan Chatrath - 1 week, 4 days ago

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Yeah; I know that; these are trajectories, but they look like phase portraits of pendulums with drag.

Krishna Karthik - 6 days, 13 hours ago

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@Karan Chatrath Yeah it worked finally.
The difference between your and my method is that , by your method we can find location of particle at any time t, and in my method I have just focused on where the particle stops.
Reply if you found it good.

Lil Doug - 1 week, 2 days ago

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Yes, this is a faster way of solving the problem. Well done.

Karan Chatrath - 1 week, 2 days ago

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@Lil Doug By the way, this is the same problem as the damping one you wanted me to solve, just with a little different final quantity to be solved for. Latex is being very buggy in the notes section, and I just lost a bunch of work trying to post the solution.

Steven Chase - 1 week ago

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@Steven Chase No problem :)

Lil Doug - 1 week ago

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@Steven Chase I am not able to understand only that last line (velocity vector turns my 2π
Can you help me to understand that line.
Thanks in advance.
Hope I am not disturbing you.

Lil Doug - 1 week ago

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The motion is described in terms of sinusoids of a certain frequency. You can find out what the time period is for the sinusoids, and evaluate the motion up until one time period has elapsed.

Steven Chase - 1 week ago

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@Steven Chase @Steven Chase sir i think the problem is whole same, except last line.
I have provided x(t)x(t) and y(t)y(t)

Can you do something from this data? Thanks in advance. Hope I am not disturbing you

Lil Doug - 1 week ago

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@Karan Chatrath Try this problem.

Lil Doug - 6 days, 15 hours ago

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@Karan Chatrath

Lil Doug - 6 days, 15 hours ago

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I will try it, but I want to see your attempt before I post mine. Also, I posted a report on your problem titled 'Spaceship Mechanics'. Could you please get back to me on that?

Karan Chatrath - 6 days, 14 hours ago

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@Karan Chatrath Yes sir, sorry I didn't reply you for that report.
The question is basically copied through the book, I checked it many times, there is nothing more than that given in question.
Can you please try to think it by seeing the graph.

Lil Doug - 6 days, 14 hours ago

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@Lil Doug t0t_0 I think is the time when the engine is turned off

Krishna Karthik - 6 days, 14 hours ago

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@Karan Chatrath

About your report, t0t_0 is when the engine is turned off, I think.

Krishna Karthik - 6 days, 14 hours ago

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Is the answer the following?

D=mvob(1e2πbqB)D = \frac{mv_o}{b}\left(1 - \mathrm{e}^{-\frac{2 \pi b}{qB}}\right)

Karan Chatrath - 6 days, 14 hours ago

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@Karan Chatrath yes it's CORRECT\Huge CORRECT

Lil Doug - 6 days, 14 hours ago

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@Lil Doug @Lil Doug Lol that's a funny way of saying correct. lmao. You gave me a fright.🤣

Krishna Karthik - 6 days, 14 hours ago

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@Krishna Karthik @Krishna Karthik he is my favorite teacher.
He is just awesome bro.
Therefore I done that. His skills of solving problem are better than even Einstein sir.

Lil Doug - 6 days, 14 hours ago

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@Lil Doug What set Einstein apart was creativity, not problem-solving. But, both do go hand in hand in some way. He is quite remarkable, Karan Chatrath.

Krishna Karthik - 6 days, 14 hours ago

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@Krishna Karthik @Karan Chatrath I have made changes in the spaceship mechanics problem, the time at which the rocket turned off is t0t_{0}

Lil Doug - 6 days, 14 hours ago

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Please tell me. How do I do the rocket problem?

Krishna Karthik - 6 days, 14 hours ago

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@Krishna Karthik I have also posted for solution 😜
I was also not able to solve that.

Lil Doug - 6 days, 14 hours ago

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@Lil Doug Here's what I got up to:

The velocity of the rocket when it is switched of can be given by:

v0=2512GM(RR0R0R)\displaystyle v_0 = \sqrt{\frac{25}{12} GM \left(\frac{R-R_0}{R_0 R} \right) }

I solved it by separating and integrating:

v015v0vdv=R0RGMR2dR\displaystyle \int_{v_0}^{\frac{1}{5}v_0} v dv = \int_{R_0}^{R} -\frac{GM}{R^2} dR

But the question is: how do I find the final distance, when the rocket reaches constant velocity?

