A parallel-plate, air-filled capacitor is being charged as in ...

Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. What is the rate at which the electric field between the plates is changing?

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to calculate the induced magnetic field strength where we will have at a given time. P A 0.75 a conduction current circulating within wires connected to a parallel round plate capacitor, the diameter of each plate is eight centimeter. And we're being asked to calculate the induced magnetic field strength between the plates at points M and N located at three centimeter and six centimeter respectively from the central axis of the capacitor. The options given are a, the magnetic field strength for M zero tesla and for N five times 10 to the power of negative six tesla B for M five times 10 to the power of negative six tesla. And for N zero tesla, C for M five times 10 to the power of negative six tesla. And for N 2.5 times 10 to the power of negative six tesla lastly D for M five times 10 to the power of negative six tesla and for N 10 times 10 to the power of negative six tesla. So we will apply, apply api's law including displacement current to a circular pad within radius R from the central axis of the capacitor api's law states that both a conduction current and a changing electric flux act as sources of the magnetic field. And the law is going to be expressed as the integral of B multiplied by DL will equals to mu knot multiplied by IC plus ID, oops ID encl or enclosed. So new knot is going to be the magnetic constant and IC is going to be the conduction current enclosed in the circular path ID is going to be the conduction current also enclosed in the circular path. And at a distance R and by symmetry, the magnetic field B is going to be constant and tangent to the path. The line integral of the magnetic field around the circular path of 3D R is then going to equals the integral of B multiplied by DL will equals to the integral of just B multiplied by DL. And since the B is constant, we can pull it out of the integral through the integration principles. So this will just equals to B multiplied by the integral of DL which uh in a circle, the circumference DL is just going to be two pi R. So this will just equals to B multiplied by two pi R. So between the plates of the capacitor, the conduction current IC will equals to zero and the displacement current between the plates has the same magnitude as the conduction current in the wire. So that means ID will equals to 0.75 amp using equation that we have written in the first place, which is this equation right here, we will be able to obtain the expression of the induced magnetic field where B multiplied by two pi R will equals to mu not multiplied by IC plus id enclosed where IC as I have described previously will equals to zero. So in this case, B multiplied by two pi R will equals to me not multiplied by ID. So we Arang this B will then equals to one divided by two pi R multiplied by mu not multiplied by ID. So this is going to be the equation that we're going to use in order for us to calculate the magnetic field strength between the plates at points M and M. So let's start with, at point M BM will then equals to one divided by two pi multiplied by Dr which is given for DM to be located at three centimeter. And um converting that into si, we will then have three times 10 to the power of negative 2 m multiplied that by mu knot, which is a constant magnetic constant to be four pi times 10 to the power of negative seven Tala meter divided by amp multiplied that by ID which is uh known or given or we just determined earlier to be 0.75 amp calculating this, that will give us uh the magnetic field at point M to be 5.0 times 10 to the power of negative six tesla. Next, at point end, we will then have B or the magnetic field at N to then be using the same to be calculated using the same formula which is two pi multiplied by R which for NS six stems stand to the power of negative 2 m multiply that with four pi times 10 to the power of negative seven tesla meter per M for the mu knot multiplied by ID which is still 0.75 amp. And that will give us the magnetic field at point N to be 2.5 times 10 to the power of negative six tesla. So there we have it, the magnetic field uh strength at point M is five times 10 to the power of negative six tesla and at point N um 2.5 times 10 to the power of negative six tesla which will correspond to option C in our answer choices. So option C will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics. But other than that, that will be it for this video. Thank you.

Key Concepts

Ampere-Maxwell Law

Displacement Current

Magnetic Field in a Capacitor

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