Question

5. Numerical Integration (15 marks] When a charged particle moves perpendicular to a magnetic field it traces out circles in the plane perpendicular to the magnetic field due to the "Lorentz force". Here we consider a small particle (e.g. dust grain or nanoparticle) with a charge to mass ratio of 1 Coulomb per kg moving perpendicular at speed v1 = 1 m/s to a 1 Tesla magnetic field that points in the z direction. In this case the motion is determined by the following set of differential equations dr dt = Ux duz dt = dy dt duy dt . =-U (a) Copy and complete the following table for two steps of Euler's method t y 1 U 1 L'y k1, kl, klex kly 0 0 0 0.1 0.2 Note that an editable table has been provided on the next page. Given that the true solution should be circular motion, compute the relative error in the distance of the particle from the origin at t = 0.2. [6 marks] (b) Copy and complete the following table for two steps of the Cauchy-Euler (a.k.a. RK2) method t Dy kl, kl, kl kl 21 0 yl ul, vly k2x k2y k2& k2wy 0 1 1 0 0.1 0.2 Note that an editable table has been provided on the next page. Repeat the error calculation from part (a) for this method. [7 marks) (c) By what factor would you expect the error from part (C) to decrease if we switch to a time step of 0.02 in the Euler-Cauchy method? Explain how you obtained this estimate. [2 marks]

Answer 1

Ans 3:

#include <iostream>
using namespace std;

int main() {

// Variables to hold operands
float f1, f2;

// Variable to hold operator
   char op;

   // Requesting user to input operands
   cout << "\nEnter a number: ";
   cin >> f1;
   cout << "Enter another number: ";
   cin >> f2;

   // Requesting user to input operator
   cout << "\nEnter a operator: ";
   cin >> op;
   cout << "\n";

   switch(op)
   {
   case '+':
   cout << f1 << " + " << f2 << " = " << f1 + f2;
   break;

   case '-':
   cout << f1 << " - " << f2 << " = " << f1 - f2;
   break;

   case '/':
   cout << f1 << " / " << f2 << " = " << f1 / f2;
   break;

   case '*':
   cout << f1 << " * " << f2 << " = " << f1 * f2;
   break;

default:
cout << "Invalid Operator!";
   }

   return 0;
}

/*********** Output ****************/

Ans 4:

#include <iostream>
#include <time.h>
#include <stdlib.h>
using namespace std;

int main()
{
srand(time(0));
cout << "\nRandom No generated (0.. 100): " << rand() % 101;

return 0;
}

/*************** below is the output *******************/

/********* Note: if you are using -std=c++11 (c++ 11) then you don't need to include stdlib.h *************/

/********** I have included stdlib.h, to just show how it will work in other versions of c++ ***********/

/************ srand(time(0)) seeds the random number generator with current system time, if you don't do this, everytime the code runs you will get the same random number ****************/

/*********** if you have any questions, you may ask in comments *****************/

/**************** thanks for reading ******************/