Krishna Karthik - 6 days, 14 hours ago

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Hi Krishna, please refrain from using swear words on Brilliant. I've updated your comment.

Brilliant Mathematics Staff - 5 days, 11 hours ago

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@Krishna Karthik it seems me that you are very interested in rocket.

Lil Doug - 6 days, 14 hours ago

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It's an interesting question.

Krishna Karthik - 6 days, 14 hours ago

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@Krishna Karthik Hope @Karan Chatrath will post a solution of Spaceship Mechanics

Lil Doug - 6 days, 14 hours ago

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@Lil Doug I really do. That one baffles the shit out of me. Btw did you see my approach on it?

The velocity of the rocket when it is switched of can be given by:

v0=2512GM(RR0R0R)\displaystyle v_0 = \sqrt{\frac{25}{12} GM \left(\frac{R-R_0}{R_0 R} \right) }

I solved it by separating and integrating:

v015v0vdv=R0RGMR2dR\displaystyle \int_{v_0}^{\frac{1}{5}v_0} v dv = \int_{R_0}^{R} -\frac{GM}{R^2} dR

But the question is: how do I find the final distance, when the rocket reaches constant velocity?

Krishna Karthik - 6 days, 14 hours ago

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@Krishna Karthik Bro let me try again the problem.

Lil Doug - 6 days, 14 hours ago

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@Lil Doug Is my approach correct? Btw did you also start with this? Also, you can do this by linking using the work energy theorem (no differential equation solving there):

Ui=GMmR\displaystyle U_i = \frac{-GMm}{R}

Uf=GMmR0\displaystyle U_f = \frac{-GMm}{R_0}

150mv0212mv02=UfUi\displaystyle \frac{1}{50}mv_0^2 - \frac{1}{2}mv_0^2 = U_f - U_i

You can get the exact same result with the above method.

Krishna Karthik - 6 days, 14 hours ago

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@Krishna Karthik @Krishna Karthik I don't know because I didn't have tried the problem with even 50%50\% of my full potential

I was feeling lazy, just posted it bro.

Lil Doug - 6 days, 14 hours ago

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@Lil Doug No worries my dude. This kind of problem; yikes. People in SpaceX working for Elon Musk be trynna solve this shit...

Krishna Karthik - 6 days, 13 hours ago

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@Krishna Karthik @Krishna Karthik is Elon Musk sir is in Brilliant.?

Lil Doug - 6 days, 13 hours ago

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@Lil Doug Lol nope. Yeah; people there would struggle solving this problem. SpaceX is extremely good with their rockets.

Krishna Karthik - 6 days, 13 hours ago

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@Krishna Karthik @Krishna Karthik what is the meaning of “people there would struggle solving this problem.“
I don't think people of spacex will difficulty to solve that problem.

Lil Doug - 6 days, 13 hours ago

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@Lil Doug Not really, but it's still super hard to do by pencil and paper. Besides, real rocket engineers probably haven't tried a problem under exam conditions in ages.

But this is a really tough one with lots of mathematically hard stuff. If you can somehow find Stephen Hawking in heaven and bring him back to Earth, he'll do it for you :)

Krishna Karthik - 6 days, 13 hours ago

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@Krishna Karthik @Krishna Karthik I agree.
Now I have some work, I am going, we will talk tomorrow regarding rocket problem .
By the way within 20min a new song of my favorite singer is coming, so I am very excited.
Bye.

Lil Doug - 6 days, 13 hours ago

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@Lil Doug Ok bud. See ya. I've gotta go to sleep anyway.

Krishna Karthik - 6 days, 13 hours ago

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@Krishna Karthik where did karan sir gone.
He was talking with us

Lil Doug - 6 days, 14 hours ago

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Let him be in peace. He should focus in order to solve the problem.

Krishna Karthik - 6 days, 14 hours ago

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@Karan Chatrath i have made necessary changes in the spaceship Mechanics.
Please delete the report.,
And within some hours I will post a note showing my attempt on that charge problem.

