Story of Chrome and Flux

Electricity and Magnetism Level pending

A particle with mass $m$ and charge $q=1C$ is placed at $(1,-80,0)$ , is launched at $t=0$ at an angle $\theta=45°$ ( $\theta$ is taken from the line $x=1m$ in the upward direction, when you see the launching of particle from $-x$ direction seeing toward $+x$ direction you will see the angle $\theta$ in the clockwise direction) with a speed of $v=40m/s$ . A chrome figure is placed in $Y-Z$ plane . The centre of the chrome has the coordinates $(0,0,40)$ . Let the flux passing through blue area is $\phi_{B}$ and through white area is $\phi_{W}$ and similarly for all surfaces. The total radius of chrome is $2m$ and total radius of blue disc $1m$ and width of white region is $0.2m$ . Evalute the value of expression at $t=2\sqrt{2}sec$ $\bigg(\frac{\phi_{B}\phi_{W}\phi_{Y}\phi_{R}\phi_{G}}{\phi_{B}+\phi_{W}+\phi_{Y}+\phi_{R}+\phi_{G}}\bigg)10^{6}$ Details and Assumtion $g=10ms^{-2}$ , $\epsilon_{0}=1$

The answer is 0.589.

1 solution

Steven Chase
Apr 15, 2020

Working out the kinematics puts the charge a distance of $1$ unit directly above the center of the disk. From there, it is a standard flux calculation. I formulated it as a double integral, although it could have been done as a single integral. Since the problem is symmetrical, we need not concern ourselves with the exact geometry of the yellow, red, and green regions. All that matters is that those three regions have identical fluxes. For convenience, I considered the disk to be in the $xy$ plane, with the charge being on the $z$ axis.

import math

N = 5000

R1 = 1.0
R2 = 1.2
R3 = 2.0

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

q = 1.0

xq = 0.0
yq = 0.0
zq = 1.0

e0 = 1.0

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

k = 1.0/(4.0*math.pi*e0)

phiB = 0.0
phiW = 0.0
phiY = 0.0
phiR = 0.0
phiG = 0.0

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

dr = R3/N
dtheta = 2.0*math.pi/N

nx = 0.0
ny = 0.0
nz = -1.0

z = 0.0

r = 0.0

while r <= R3:

    theta = 0.0

    while theta <= 2.0*math.pi:

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

        dS = r*dr*dtheta

        Dx = x - xq
        Dy = y - yq
        Dz = z - zq

        D = math.sqrt(Dx**2.0 + Dy**2.0 + Dz**2.0)

        ux = Dx/D
        uy = Dy/D
        uz = Dz/D

        E = k*q/(D**2.0)

        Ex = E * ux
        Ey = E * uy
        Ez = E * uz

        dot = Ex*nx + Ey*ny + Ez*nz

        dphi = dot * dS

        if r < R1:
            phiB = phiB + dphi

        if (r >= R1) and (r < R2):
            phiW = phiW + dphi

        if r >= R2:
            phiY = phiY + dphi/3.0
            phiR = phiR + dphi/3.0
            phiG = phiG + dphi/3.0

        theta = theta + dtheta

    r = r + dr

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

print N
print ""

print phiB
print phiW
print phiY
print phiR
print phiG

print ""

num = phiB * phiW * phiY * phiR * phiG * (10.0**6.0)
denom = phiB + phiW + phiY + phiR + phiG

print (num/denom)

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

# Results

#>>> 
#1000

#0.146415906218
#0.0335140213742
#0.0322166493246
#0.0322166493246
#0.0322166493246

#0.593246705123
#>>> ================================ RESTART ================================
#>>> 
#2000

#0.146431351113
#0.0336451033926
#0.0321665470528
#0.0321665470528
#0.0321665470528

#0.59286338089
#>>> ================================ RESTART ================================
#>>> 
#4000

#0.14652741402
#0.0334647160092
#0.0321641750222
#0.0321641750222
#0.0321641750222

#0.590136186535
#>>> ================================ RESTART ================================
#>>> 
#5000

#0.146481956899
#0.0334573197314
#0.0321572688868
#0.0321572688868
#0.0321572688868

#0.589599759953
#>>>

@Steven Chase Sir Fast and Accurate solution. But sir if the position of charge is not symmetrical. Then how can we find the flux??

A Former Brilliant Member - 1 year, 1 month ago

It can still be done, but it is much less trivial. The geometry of those three regions would have to be explicitly given. Probably the best way would be to specify the start and end points of the three line segments.

Steven Chase - 1 year, 1 month ago

@Steven Chase Sir your new question of RLC. Can I find $I_{min}$ with pen and page. Does it need numerical resolution which I don't know how to do?

A Former Brilliant Member - 1 year, 1 month ago

@A Former Brilliant Member – It probably requires numerical simulation. But if you derive the state-space (which isn't too hard to do), the numerical integration part is easy. There are many examples of it in the solutions that have already been posted.