Average Magnetic Field

A bounded volume has the following equation x 2 4 + y 2 25 + z 2 9 1 \frac{x^{2}}{4}+\frac{y^{2}}{25}+\frac{z^{2}}{9}≤1 .
A charge ( q = + 10 C ) q=+10C) is placed at a random postion in this whole volume. Calculate the magnitude of B a v e r a g e B_{average} magnetic field generated at ( 67 , 68 , 69 ) (67, 68,69)
Details and Assumptions
1 ) ϵ 0 = 1 1) \epsilon_{0}=1
2 ) μ 0 = 1 2) \mu_{0}=1
The problem is purely original.


The answer is 0.

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1 solution

Steven Chase
May 3, 2020

Since the charge is not moving, there is no magnetic field

@Steven Chase I suggest you to make this type of problem

A Former Brilliant Member - 1 year, 1 month ago

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I'll post one like this soon. The first part will just ask for the steady state speed. The next will ask about dynamics

Steven Chase - 1 year, 1 month ago

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@Steven Chase Sir for the new problem I am getting 2 equations B v l i R L d i d t = 0 Bvl-iR-L\frac{di}{dt}=0 F B i l = m a F-Bil=ma

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member That looks right

Steven Chase - 1 year, 1 month ago

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@Steven Chase @Steven Chase after simplyfying I am getting 5 v i = i ˙ 5v-i=\dot{i} 10 5 i = v ˙ 10-5i=\dot{v}

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member In steady state, what happens to the time derivatives?

Steven Chase - 1 year, 1 month ago

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@A Former Brilliant Member Check your numerical substitution also

Steven Chase - 1 year, 1 month ago

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@Steven Chase @Steven Chase I will be very happy if you post a nice follow up. To electromechanics problem. And sorry for the delayed solution

A Former Brilliant Member - 1 year, 1 month ago

@Steven Chase are you posting part 3 today???

A Former Brilliant Member - 1 year, 1 month ago

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