Member Avatar for imthiyas92

PLS HELP ME TO CODE THE MAGNETIC FIELD AT A POINT FROM A CIRCULAR CURRENT LOOP BY BIOT-SAVART LAW.PLS HELP NEED TO SUBMIT COMPUTER PROJECT WILL DUE TOMMORROW PLSSSSS
HERE IS CODE BY MY OWN WAY


#include <iostream>
#include <conio.h>
#include <stdlib.h>
#define M 4*3.14*10^-7
main()
{
float B,N,I,R;
cout<<"Enter the no of the turns,N: /n";
cin>>N;
cout<<"Enter the intensity of current,I: /n";
cin>>I;
cout<<"Enter the magnitude of Resistance,R: /n";
cin>>R;
int x;
cout<<"Distance from the center of the coil to the point where field is to be calculated,x: /n";
cin>>x;
const float PI=3.14;
B=N*I*R*R/2[R*R+x+x]^3/2;
cout<<"The Magnetic field at the point is "<<B<<"T";
return0;
}

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  • Latest Post Latest Post by mike_2000_17
sergent commented: CAPITAL LETTERS + HOMEWORK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! -1

Dani AI

Generated

A few focused notes that build on the replies already here. correctly pointed out the caret (^) mistake and gave the on‑axis closed form; that closed form is the fastest, most accurate route when the observation point lies on the loop axis. For a general point the Biot–Savart line integral must be evaluated numerically.

Practical implementation tips: use SI units (I in A, distances in m, B in T) and double precision. Use mu0 = 4*pi*1e-7 (H/m). In C/C++ do not use ^ for power; use pow() (or std::pow) and in Python use **. Keep angles in radians. Prefer descriptive names (e.g. R_radius not just R) and validate inputs. Avoid nonportable headers like conio.h and guard against division by zero when the observation point lies very near a wire element.

Algorithm outline for the general (off-axis) field:

  • Discretize the loop into N segments of angle dtheta = 2*pi/N.
  • For each segment compute the element position r'(theta) and the element vector dl = (dLx,dLy,dLz).
  • Compute r = r_obs - r' and accumulate dB = (mu0/4/pi) I (dl x r) / |r|^3.
  • Sum all dB contributions and multiply by the number of turns.
  • Verify convergence by increasing N until the result stabilizes; compare axis results to the closed form as a check.

Example Python integration (straightforward, easy to verify):

import math

mu0 = 4*math.pi*1e-7

def B_loop(I, R, obs, Nturns=1, Nseg=2000):
    bx = by = bz = 0.0
    dtheta = 2*math.pi / Nseg
    for k in range(Nseg):
        theta = k * dtheta
        xP, yP, zP = R*math.cos(theta), R*math.sin(theta), 0.0
        dlx = -R*math.sin(theta) * dtheta
        dly =  R*math.cos(theta) * dtheta
        dlz = 0.0
        rx, ry, rz = obs[0]-xP, obs[1]-yP, obs[2]-zP
        r3 = (rx*rx + ry*ry + rz*rz) ** 1.5
        cx = dly*rz - dlz*ry
        cy = dlz*rx - dlx*rz
        cz = dlx*ry - dly*rx
        bx += cx / r3
        by += cy / r3
        bz += cz / r3
    const = mu0 * I / (4*math.pi)
    return (const * Nturns * bx, const * Nturns * by, const * Nturns * bz)

Validation and troubleshooting: compare the numeric result on the axis with the closed form (the result given earlier by ). Increase Nseg until changes are negligible. For observation points extremely close to the wire use adaptive refinement or analytic limits to avoid large numerical error.

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Member Avatar for jonsca

Please don't type in all caps no need to YELL lol.

What's the problem with the program? Is it not giving you the result you expect or are there errors? Please be more specific.

Please place your code in [code] //code here [/code] (it's not too late to edit your post).

Member Avatar for mike_2000_17

B=N*I*R*R/2[R*R+x+x]^3/2;

Is not a valid C/C++ syntax. the ^ operator is not for power (it's binary XOR which I doubt is what you intended). C/C++ is not matlab or mathematica or a hand calculator.

If the function is, as I understand it: B = (N * I * R * R) / (2 * (R^2 + x^2)^(3/2))
Then in C++ it is:

B = N * I * R * R / (2.0 * pow(R*R + x*x, 3.0 / 2.0));

BTW: the value R is the radius of the coil, not the resistance, and x should be declared as a float. Oh and PI = 3.14159.. not 3.14.

Member Avatar for imthiyas92

Is not a valid C/C++ syntax. the ^ operator is not for power (it's binary XOR which I doubt is what you intended). C/C++ is not matlab or mathematica or a hand calculator.

If the function is, as I understand it: B = (N * I * R * R) / (2 * (R^2 + x^2)^(3/2))
Then in C++ it is:

B = N * I * R * R / (2.0 * pow(R*R + x*x, 3.0 / 2.0));

BTW: the value R is the radius of the coil, not the resistance, and x should be declared as a float. Oh and PI = 3.14159.. not 3.14.

thanks

Member Avatar for imthiyas92 
pls give me the complete source code for the mentioned topic...plsssssss
Member Avatar for mike_2000_17

Are you kidding.. it's been twenty days and you still haven't had to hand-in your assignment, such a trivial one too. Man... talk about easy grades..

here it is.. so that you stop crying about it:

#include <iostream>
#include <cmath>

int main()
{
  float B,N,I,R;
  cout<<"Enter the no of the turns,N: \n";
  cin>>N;
  cout<<"Enter the intensity of current,I: \n";
  cin>>I;
  cout<<"Enter the magnitude of Resistance,R: \n";
  cin>>R;
  int x;
  cout<<"Distance from the center of the coil to the point where field is to be calculated,x: \n";
  cin>>x;

  B = N * I * R * R / (2.0 * pow(R*R + x*x, 3.0 / 2.0));

  cout<<"The Magnetic field at the point is "<<B<<"T";
  return 0;
}

see... all I had to do is put that one line I posted earlier into your code. How hard was that for you to do?
In the future, you should know that putting many sss after pls only convinces people that for you, begging is easier than reading, thinking and learning. In other words, it tells people: I'm lazy and stupid, please pity me enough to give me a free-pass. Well I do pity you, and the above is the only pocket-change I can give you.

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