Two Magnetic Loops (Part 3)

There are two circular loops of wire, each carrying a current of magnitude 1 1 .

Loop 1 1 has a radius of 1 1 and center ( x , y , z ) = ( 0 , 0 , 0 ) (x,y,z) = (0,0,0) . The vector ( 1 , 2 , 3 ) (1,-2,3) is normal to the plane of loop 1.
Loop 2 2 has a radius of 2 2 and center ( x , y , z ) = ( 1 , 2 , 3 ) (x,y,z) = (1,2,3) . The vector ( 5 , 4 , 1 ) (-5,4,1) is normal to the plane of loop 2.

What is the magnitude of the magnetic force exerted by one loop on the other?

Details and Assumptions:
1) Magnetic permeability μ 0 = 1 \mu_0 = 1 , for simplicity


The answer is 0.0728.

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1 solution

Steven Chase
Oct 28, 2019

Simulation code below. This one can actually solve any of the three from this set of problems, depending on the "opt" setting.

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import math

N = 5000

I = 1.0
u0 = 1.0

Bmult = u0*I/(4.0*math.pi)

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

opt = 3

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

if opt == 1:

    R1 = 1.0
    R2 = 2.0

if opt == 2:

    R1 = 1.0
    R2 = 2.0

if opt == 3:

    R1 = 1.0
    R2 = 2.0

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

if opt == 1:

    C1x = 0.0
    C1y = 0.0
    C1z = 0.0

    C2x = 0.0
    C2y = 0.0
    C2z = 1.0

if opt == 2:

    C1x = 0.0
    C1y = 0.0
    C1z = 0.0

    C2x = 1.0
    C2y = 0.0
    C2z = 0.0

if opt == 3:

    C1x = 0.0
    C1y = 0.0
    C1z = 0.0

    C2x = 1.0
    C2y = 2.0
    C2z = 3.0

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

if opt == 1:

    N1x = 0.0
    N1y = 0.0
    N1z = 1.0

    N2x = 0.0
    N2y = 0.0
    N2z = 1.0

if opt == 2:

    N1x = 0.0
    N1y = 0.0
    N1z = 1.0

    N2x = 1.0
    N2y = 0.0
    N2z = 10.0**(-6.0)

if opt == 3:

    N1x = 1.0
    N1y = -2.0
    N1z = 3.0

    N2x = -5.0
    N2y = 4.0
    N2z = 1.0


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

# Nx*ux + Ny*uy + Nz*uz = 0

u1x = 1.0
u1y = 1.0
u1z = -(N1x*u1x + N1y*u1y)/N1z

u2x = 1.0
u2y = 1.0
u2z = -(N2x*u2x + N2y*u2y)/N2z

u1 = math.sqrt(u1x**2.0 + u1y**2.0 + u1z**2.0)
u2 = math.sqrt(u2x**2.0 + u2y**2.0 + u2z**2.0)

u1x = u1x/u1
u1y = u1y/u1
u1z = u1z/u1

u2x = u2x/u2
u2y = u2y/u2
u2z = u2z/u2

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

v1x = N1y*u1z - N1z*u1y
v1y = -(N1x*u1z - N1z*u1x)
v1z = N1x*u1y - N1y*u1x

v2x = N2y*u2z - N2z*u2y
v2y = -(N2x*u2z - N2z*u2x)
v2z = N2x*u2y - N2y*u2x

v1 = math.sqrt(v1x**2.0 + v1y**2.0 + v1z**2.0)
v2 = math.sqrt(v2x**2.0 + v2y**2.0 + v2z**2.0)

v1x = v1x/v1
v1y = v1y/v1
v1z = v1z/v1

v2x = v2x/v2
v2y = v2y/v2
v2z = v2z/v2

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

print (u1x**2.0 + u1y**2.0 + u1z**2.0)
print (u2x**2.0 + u2y**2.0 + u2z**2.0)

print ""

print (v1x**2.0 + v1y**2.0 + v1z**2.0)
print (v2x**2.0 + v2y**2.0 + v2z**2.0)

print ""

print (N1x*u1x + N1y*u1y + N1z*u1z)
print (N1x*v1x + N1y*v1y + N1z*v1z)
print (v1x*u1x + v1y*u1y + v1z*u1z)

print ""

print (N2x*u2x + N2y*u2y + N2z*u2z)
print (N2x*v2x + N2y*v2y + N2z*v2z)
print (v2x*u2x + v2y*u2y + v2z*u2z)

print ""


