Mastering Physics

1 January 2017

To learn more, read your instructor’s Grading Policy [Switch to Standard Assignment View] [ Print ] Electric Field Due to Increasing Flux Learning Goal: To work through a straightforward application of Faraday’s law to find the EMF and the electric field surrounding a region of increasing flux Faraday’s law describes how electric fields and electromotive forces are generated from changing magnetic fields.

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This problem is a prototypical example in which an increasing magnetic flux generates a finite line integral of the electric field around a closed loop that surrounds the changing magnetic flux through a surface bounded by that loop. A cylindrical iron rod with cross-sectional area is oriented with its symmetry axis coincident with the z axis of a cylindrical coordinate system as shown. It has a uniform magnetic field inside that varies according to . In other words, the magentic field is always in the positive z direction, and it has no other components.

For your convenience, we restate Faraday’s law here: , where is the line integral of the electric field, and the magnetic flux is given by , where is the angle between the magnetic field and the local normal to the surface bounded by the closed loop. Direction: The line integral and surface integral reverse their signs if the reference direction of or is reversed. The right-hand rule applies , then the fingers point along .

You are free to here: If the thumb of your right hand is taken along take the loop anywhere you choose, although usually it makes sense to choose it to lie along the path of the circuit you are considering. Part A Find ositive. Hint A. 1 Selecting the loop Hint not displayed , the electromotive force (EMF) around a loop that is at distance from the z axis, where is restricted to the region outside the iron rod as shown. Take the direction shown in the figure as session. masteringphysics. com/myct/courseHome?

This law was extended by Maxwell to include a new type of “current” that is due to changing electric fields: The first term on the right-hand side, . , describes the effects of the usual electric current due to as usual. The second term, moving charge. In this problem, that current is designated , is called the displacement current; it was recognized as necessary by Maxwell. His motivation was largely to make Ampere’s law symmetric with Faraday’s law of induction when the electric fields and magnetic fields are reversed.

By calling for the production of a magnetic field due to a change in electric field, this law lays the groundwork for electromagnetic waves in which a changing magnetic field session. masteringphysics. com/myct/courseHome? start=1 2/23 12/3/11 MasteringPhysics: Course Home generates an electric field whose change, in turn, sustains the magnetic field. We will discuss these issues later. (Incidentally, a third type of “current,” called magnetizing current, should also be added to account for the presence of changing magnetic materials, but it will be neglected, as it has been in the equation above.

The purpose of this problem is to consider a classic illustration of the need for the additional displacement current term in Ampere’s law. Consider the problem of finding the magnetic field that loops around just outside the circular plate of a charging capacitor. The cone-shaped surface shown in the figure has a current passing through it, so Ampere’s law indicates a finite value for the field integral around this loop. However, a slightly different surface bordered by the same loop passes through the center of the capacitor, where there is no current due to moving charge.

To get the same loop integral independent of the surface it must be true that either a current or an increasing electric field that passes through the Amperean surface will generate a looping magnetic field around its edge. The objective of this example is to introduce the displacement current, show how to calculate it, and then to show that the displacement current is identical to the conduction current . Assume that the capacitor has plate area and an electric field between the plates. Take to be the permeability of free space and to be the permittivity of free space.

Part A First find , the line integral of around a loop of radius located just outside the left capacitor plate. This can be found from the usual current due to moving charge in Ampere’s law, that is, without the displacement current. Find an expression for this integral involving the current the introduction. Correct and any needed constants given in Part B Now find an expression for , the same line integral of around the same loop of radius located just outside the left capacitor plate as before. Use the surface that passes between the plates session. masteringphysics. com/myct/courseHome? start=1 /23 12/3/11 MasteringPhysics: Course Home of the capacitor, where there is no conduction current.

This should be found by evaluating the amount of displacement current in the Ampere-Maxwell law above. Hint B. 1 Find the electric flux Hint not displayed Hint B. 2 Express in terms of Hint not displayed Express your answer in terms of the electric field between the plates area , and any needed constants given in the introduction. , , the plate ANSWER: = Correct A necessary consistency check Part C We now have two quite different expressions for the line integral of the magnetic field around the same loop.

The point here is to see that they both are intimately related to the charge on the left capacitor plate. First find the displacement current Hint C. 1 Find the flux using Gauss’s law Hint not displayed Hint C. 2 Find the displacement current Hint not displayed in terms of . session. masteringphysics. com/myct/courseHome? start=1 4/23 12/3/11 MasteringPhysics: Course Home Express your answer in terms of introduction. ANSWER: = , , and any needed constants given in the Correct Part D Now express the normal current Express your answer in terms of introduction.

ANSWER: = Correct Using Gauss’s law, you have shown that the displacement current from the changing electric field between the plates equals the current from the flow of charge through the wire onto that plate. This means that the Ampere-Maxwell law can consistently treat cases in which the normal current due to the flow of charge is not continuous. This realization was a great boost to Maxwell’s confidence in the physical validity of his new displacement-current term. , in terms of the charge on the capacitor plate.

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