3D B and E Fields (part 2)

Electricity and Magnetism Level pending

A particle of mass $m$ and charge $q$ is positioned at the origin at time $t=0$ .Its initial velocity and initial acceleration are $\vec{v}$ and $\vec{a}$ respectively. There are electric and magnetic fields $\vec{E}$ and $\vec{B}$ throughout space.

How far from the origin is the particle at time $t=15?$

Inspiration

Bonus: Plot the trajectories in the $xy$ , $yz$ and $zx$ planes

Details and Assumptions
1) $m=1$
2) $q=+2$
3) $\vec{E} = (E_{x},E_{y},E_{z})=(3,4,5)$
4) $\vec{B} = (B_{x},B_{y},B_{z})=(6,7,8)$
5) $\vec{v}=(v_{x},v_{y},v_{z})=(9,10,11)$
6) $\vec{a}=(a_{x},a_{y},a_{z})=(12,13,14)$

The answer is 1845.732.

1 solution

Karan Chatrath
Jul 26, 2020

Code-based solution. Done using MATLAB:

clear all
clc

% Initialisations:
m         = 1;
q         = 2;
E         = [3;4;5];
B         = [6;7;8];

% Initial conditions:
dx(:,1)   = [9;10;11];
x(:,1)    = [0;0;0];

% Time step:
dt        = 1e-5;

% Final time:
tf        = 15;

% Time vector:
t         = 0:dt:15;

% Loop for numerical integration:
for k = 1:length(t)-1

    % Acceleration of particle:
    ddx = (q/m)*(E + cross(dx(:,k),B));  

    % Numerical integration to obtain velocities:
    dx(:,k+1) = dx(:,k) + dt*ddx;

    % Numerical integration to obtain position:
    x(:,k+1)  = x(:,k)  + dt*dx(:,k+1);
end

% Distance from origin at t = 15:
ANSWER = norm(x(:,end));

% ANSWER = 1845.73

@Karan Chatrath sir this problem is clearly stated or not??

Thanks in advance.

Talulah Riley - 10 months, 2 weeks ago

Apart from one typo in the problem, $a$ is not clearly defined. I would guess that $a$ is the distance between the circle centre and the point charge along its axis.

Karan Chatrath - 10 months, 2 weeks ago

@Karan Chatrath hello sir, I am Neeraj only. Due to some reasons I have changed my name.
Recently, I have posted a new problem Complex flux,
Two members have posted the solution, the 2nd one is Novak sir,
But I am not able to understand his solution and that solution seems me interesting and I am curious about it.
Can you help me to understand that solution's first 2 paragraphs.
Thanks in advance.

Talulah Riley - 10 months, 2 weeks ago

I have not yet attempted that problem. When I do, I will take a look at the solutions.

Karan Chatrath - 10 months, 1 week ago

@Karan Chatrath I have now seen your report sincerely.
I would talk to you before what is your mistake, but I am such a lazy person, even I didn't look at your solution.
I just go to Mr chase.
So, the mistake in your solution is The particle doesn't have to go to that corner of capacitor.
It have to lift up. My English is not good so let me explain in hindi.
Jaise electron niche jaayega to gravity lagega, par thode time baad usme inta electric force lag jaayega ki vo fir se upar chadh jaayega.

Talulah Riley - 10 months, 1 week ago

Like this

Talulah Riley - 10 months, 1 week ago

Okay, I understand your point. I will attempt it again later. Thanks for the clarification

Karan Chatrath - 10 months, 1 week ago

3D B and E Fields (part 2)

The answer is 1845.732.

1 solution

0 pending reports