Magnetic Field Induced by Nonuniform Electric Flux

AI Thread Summary
The discussion focuses on calculating the induced magnetic field between two circular plates in a Doppler radar unit design, with a specific electric field equation provided. The user attempts to derive the magnetic field using the time derivative of the electric field but encounters issues with the software rejecting their answer. It is clarified that the electric field is not uniform across the area defined by the radius, necessitating integration to find the electric flux. The conversation suggests two methods for calculating the flux, emphasizing the need for a double integral in polar coordinates. Understanding the integration process is crucial for solving the problem correctly.
Badre
Messages
11
Reaction score
0

Homework Statement



During the design of a new doppler radar unit, you have two circular plates with radius R = 3 cm and a plate separation of d = 5 mm. The magnitude of the electric field between the plates is given as:

E = (700 V/(m sec2))*(1-r/Rp)*t 2

where Rp is the radius of the plates, t is the time in sec and r is the distance from the axis of the plates (for r < Rp ).

a) What is amplitude of the induced magnetic field a distance 1.5 cm from the center axis joining the plates at time t = 2.9 sec?

Homework Equations



∫B.ds = \mu_{0}\epsilon_{0} d\Phie/dt

The Attempt at a Solution



I did this problem as I worked out previous ones, taking the time derivative of the electric field to find the changing electric flux.

B=\mu_{0}\epsilon_{0} pi*r^2/2*pi*r * dE/dt
B=\mu_{0}\epsilon_{0} (.015/2) * 2*700t(1-.015/.03)
B = 1.69E-16

This answer is rejected by the software. I'm at a loss for what I'm missing here. I don't understand why I'm given the separation between the plates, I suspect that may be involved in the solution but I'm not seeing how. Thanks for any help.
 
Physics news on Phys.org
Badre said:

Homework Statement



During the design of a new doppler radar unit, you have two circular plates with radius R = 3 cm and a plate separation of d = 5 mm. The magnitude of the electric field between the plates is given as:

E = (700 V/(m sec2))*(1-r/Rp)*t 2

where Rp is the radius of the plates, t is the time in sec and r is the distance from the axis of the plates (for r < Rp ).

a) What is amplitude of the induced magnetic field a distance 1.5 cm from the center axis joining the plates at time t = 2.9 sec?


Homework Equations



∫B.ds = \mu_{0}\epsilon_{0} d\Phie/dt



The Attempt at a Solution



I did this problem as I worked out previous ones, taking the time derivative of the electric field to find the changing electric flux.

B=\mu_{0}\epsilon_{0} pi*r^2/2*pi*r * dE/dt
Okay, I think I see what you are doing here (it took me awhile to figure out your above result).

When you tried to calculate the flux, you just multiplied the electric field E by the area, πr2. That would work if the electric field was [STRIKE]constant[/STRIKE] uniform within the entire circle defined by r (it can be time changing, but at a particular point in time it must be uniform across the surface defined within r). But it's not [STRIKE]constant[/STRIKE] uniform for this problem. (Although due to symmetry, the magnetic field strength is [STRIKE]constant[/STRIKE] uniform around the perimeter 2πr, which is why the ∫B·ds = B[2πr] part worked okay.)

You're going to have to do an integration, one way or the other. You have two choices. You can calculate the flux directly via

\Phi_E = \int_S \vec E \cdot \vec{dA},

and then take the time derivative to find \frac{\partial \Phi_E}{\partial t}. Or, you can take the time derivative first to obtain \frac{\partial E}{\partial t}, and then evaluate

\frac{ \partial \Phi_E}{\partial t} = \int_S \frac{\partial \vec E}{\partial t} \cdot \vec{dA}.

Either way works fine.

It may be useful to note that in polar coordinates, dA = r \ dr \ d\theta.
 
Last edited:
Thanks! Haven't had to do a double integral before now.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top