Sinusoidal Wire Magnetics (part 4 4 )

Electricity and Magnetism Level pending

A piece of wire has the following shape 0 ≤ x ≤ 2 π 0≤x≤2π y = sin ⁡ x y=\sin x z = 0 z=0 from position A A to B B . A circle of radius 1 1 is placed in Y − Z Y-Z plane. Both carries current of 2 π 2\sqrt{π} .What is the magnitude of magnetic force exerted between them.I have provided the view of system from + Y +Y axis.

Details and Assumptions 1) Magnetic permeability μ 0 = 1 \mu_{0} =1


The answer is 1.826.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Steven Chase
Mar 19, 2020

Nice problem. Solution code attached: 

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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)

dx1 = 2.0*math.pi/N
dtheta = 2.0*math.pi/N

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

# Wire 1 is the sinusoid
# Wire 2 is the circle

# 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

x1 = 0.0

while x1 <= 2.0*math.pi:

    y1 = math.sin(x1)
    z1 = 0.0

    dy1 = math.cos(x1) * dx1
    dz1 = 0.0

    Bx = 0.0
    By = 0.0
    Bz = 0.0

    theta = 0.0
    x2 = 0.0

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

        dx2 = 0.0
        dy2 = -math.sin(theta) * dtheta        
        dz2 = math.cos(theta) * dtheta

        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

        theta = theta + dtheta

    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

    x1 = x1 + dx1

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

# 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

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

#>>> 
#1000

#-1.77485676612e-14
#9.72537644059e-14
#-1.8408051273

#1.8408051273
#>>> ================================ RESTART ================================
#>>> 
#2000

#-1.82758909625e-14
#8.9307827061e-14
#-1.8335305663

#1.8335305663
#>>> ================================ RESTART ================================
#>>> 
#5000

#4.8616910172e-14
#-2.76391677998e-13
#-1.83019813904

#1.83019813904
#>>>


#10000

#6.04380868837e-14
#-3.45792702307e-13
#-1.82823188453

#1.82823188453

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Amazing. Thanks for the solution. Can you please upload solution of infinite wire Magnetics??

A Former Brilliant Member - 1 year, 2 months ago

The infinite wire problem is similar to the sine magnetics Part 2 problem, just with a wider integration range. For the disk problem, we need more information about the current distribution.

Steven Chase - 1 year, 2 months ago

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@Steven Chase Current is distributed uniformly.

A Former Brilliant Member - 1 year, 2 months ago

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How does it flow? Does it circulate around the center?

Steven Chase - 1 year, 2 months ago

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@Steven Chase – Yes it circulate around the center . Sir i have taken a element at a distance x there multiply element's area with current per unit area in the whole disk.

A Former Brilliant Member - 1 year, 2 months ago

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@A Former Brilliant Member – I am not 100% sure that my answer of magnetic disc is correct. Can you check it by code??

A Former Brilliant Member - 1 year, 2 months ago

@Steven Chase Sir are you a Lead power engineer at Schweitzer Engineering Laboratories??

A Former Brilliant Member - 1 year, 2 months ago

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Yes, I am. You must have seen my profile on LinkedIn

Steven Chase - 1 year, 2 months ago

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