Lil Doug - 6 days, 13 hours ago

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I have tried that problem and I do not know how to solve it. In my opinion, the available information is inadequate, but I may be wrong.

Karan Chatrath - 6 days, 6 hours ago

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I thought the problem was telling us that 5t05 t_0 was the time it took for the velocity to drop to e5v0 e^{-5} v_0 . Like a "five time constant" thing. Which doesn't quite make sense, because the solution to the diff eq shouldn't be an exponential. Anyway, it rejected my answer too (I tried typing in 0.053)

Steven Chase - 6 days, 6 hours ago

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I agree as well. I think the question should tell what the final time is, at least. There's no way there's enough information. The most I got up to was this:

The velocity of the rocket when it is switched of can be given by:

v0=2512GM(RR0R0R)\displaystyle v_0 = \sqrt{\frac{25}{12} GM \left(\frac{R-R_0}{R_0 R} \right) }

I solved it by separating and integrating:

v015v0vdv=R0RGMR2dR\displaystyle \int_{v_0}^{\frac{1}{5}v_0} v dv = \int_{R_0}^{R} -\frac{GM}{R^2} dR

But the question is: how do I find the final distance, when the rocket reaches constant velocity?

Edit: I managed to write the final distance as a function of time when the rocket reaches a fifth of its initial velocity. But if you aren't given the anything else, the question is impossible to solve.

Would it be right in saying that the distance from Earth as a function of time would be R=e2GMt\displaystyle R = e^{\sqrt{-2GM}t}?

The golden question is: How is the final time related to t0t_0?

@Steven Chase @Karan Chatrath

Krishna Karthik - 6 days, 1 hour ago

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@Steven Chase @Karan Chatrath I have posted the problem again with right answer here

Lil Doug - 6 days, 3 hours ago

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@Karan Chatrath Does your answer of spaceship Mechanics problem matches with @Steven Chase sir's answer.
His answer is around 0.0530.053 ?

Lil Doug - 6 days, 1 hour ago

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@Steven Chase @Karan Chatrath @Krishna Karthik i have changed the graph little bit and added Note on the problem spaceship Mechanics.
I hope the problem is clearly stated now.

Lil Doug - 6 days ago

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I still don't understand. Finding the time as a function of initial time is hard, but necessary. It must be done in order to achieve a final expression!

Krishna Karthik - 6 days ago

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@Karan Chatrath sir if you don't mind, there is an error in 15th line of solution it should be a2+b2a^{2}+b^{2}

Lil Doug - 5 days, 8 hours ago

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Good catch. It is corrected now.

Karan Chatrath - 5 days, 7 hours ago

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@Karan Chatrath at t=mπqBt=\frac{mπ}{qB} the velocity vector turns by 2π
I can find the displacement from origin but cannot able to find distance. The particle has a weird trajectory, how to calculate distance travelled , Give me some clue /hint?

Lil Doug - 5 days, 7 hours ago

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No, the time you have computed is incorrect.

For distance:

ds=V dtds = V \ dt Where VV is the speed.

Karan Chatrath - 5 days, 7 hours ago

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@Karan Chatrath after 20 minutes I am going to post a note sharing my attempt Please be active after 20 minutes.

Lil Doug - 5 days, 6 hours ago

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@Karan Chatrath the attempt is posted.

Lil Doug - 5 days, 6 hours ago

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@Karan Chatrath
Sir i want to see your solution. Share whenever you will be free.
Thanks in advance.

Lil Doug - 3 days, 15 hours ago

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@Karan Chatrath the question in the book is asking for (velocity of the particle A when angle between the rods becomes θ\theta)
Can you show your attempt by posting a discussion ?
Thanks in advance

Lil Doug - 2 days, 18 hours ago

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Which question? I would use Lagrangian mechanics.

Krishna Karthik - 2 days, 18 hours ago

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Mechanics |17-09-2020|

Lil Doug - 2 days, 18 hours ago

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@Lil Doug I'm trying it out right now. I'm trying to create a time-domain simulation.

Krishna Karthik - 2 days, 18 hours ago

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I will create a note after some time.

Karan Chatrath - 2 days, 17 hours ago

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