Flat Loops and Magnetism

Electricity and Magnetism Level 5

Two identical loops of R = 1 R=1 both placed in the X − Y X-Y plane. A current of I = 2 π A I=2\sqrt{π}A is passing through both the loops. A B = C D = E F = 1 AB=CD=EF=1 Find the magnitude of Magnetic force between them. Details and Assumptions Magnetic Permeability μ 0 = 1 \mu_{0}=1


The answer is 0.622.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Steven Chase
Mar 23, 2020 
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import math

# Scan resolution

N = 10000

######################################################

# Constants

u0 = 1.0
I1 = 2.0*math.sqrt(math.pi)
I2 = 2.0*math.sqrt(math.pi)

dtheta1 = 2.0*math.pi/N
dtheta2 = 2.0*math.pi/N

######################################################
######################################################

# Wire 1 is on the right (+x)
# Wire 2 is on the left (-x)

# For every point on wire 1, integrate over wire 2 to find the B field
# Then calculate the infinitesimal force contribution to wire 1
# Add up the infinitesimal force contributions

Fx = 0.0
Fy = 0.0
Fz = 0.0

theta1 = 0.0

while theta1 <= 2.0*math.pi:

    x1 = 1.5 + math.cos(theta1)
    y1 = math.sin(theta1)
    z1 = 0.0

    dx1 = -math.sin(theta1) * dtheta1
    dy1 = math.cos(theta1) * dtheta1
    dz1 = 0.0

    Bx = 0.0
    By = 0.0
    Bz = 0.0

    theta2 = 0.0

    while theta2 <= 2.0*math.pi:     # Determine B field at point on wire 1....
                                    # by integrating over wire 2
                                    # Biot Savart
        x2 = -1.5 + math.cos(theta2)
        y2 = math.sin(theta2)
        z2 = 0.0

        dx2 = -math.sin(theta2) * dtheta2
        dy2 = math.cos(theta2) * dtheta2
        dz2 = 0.0

        rx = x1 - x2                 
        ry = y1 - y2
        rz = z1 - z2

        r = math.sqrt(rx**2.0 + ry**2.0 + rz**2.0)  

        Bcrossx = dy2*rz - dz2*ry      
        Bcrossy = -(dx2*rz - dz2*rx)
        Bcrossz = dx2*ry - dy2*rx

        dBx = (u0*I2/(4.0*math.pi))*Bcrossx/(r**3.0)  
        dBy = (u0*I2/(4.0*math.pi))*Bcrossy/(r**3.0)
        dBz = (u0*I2/(4.0*math.pi))*Bcrossz/(r**3.0)

        Bx = Bx + dBx
        By = By + dBy
        Bz = Bz + dBz

        theta2 = theta2 + dtheta2

    Fcrossx = dy1*Bz - dz1*By        # Add up infinitesimal forces on wire 1
    Fcrossy = -(dx1*Bz - dz1*Bx)     # dF = I*(dL x B)
    Fcrossz = dx1*By - dy1*Bx

    dFx = I1 * Fcrossx
    dFy = I1 * Fcrossy
    dFz = I1 * Fcrossz

    Fx = Fx + dFx            
    Fy = Fy + dFy
    Fz = Fz + dFz

    theta1 = theta1 + dtheta1

######################################################
######################################################

# Print results

print N
print ""
print Fx
print Fy
print Fz
print ""
F = math.sqrt(Fx**2.0 + Fy**2.0 + Fz**2.0)

print F

######################################################
######################################################

#Python 2.7.4 (default, Apr  6 2013, 19:54:46) [MSC v.1500 32 bit (Intel)] on win32
#Type "copyright", "credits" or "license()" for more information.
#>>> ================================ RESTART ================================
#>>> 
#1000

#0.628412076704
#3.73781204203e-14
#0.0

#0.628412076704
#>>> ================================ RESTART ================================
#>>> 
#2000

#0.625507730558
#3.41809805287e-14
#0.0

#0.625507730558
#>>> ================================ RESTART ================================
#>>> 
#4000

#0.624054735018
#1.97704275719e-14
#0.0

#0.624054735018
#>>> ================================ RESTART ================================
#>>> 
#10000

#0.622601191168
#-1.20999490363e-13
#0.0

#0.622601191168
>>>

4 Helpful 4 Interesting 3 Brilliant 1 Confused

@Steven Chase Nice solution. post more question on gauss law.

A Former Brilliant Member - 1 year, 2 months ago

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