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

count = 0

Fx = 0.0
Fy = 0.0
Fz = 0.0

theta1 = 0.0

while theta1 <= 2.0*math.pi:

    x1 = C1x + R1*math.cos(theta1)*u1x + R1*math.sin(theta1)*v1x
    y1 = C1y + R1*math.cos(theta1)*u1y + R1*math.sin(theta1)*v1y
    z1 = C1z + R1*math.cos(theta1)*u1z + R1*math.sin(theta1)*v1z

    dx1 = (-R1*math.sin(theta1)*u1x + R1*math.cos(theta1)*v1x)*dtheta1
    dy1 = (-R1*math.sin(theta1)*u1y + R1*math.cos(theta1)*v1y)*dtheta1
    dz1 = (-R1*math.sin(theta1)*u1z + R1*math.cos(theta1)*v1z)*dtheta1

    Bx = 0.0
    By = 0.0
    Bz = 0.0

    theta2 = 0.0

    while theta2 <= 2.0*math.pi:

        x2 = C2x + R2*math.cos(theta2)*u2x + R2*math.sin(theta2)*v2x
        y2 = C2y + R2*math.cos(theta2)*u2y + R2*math.sin(theta2)*v2y
        z2 = C2z + R2*math.cos(theta2)*u2z + R2*math.sin(theta2)*v2z

        dx2 = (-R2*math.sin(theta2)*u2x + R2*math.cos(theta2)*v2x)*dtheta2
        dy2 = (-R2*math.sin(theta2)*u2y + R2*math.cos(theta2)*v2y)*dtheta2
        dz2 = (-R2*math.sin(theta2)*u2z + R2*math.cos(theta2)*v2z)*dtheta2

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

        rcube = (rx**2.0 + ry**2.0 + rz**2.0)**(3.0/2.0)

        crossBx = dy2*rz - dz2*ry
        crossBy = -(dx2*rz - dz2*rx)
        crossBz = dx2*ry - dy2*rx

        dBx = Bmult*crossBx/rcube
        dBy = Bmult*crossBy/rcube
        dBz = Bmult*crossBz/rcube

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

        theta2 = theta2 + dtheta2

        count = count + 1

        if count%10**6 == 0:
            print count/10**6

    crossFx = dy1*Bz - dz1*By
    crossFy = -(dx1*Bz - dz1*Bx)
    crossFz = dx1*By - dy1*Bx

    dFx = I * crossFx
    dFy = I * crossFy
    dFz = I * crossFz

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

    theta1 = theta1 + dtheta1

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

print ""
print ""

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

@Steven Chase sir i want to know how do you make this type of solution. What's this import math???? Can you bit explain Please

A Former Brilliant Member - 1 year, 7 months ago

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This is the Python programming language. The "math" library allows you to do things like trig functions and square roots

Steven Chase - 1 year, 7 months ago

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@Steven Chase sir How can i do this??? Did I can do this in mobile also? What input did you give to python programming?? I have don't studied python programming language yet. Due to that i solve all questions with pen and page. Did phython programming language makes the question easier OR the calculation of question easier??? Please

A Former Brilliant Member - 1 year, 7 months ago

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@A Former Brilliant Member Python is just a specific programming language. There are many others (C, Java, Matlab, etc.). Programming languages allow complicated physics problems to be solved using numerical integration. All inputs to the program are contained within the code.

Steven Chase - 1 year, 7 months ago

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@Steven Chase @steven chase sir please please please upload more questions like statics (11/12/2019) please.?? I am waiting and hungry for such beautiful questions.

A Former Brilliant Member - 1 year, 7 months ago

@Steven Chase sir can you please upload the solution of https://brilliant.org/problems/effect-of-circuit-tolerances/

A Former Brilliant Member - 1 year, 7 months ago

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@A Former Brilliant Member It's up now

Steven Chase - 1 year, 7 months ago

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@Steven Chase @Steven Chase how can i post a photo of my solution in brilliant. Using mobile

A Former Brilliant Member - 1 year, 7 months ago

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@A Former Brilliant Member I don't know about mobile, but if you have it in a common image format (.png, .jpg. etc.), you can use the image upload feature on the utility bar in the solution.

Steven Chase - 1 year, 7 months ago

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@Steven Chase @Steven Chase sir i want to upload the solution of statics 11/12/2019 (beautiful question) . So where is the utility bar??

A Former Brilliant Member - 1 year, 7 months ago

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@A Former Brilliant Member It's at the top of the solution workspace

Steven Chase - 1 year, 7 months ago

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@Steven Chase @Steven Chase sir in my solution workspace I can't see utility bar. I can send this to your email id and then you can post it in the brilliant. Can you please?

A Former Brilliant Member - 1 year, 7 months ago

@Steven Chase sir i have some doubts in physics. Don't mind. Can i have your phone number? Please?

A Former Brilliant Member - 1 year, 7 months ago

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@A Former Brilliant Member I won't provide my phone number, but we can discuss on this site. Perhaps you can post in the community discussion section.

Steven Chase - 1 year, 7 months ago

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