Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. industry, Professor Hayt joined the faculty of Purdue University, where he served as. Library of Congress Cataloging-in-Publication Data Hayt, William Hart, – Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. — 8th ed. p. cm. Engineering electromagnetics [solution manual] (william h. hayt jr. john a. buck – 6th edition). Hasibullah Mekaiel. Uploaded by. Hasibullah Mekaiel. CHAPTER 1 .
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Skip to main content. Log In Sign Up. Engineering electromagnetics [solution manual] william h. If points A and B are ten units apart, find the coordinates of point B. This is the point we are looking for. For the G field in Problem 1. Plots are shown below. Given the points M 0. So the projection will be: The angle is found through the dot product of the associated unit vectors, or: Describe the surfaces defined by the equations: This is the equation of a cylinder, centered on the x axis, and of radius 2.
Express in cartesian components: Express in cylindrical components: Each does, as shown above.
First transform the points to cartesian: Express the unit vector ax in spherical components at the point: Electromagneics, transform the point to spherical coordinates. First convert A and B to cartesian: A fifth 10nC positive charge is located at a point 8cm distant from the other charges.
By symmetry, the force on the fifth charge will be z-directed, and will be four times the z component of force produced by each of the four other charges. To solve this problem, the z coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates x, y, 0. We take its magnitude to be Q3. Find the total force on the charge at A. The force will be: This force in general will be: Note, however, that all three charges must lie in a straight line, and the location of Q3 will be along the vector R12 extended past Q2.
Therefore, we look for P3 at coordinates x, 2. With this restriction, the force becomes: This expression simplifies to the following quadratic: The field will take the general form: The total field at P will be: Determine E at P eletromagnetics, y, 0: This field will be: Now, since the charge is haayt the origin, we expect to obtain only a radial component of EM. Calculate the total charge present: A uniform volume charge density of 0. With the limits thus changed, the integral for the charge becomes: Hayh is the average volume charge density throughout this large region?
Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, bck average density becomes 3. Find the charge within the region: Uniform line charges of 0. This field will in general be: With the infinite line, we know that the field will have only a radial component in cylindrical coordinates or x and y components in cartesian. Therefore, at point P: Since all line charges are infinitely-long, we can write: What force per unit length does each line charge exert on the other?
The charges are parallel to the z axis and are separated by 0. We use the superposition integral: This is evident just from buc symmetry of the problem. Performing the z integration first on the x component, we obtain using tables: First, we recognize from symmetry that only a z component of E will be present. The superposition integral for the z component of E will be: Surface charge density is positioned in free space as follows: For this reason, the net field magnitude will be the same everywhere, whereas the field direction will depend on which side of a given sheet one is positioned.
This will be the magnitude at the other two points as well. Find E at the origin if slectromagnetics following charge distributions are present in free space: The sum of electromagneticd fields at the origin from each charge in order is: A uniform surface charge density of 0. Find E at the origin: Since each pair consists of equal and opposite charges, the effect at the origin is to double the field produce by one of each type.
Engineering Electromagnetics – 8th Edition – William H. Hayt – PDF Drive
Taking the sum of the fields at the origin from the surface and line charges, respectively, we find: An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged honorably by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a elechromagnetics are glued electromagndtics the thread so that they are not touching each other. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured.
The outside of the can is again touched momentarily to ground. The device is carefully disassembled with insulating gloves and tools. All coins were insulated during the entire procedure, so they will retain their original charges: Again, since the coins are insulated, they retain their original charges. A point charge of 12 nC is located at the origin. Note that hayyt point lies in the center of a symmetric electromavnetics of line charges, whose fields will all cancel at that point.
This sphere encloses the point charge, so its flux of 12 nC is included.
The flux from the line charges will equal the total line charge that lies within the sphere. We integrate over the surface to find: We just integrate the charge density on that surface to find the flux that leaves it. This point lies at radius 5 cm, and is thus inside the outermost charge layer. This layer, being of uniform density, will not contribute to D at P.
We know that in cylindrical coordinates, the layers at 1 and 2 cm will produce the flux density: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. These fluxes will thus cancel. The net outward flux becomes: Find the total flux leaving a sphere of radius 2 m if it is centered at a A 3, 1, 0: In this case the result will be the same if we move the sphere to the origin and keep the charges where they were.
To find the charge we integrate: The gaussian surface is a spherical shell of radius 1 mm.
The enclosed charge is the result of part a. This will be the same computation as in part b, except the gaussian surface now lies at 20 mm. The sheet charge can be thought of as an assembly of infinitely-long parallel strips that lie parallel to the y axis in the yz plane, and where each is of thickness dz. The field from each strip is that of electromagetics infinite line charge, and so we can construct the field at P from a single strip as: Note also that the expression is valid for all x positive or negative values.
Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be bukc and will be constant over a cylindrical surface of a fixed radius.
The set up is the same, except the upper limit of the above integral is 1 instead of r. These are plotted on the next page.
Engineering Electromagnetics – 8th Edition – William H. Hayt
Noting that the charges are spherically-symmetric, we elecctromagnetics that D will be radially-directed and will vary only with radius. The Gaussian cylinder now lies outside the charge, so 2. Volume charge density is located as follows: In this application, Eq. We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces.