# Magnetic Field of Charged Capacitor: Calculating E & B Fields

• fedderenator
In summary, we discussed the problem of determining the time rate of increase of electric field between circular plates with a 0.450 A current and a separation of 4.00 mm. We used Maxwell's equations, Ampere's Law, and the Biot-Savart Law to find the magnitude of the magnetic field between the plates 5.00 cm from the center. After some trial and error, we were able to find the correct equation to use and ultimately calculated the magnetic field to be 0.230 microT.
fedderenator

## Homework Statement

A 0.450 A current is charging a capacitor that has circular plates 14.0 cm in radius.

(a) If the plate separation is 4.00 mm, what is the time rate of increase of electric field between the plates? Answer: 8.26e5 MV/m/s

(b) What is the magnitude of the magnetic field between the plates 5.00 cm from the center?

## Homework Equations

Maxwell's Equations, Ampere's Law, Biot-Savart Law

## The Attempt at a Solution

(a) To solve this, I used the following process:

V = Ed = $$\frac{Q}{C}$$ $$\Rightarrow$$ E = $$\frac{Q}{Cd}$$

$$\frac{dE}{dt}$$ = $$\frac{I}{Cd}$$ = $$\frac{I}{(\frac{\epsilon_{0}A}{d})d}$$ = $$\frac{I}{\epsilon_{0}A}$$

(b) I really have no clue, but I would appreciate some help...

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Which of Maxwell's equations gives B in terms of dE/dt?

Well, none of them really give B in terms of dE/dt, but the equation
$$\oint(Bds)$$ = $$\mu_{0}I + \epsilon_{0}\mu_{0}\frac{d\Phi_{E}}{dt}$$...
gives B in terms of electric flux...

and then also, the Lorentz force law also describes a relationship between B and E fields...
F = q(E + vB)...

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fedderenator said:
Well, none of them really give B in terms of dE/dt, but the equation
$$\oint(Bds)$$ = $$\mu_{0}I + \epsilon_{0}\mu_{0}\frac{d\Phi_{E}}{dt}$$...
gives B in terms of electric flux...

That's the one you want. If you imagine a small circle perpendicular to the plates of the capacitor, can you write the flux term in terms of dE/dt? You'll have to use the symmetry of the situation to get rid of the integral.

ideasrule said:
That's the one you want. If you imagine a small circle perpendicular to the plates of the capacitor, can you write the flux term in terms of dE/dt? You'll have to use the symmetry of the situation to get rid of the integral.

so would it be dE/dt*A where A is the circular area of the plates...and then int(ds) would be the radius away from the center given...

B = (mu_0*I + epsilon_0*mu_0*dE/dt*A)/r ? where r is the "radius" away from the center of the parallel plate configuration (5.00 cm)

No, that's not right. The circle on the integral sign means you have to chose a closed curve. What's the integral of B*dl around a circle?

ideasrule said:
No, that's not right. The circle on the integral sign means you have to chose a closed curve. What's the integral of B*dl around a circle?

I understand it would be 2pir for a circle, but that makes absolutely no sense with the value that they gave us because they are asking the magnetic field 5 cm from the center of the plate separation, not from the center of the actual plates themselves...and by the way, is everything else right in my equation?

fedderenator said:
I understand it would be 2pir for a circle, but that makes absolutely no sense with the value that they gave us because they are asking the magnetic field 5 cm from the center of the plate separation, not from the center of the actual plates themselves...

You can imagine a circle 5 cm in radius encircling the center of the plate separation. The integral of B*dl would then be 2*pi*r*B while d(phi)/dt=A*dE/dt, as you said.

I submitted that and I got it wrong...when I calculated, I got 3.60 microT

I get 1.8 microT.

Ok, I figured the answer out (its not 1.80 microT or 3.60 microT because I had tried both of those before)...it is completely off base from what we were attempting to do...first of all you were on the right track with half of my answer, 1.80 microT because you didn't include the mu_0*I term from Maxwell's Equation, and just include the term with the flux (mu_0*epsilon_0*A*dE/dt)/(2*pi*r), but since dE/dt = I/(epsilon_0*A), we end up with the classic magnetic field equation (mu_0*I)/(2*pi*r), but I knew from the beginning, we had to modify that equation...the only things that didnt make sense were the "I" and "r"...
So what I did, is assume a new current through the center of the circular capacitor (and then after it passes through, it returns back to the initial current "I" from the voltage source)...so since the initial effective radius, which I'll call "r" is shrunk down into the capacitor plates' circular radius, which I'll call "R", we have (with a new current through the capacitor of " I' ")
I'/I = (pi*r^2)/(pi*R^2) --------> and therefore, I' = (r^2/R^2)I
now if we apply Ampere's law, with I', we have lineint(B*ds) = mu_0*I'
B is constant, so you can pull it out of the integral and lineint(ds) = 2*pi*r, where "r" is the radial distance away from the center of the capacitor separation. Now, since
I' =(r^2/R^2)I we have B(2*pi*r) =mu_0*(r^2/R^2)I...So therefore,

B = $$\left(\frac{\mu_{0}I}{2\pi R^{2}}\right)r$$

= ((4pie-7)(.45)/(2*pi*(14e-2)^2))(5e-2) = 0.230 microT

## What is a charged capacitor?

A charged capacitor is an electrical component that stores energy in the form of an electric field. It is made up of two conductive plates separated by a dielectric material.

## How is the magnetic field of a charged capacitor calculated?

The magnetic field of a charged capacitor can be calculated using the formula B = μ0I/2πr, where μ0 is the permeability of free space, I is the current flowing through the capacitor, and r is the distance from the capacitor.

## What is the relationship between the electric and magnetic fields of a charged capacitor?

The electric and magnetic fields of a charged capacitor are perpendicular to each other and are both directly proportional to the charge and inversely proportional to the distance from the capacitor.

## How does the magnetic field of a charged capacitor change over time?

The magnetic field of a charged capacitor changes over time as the charge on the capacitor decreases due to the discharge of energy. This results in a decrease in the strength of the magnetic field.

## What are some real-world applications of the magnetic field of a charged capacitor?

The magnetic field of a charged capacitor is used in devices such as generators, motors, and speakers. It is also used in magnetic resonance imaging (MRI) machines in the medical field.

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