Induced Voltage

Electricity and Magnetism Level 3

$\large{\frac{x^{2}}{4t^{2}} + \frac{y^{2}}{9t}} = 1$

A conducting loop in the $xy$ plane takes the form of the curve given above. The parameter $t$ denotes time. Suppose that there is a uniform magnetic flux density $B$ which is normal to the $xy$ plane.

If the magnitude of the voltage induced in the loop at time $t = 9$ can be expressed as $\alpha \pi B$ , determine the value of $\alpha$ .

Details and Assumptions: Neglect units and assume that all physical expressions take their simplest forms.

The answer is 27.

1 solution

Steven Chase
Sep 22, 2016

The curve is an ellipse with semi-axes of length $2t$ and $3\sqrt{t}$ .
The area of the ellipse is therefore $6\pi t^{\frac{3}{2}}$ .
The flux linkage is $6\pi B t^{\frac{3}{2}}$ , and the time derivative of the flux linkage (the voltage) is $9 \pi B \sqrt{t}$ .
At $t = 9$ , the voltage is therefore $27\pi B$ .

this question is quite over-rated in my opinion

Rishi Sharma - 4 years, 7 months ago

In what sense?

Steven Chase - 4 years, 7 months ago

It is more like a formula based question. find the area and just plug it in the definition and voila you have the answer. I guess it should be a level 3 problem .

Rishi Sharma - 4 years, 7 months ago

@Rishi Sharma – Yeah, I think you're right

Steven Chase - 4 years, 7 months ago

@Steven Chase – right level 3 problem

rajdeep brahma - 2 years, 11 months ago

@Steven Chase Sir how to solve the below integral using python $\int_{x=1}^{x=3} \int_{y=5}^{y=7} \int_{z=11}^{z=19} x^{2}y^{3}z^{5} dzdydx$
Hope I am not disturbing you.
Thanks in advance.

NJ STAR - 10 months, 3 weeks ago

Hello. You can evaluate the three integrals separately and then multiply the results together.

Steven Chase - 10 months, 3 weeks ago

The code below shows how to do it by multiplying three single integrals together, and by evaluating a triple integral. Multiplying three single integrals is more efficient and accurate, but the triple integral yields an answer that is reasonably close. The difference comes from the fact that nesting loops is very computationally expensive. To get an answer of comparable accuracy, a huge amount of computation has to go into the triple integral.

import math

x1 = 1.0
x2 = 3.0

y1 = 5.0
y2 = 7.0

z1 = 11.0
z2 = 19.0

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

x = x1
dx = (x2-x1)/(10.0**6.0)

Ix = 0.0

while x <= x2:

    Ix = Ix + (x**2.0)*dx

    x = x + dx

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

y = y1
dy = (y2-y1)/(10.0**6.0)

Iy = 0.0

while y <= y2:

    Iy = Iy + (y**3.0)*dy

    y = y + dy

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

z = z1
dz = (z2-z1)/(10.0**6.0)

Iz = 0.0

while z <= z2:

    Iz = Iz + (z**5.0)*dz

    z = z + dz

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

dx = (x2-x1)/(400.0)
dy = (y2-y1)/(400.0)
dz = (z2-z1)/(400.0)

Ixyz = 0.0

x = x1

while x <= x2:

    y = y1

    while y <= y2:

        z = z1

        while z <= z2:

            Ixyz = Ixyz + (x**2.0)*(y**3.0)*(z**5.0)*dx*dy*dz

            z = z + dz

        y = y + dy

    x = x + dx

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

print (Ix*Iy*Iz)
print Ixyz

#29035914173.6
#29298573361.4

Steven Chase - 10 months, 3 weeks ago

@Steven Chase Thank you so much.
I think this is explicit euler integration ,BTW can we do it with Midpoint method. ??
Which method gives more accurate answer?

By the way, in your new problem does $Z_{B}$ and $Z_{C}$ are also purely resistive ??
Thanks in advance
Hope I am not disturbing you.

NJ STAR - 10 months, 3 weeks ago

@Nj Star – The midpoint method is more accurate for a given time step. In general, you can use any numerical integration method you want. Accuracy is generally not a concern for me, since I run explicit Euler with very small steps and look for converging results. In the new problem, $Z_B$ and $Z_C$ are generally not purely resistive.

Steven Chase - 10 months, 3 weeks ago

Induced Voltage

The answer is 27.

1 solution

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