Hayt, John A. Morris 3rd ed. EMBED for wordpress. Advanced embedding details, examples, and help! Flag this item for. Graphic Violence ; Graphic Sexual Content ; texts. How is Chegg Study better than a printed Engineering Electromagnetics student solution manual from the bookstore?
Hayt] with solution manual Thanks a lot for providing this book. Library of Congress Cataloging-in-Publication Data. The x and p y coordinates of the nC charge will both be equal in magnitude to 4. The coodinates of the nC charge are then 3. Point charges of 1nC and -2nC are located at 0,0,0 and 1,1,1 , respectively, in free space. Determine the vector force acting on each charge. Point charges of 50nC each are located at A 1, 0, 0 , B 1, 0, 0 , C 0, 1, 0 , and D 0, 1, 0 in free space.
Find the total force on the charge at A. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with one charge at the origin, and with the three nearest charges at a, 0, 0 , 0, a, 0 , and 0, 0, a. Find an expression for the total vector force on the charge at P a, a, a , assuming free space: The total electric field at P a, a, a that produces a force on the charge there will be the sum of the fields from the other seven charges.
This expression simplifies to the following quadratic: 0. A crude device for measuring charge consists of two small insulating spheres of radius a, one of which is fixed in position.
The other is movable along the x axis, and is subject to a restraining force kx, where k is a spring constant. This will occur at location x for the movable sphere. No further motion is possible, so nothing happens.
Connections made to the two rings show a resistance of ohms between them. Using the two radii 1. The square washer shown in Fig. The inside and outside surfaces are perfectly-conducting. Having found this, we can construct the total resistance by using the fundamental square as a building block. These numbers are found from the curvilinear square plot shown. A two-wire transmission line consists of two parallel perfectly-conducting cylinders, each having a radius of 0.
A V battery is connected between the wires. A coaxial transmission line is modelled by the use of a rubber sheet having horizontal dimensions that are times those of the actual line. The model is 8 cm in height at the inner conductor and zero at the outer. No, since the charge density is not zero. It is known that both Ex and V are zero at the origin.
The problem did not provide information necessary to determine this. So they apply to different situations. No for V1 V2. Only V2 is, since it is given as satisfying all the boundary conditions that V1 does. The others, not satisfying the boundary conditions, are not the same as V1. B is then found through either equation; e. The region between the cylinders is filled with a homogeneous perfect dielectric.
If the inner cylinder is at V and the outer at 0V, find: a the location of the 20V equipotential surface: From Eq. A narrow insulating strip separates them along the z axis. The two conducting planes illustrated in Fig. The medium surrounding the planes is air. For region 1, 0. Then The capacitance will be Qnet The region between the spheres is filled with a perfect dielectric.
Concentric conducting spheres have radii of 1 and 5 cm. The potential of the inner sphere is 2V and that of the outer is -2V. Find: a V r : We use the general expression derived in Problem 7. Two coaxial conducting cones have their vertices at the origin and the z axis as their axis.
Cone A has the point A 1, 0, 2 on its surface, while cone B has the point B 0, 3, 2 on its surface. Integrate again to find: Let the volume charge density in Fig. Use a development similar to that of Sec.
The solution is found from Eq. Using thirteen terms, and. The series converges rapidly enough so that terms after the sixth one produce no change in the third digit. Thus, quoting three significant figures, The four sides of a square trough are held at potentials of 0, 20, , and 60 V; the highest and lowest potentials are on opposite sides.
Find the potential at the center of the trough: Here we can make good use of symmetry. The solution for a single potential on the right side, for example, with all other sides at 0V is given by Eq. Functions of this form are called circular harmonics. The Biot- Savart method was used here for the sake of illustration. A current filament of 3ax A lies along the x axis.
Each carries a current I in the az direction. The parallel filamentary conductors shown in Fig. For the finite-length current element on the z axis, as shown in Fig. Since the limits are symmetric, the integral of the z component over y is zero. Let a filamentary current of 5 mA be directed from infinity to the origin on the positive z axis and then back out to infinity on the positive x axis. The problem asks you to find H at various positions. Before continuing, we need to know how to find H for this type of current configuration.
The sketch below shows one of the slabs of thickness D oriented with the current coming out of the page. The problem statement implies that both slabs are of infinite length and width. For example, if the sketch below shows the upper slab in Fig. Thus H will be in the positive x direction above the slab midpoint, and will be in the negative x direction below the midpoint. Reverse the current, and the fields, of course, reverse direction. We are now in a position to solve the problem.
This point lies within the lower slab above its midpoint. Thus the field will be oriented in the negative x direction. Referring to Fig. Since 0. There sec. The only way to enclose current is to set up the loop which we choose to be rectangular such that it is oriented with two parallel opposing segments lying in the z direction; one of these lies inside the cylinder, the other outside.
The loop is now cut by the current sheet, and if we assume a length of the loop in z of d, then the enclosed current will be given by Kd A. If we assume an infinite cylinder length, there will be no z dependence in the field, since as we lengthen the loop in the z direction, the path length over which the integral is taken increases, but then so does the enclosed current — by the same factor.
