Flux Calculation - Qualitative

Electricity and Magnetism Level 3

You can calculate the magnetic flux through an open surface by taking the surface integral of the magnetic flux density, or by taking the line integral of the magnetic vector potential over the boundary curve. Suppose that the $\vec{B}$ and/or $\vec{A}$ fields are produced by a thin current-carrying wire located some distance away from the open surface.

$\int \int_S \vec{B} \cdot \vec{dS} = \oint_C \vec{A} \cdot \vec{d \ell}$

To evaluate the left side, you effectively must evaluate a (single ?, double ?, triple ?) integral, and to evaluate the right side, you effectively must evaluate a (single ?, double ?, triple ?) integral.

Give your answer as (left side integral type, right side integral type)

Note: The fields are not known in advance and must be calculated
Bonus: What are the implications for numerical integration?

triple, double single, double double, single double, triple

1 solution

Talulah Riley
Jul 26, 2020

I don't know my solution is correct or not, but here is my attempt :)

When we evaluate the left side ,double integral will work for area and the magnetic field is due to a current carrying wire so it will not uniform everywhere.
It will change according to it's perpendicular distance. So we have to add one Integral for that also. So left side will be a triple integral.

And by little bit common sense (double integral) is included in all 4 options.

Bonus: I think the implications for numerical integration are that
It is difficult to parameterize complex surfaces, therefore we find complex fluxes by integrating numerically.

Please correct me if anything is wrong in my solution.
Thanks in advance.

Thanks for the solution. You're right about why you have to add another integral to each. Regarding numerical integration, a triple integral consists of three nested loops, while a double integral consists of two nested loops. So triple integration is generally much more computationally expensive than double integration.

Steven Chase - 10 months, 2 weeks ago

@Steven Chase Good Morning LPE .
Please check my new problem.
And reply it is correct or not.
Thanks in advance. .
Hope I am not disturbing you.

Talulah Riley - 10 months, 2 weeks ago

@Steven Chase sir what answer you are getting for my new problem? One person has already posted a report.

Talulah Riley - 10 months, 2 weeks ago

I tried it and got about 0.664

Steven Chase - 10 months, 2 weeks ago

@Steven Chase – @Steven Chase sir i have again posted the problem with correct answer. Thanks.

Talulah Riley - 10 months, 2 weeks ago

@Steven Chase is it OK with you if I use the image of vector calculus for making a new problem?
By the way it will be dedicated to you only.

Talulah Riley - 10 months ago

Yeah, that's fine. Here is the numerical code for both problems. The second print statement only applies to the first problem.

import math

dtheta = 10.0**(-5.0)

theta = 0.0

I = 0.0

opt = 1   # opt = 1 for first problem and 2 for second

while theta <= 4.0*math.pi:

    x = theta*math.cos(theta)
    y = theta*math.sin(theta)

    if opt == 1:

        Fx = 2.0*x
        Fy = 2.0*y

    if opt == 2:

        Fx = -y
        Fy = x

    dx = -theta*math.sin(theta) + math.cos(theta)
    dy = theta*math.cos(theta) + math.sin(theta)

    dx = dx*dtheta
    dy = dy*dtheta

    dI = Fx*dx + Fy*dy

    I = I + dI

    #print x,y

    theta = theta + dtheta

print ""
print ""

print dtheta
print I
print ""

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

x1 = 0.0
y1 = 0.0

x2 = 4.0*math.pi
y2 = 0.0

U1 = x1**2.0 + y1**2.0
U2 = x2**2.0 + y2**2.0

print (U2-U1)

Steven Chase - 10 months ago

Flux Calculation - Qualitative

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