Magnetic Field Induced by Nonuniform Electric Flux

In summary, the amplitude of the induced magnetic field at a distance 1.5 cm from the center axis joining the plates at time t = 2.9 sec is 1.69E-16
  • #1
Badre
11
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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 = [itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] d[itex]\Phi[/itex]e/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=[itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] pi*r^2/2*pi*r * dE/dt
B=[itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] (.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.
 
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  • #2
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 = [itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] d[itex]\Phi[/itex]e/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=[itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] 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

[tex] \Phi_E = \int_S \vec E \cdot \vec{dA}, [/tex]

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

[tex] \frac{ \partial \Phi_E}{\partial t} = \int_S \frac{\partial \vec E}{\partial t} \cdot \vec{dA}. [/tex]

Either way works fine.

It may be useful to note that in polar coordinates, [itex] dA = r \ dr \ d\theta [/itex].
 
Last edited:
  • #3
Thanks! Haven't had to do a double integral before now.
 

FAQ: Magnetic Field Induced by Nonuniform Electric Flux

What is a nonuniform electric flux?

A nonuniform electric flux refers to a situation where the electric field is not constant in magnitude and direction throughout a given area. This can occur when there are varying charges or electric fields present within the area.

Why does a nonuniform electric flux induce a magnetic field?

According to Maxwell's equations, a changing electric field will produce a magnetic field. In the case of a nonuniform electric flux, the changing electric field creates a circulating magnetic field, known as an induced field.

How is the magnetic field strength affected by the nonuniform electric flux?

The strength of the magnetic field induced by a nonuniform electric flux depends on the rate of change of the electric flux. In general, a faster rate of change will result in a stronger induced magnetic field.

Can a nonuniform electric flux be used to control the strength of a magnetic field?

Yes, by manipulating the electric flux, the induced magnetic field can also be controlled. This can be useful in applications such as electromagnets, where the strength of the magnetic field needs to be adjusted.

How is the direction of the induced magnetic field determined?

The direction of the induced magnetic field is determined by the right-hand rule. If the fingers of your right hand point in the direction of the changing electric flux, your thumb will point in the direction of the induced magnetic field.

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