Thus H would not change with z. There would also be no change if the loop was simply moved along the z direction. Again, using the Biot-Savart law, we note that radial field components will be produced by individual current elements, but such components will cancel from two elements that lie at symmetric distances in z on either side of the observation point. We would expect Hz outside to decrease as the Biot-Savart law would imply but the same amount of current is always enclosed no matter how far away the outer segment is.
We therefore must conclude that the field outside is zero. Between the cylinders, we are outside the inner one, so its field will not contribute. Inner and outer currents have the same magnitude. We can now proceed with what is requested: a PA 1. We obtain 2. A current filament on the z axis carries a current of 7 mA in the az direction, and current sheets of 0.
We require that the total enclosed current be zero, and so the net current in the proposed cylinder at 4 cm must be negative the right hand side of the first equation in part b. Symmetry does help significantly in this problem. As a consequence of this, we find that the net current in region 1, I1 see the diagram on the next page , is equal and opposite to the net current in region 4, I4. Also, I2 is equal and opposite to I3. H from all sources should completely cancel along the two vertical paths, as well as along the two horizontal paths.
Assuming the height of the path is. Therefore, H will be in the opposite direction from that of the right vertical path, which is the positive x direction. Answer: No. Reason: the limit of the area shrinking to zero must be taken before the results will be equal. The value of H at each point is given. Each curl component is found by integrating H over a square path that is normal to the component in question.
Along each segment, the field is assumed constant, and so the integral is evaluated by summing the products of the field and segment length 4 mm over the four segments.
If so, what is its value? Their centers are at the origin. This problem was discovered to be flawed — I will proceed with it and show how. The reader is invited to explore this further. Integrals over x, to complete the loop, do not exist since there is no x component of H. The path direction is chosen to be clockwise looking down on the xy plane.
A long straight non-magnetic conductor of 0. A solid nonmagnetic conductor of circular cross-section has a radius of 2mm. All surfaces must carry equal currents. Use an expansion in cartesian coordinates to show that the curl of the gradient of any scalar field G is identically equal to zero. Thus, using the result of Section 8. The solenoid shown in Fig. Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Fig.
By expanding Eq. Use Eq. The initial velocity in x is constant, and so no force is applied in that direction. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7.
A rectangular loop of wire in free space joins points A 1, 0, 1 to B 3, 0, 1 to C 3, 0, 4 to D 1, 0, 4 to A. The wire carries a current of 6 mA, flowing in the az direction from B to C. A filamentary current of 15 A flows along the entire z axis in the az direction. Note that by symmetry, the forces on sides AB and CD will be equal and opposite, and so will cancel. Find the total force on the rectangular loop shown in Fig.
We wish to find the force acting to split the outer cylinder, which means we need to evaluate the net force in one cartesian direction on one half of the cylinder. Since the outer cylinder is a two-dimensional current sheet, its field exists only just outside the cylinder, and so no force exists.
If this cylinder possessed a finite thickness, then we would need to include its self-force, since there would be an interior field and a volume current density that would spatially overlap. Two infinitely-long parallel filaments each carry 50 A in the az direction. Find the force exerted on the: a filament by the current strip: We first need to find the field from the current strip at the filament location.
A current of 6A flows from M 2, 0, 5 to N 5, 0, 5 in a straight solid conductor in free space. An infinite current filament lies along the z axis and carries 50A in the az direction. The rectangular loop of Prob. Find the vector torque on the loop, referred to an origin: a at 0,0,0 : The fields from both current sheets, at the loop location, will be negative x-directed.
This filament carries a current of 3 A in the ax direction. An infinite filament on the z axis carries 5 A in the az direction. Assume that an electron is describing a circular orbit of radius a about a positively-charged nucleus.
The hydrogen atom described in Problem 16 is now subjected to a magnetic field having the same direction as that of the atom. What are these decreases for the hydrogen atom in parts per million for an external magnetic flux density of 0. We first write down all forces on the electron, in which we equate its coulomb force toward the nucleus to the sum of the centrifugal force and the force associated with the applied B field.
With the field applied in the same direction as that of the atom, this would yield a Lorentz force that is radially outward — in the same direction as the centrifugal force. Finally, 1m e2 a 2 B 2.
Calculate the vector torque on the square loop shown in Fig. So we must use the given origin. Then M 0. At radii between the currents the path integral will enclose only the inner current so, 3. Find the magnitude of the magnetization in a material for which: a the magnetic flux density is 0. Let its center lie on the z axis and let a dc current I flow in the az direction in the center conductor.
Find a H everywhere: This result will depend on the current and not the materials, and is: I 1. Point P 2, 3, 1 lies on the planar boundary boundary separating region 1 from region 2. The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg. We now have mmf The flux in the center leg is now In Problem 9. Using this value of B and the magnetization curve for silicon steel,. Using Fig. A toroidal core has a circular cross section of 4 cm2 area.
The mean radius of the toroid is 6 cm. There is a 4mm air gap at each of the two joints, and the core is wrapped by a turn coil carrying a dc current I1.
The reluctance of each gap is now 0. From Fig. Then, in the linear material, 1. This is still larger than the given value of. The result of 0. A toroid is constructed of a magnetic material having a cross-sectional area of 2. There is also a short air gap 0. This is d 0. I will leave the answer at that, considering the lack of fine resolution in Fig.
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