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Need an account? Click here to sign up. Download Free PDF. Engineering electromagnetics [solution manual] william h. Hasibullah Mekaiel. A short summary of this paper. 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. This is the equation of a cylinder, centered on the x axis, and of radius 2. Thus 1. Each does, as shown above. 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. 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. This expression simplifies to the following quadratic: 0. Now, since the charge is at the origin, we expect to obtain only a radial component of EM. A uniform volume charge density of 0. What 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, the average density becomes 3. Uniform line charges of 0. 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.
This is evident just from the symmetry of the problem. 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.
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. Taking the sum of the fields at the origin from the surface and line charges, respectively, we find: 0. 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 dime are glued to 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. Again, since the coins are insulated, they retain their original charges. A point charge of 12 nC is located at the origin. 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 just integrate the charge density on that surface to find the flux that leaves it.
This layer, being of uniform density, will not contribute to D at P. These fluxes will thus cancel. The enclosed charge is the result of part a. Note also that the expression is valid for all x positive or negative values.
Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed 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.
We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces. Evaluate both sides of Eq. Evaluate the partial derivatives at the center of the volume. A spherical surface of radius 3 mm is centered at P 4, 1, 5 in free space. Use the. A cube of volume a 3 has its faces parallel to the cartesian coordinate surfaces.
Show that div D is zero everywhere except at the origin. Using the formula for divergence in spherical coordinates see problem 3. We note that D has only a radial component, and so flux would leave only through the cylinder sides. The total surface charge should be equal and opposite to the total volume charge. The plot is zero at larger radii.
Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other. The same is true for the left and right surfaces, since Dy does not vary with y. We could just as well position the two points at the same z location and the problem would not change. Halfway along this line is a point of symmetry in the field make a sketch to see this. This means that when starting from either point, the initial force will be the same. This is also found by going through the same procedure as in part a, but with the direction roles of P and Q reversed.
Repeat Problem 4. A point charge Q1 is located at the origin in free space. Three point charges, 0. The sketch will show that V maximizes to a value of 8. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them?
We keep in mind the definition of absolute potential as the work done in moving a unit positive charge from infinity to location r. This will be just How much charge lies within the cylinder?
Two point charges, 1 nC at 0, 0, 0. From the above field expression, the radial component magnitude is twice that of the theta component. Therefore, the given surface cannot be an equipotential the problem was ill-conceived. Only a surface of constant r could be an equipotential in this field. Four 0. Again find the total stored energy: This will be the energy found in part a plus the amount of work done in moving the fifth charge into position from infinity.
The latter is just the potential at the square center arising from the original four charges, times the new charge value, or 4.
Make the assumption that the electrons are emitted continuously as a beam with a 0. In part c, the net outward flux was found to be zero, and in part b, the divergence of J was found to be zero as will be its volume integral. Therefore, the divergence theorem is satisfied. Assuming that there is no transformation of mass to energy or vice-versa, it is possible to write a continuity equation for mass. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center.
The continuity equation for mass equates the divergence of the mass rate of flow mass per second per square meter to the negative of the density mass per cubic meter.
After setting up a cartesian coordinate system inside a star, Captain Kirk and his intrepid crew make measurements over the faces of a cube centered at the origin with edges 40 km long and parallel to the coordinate axes. Therefore, the rate of change of density at the origin will be just the negative. Thus 3. Let the total current carried by this hybrid conductor be 80 A dc. Find: a Jst.
We begin with the fact that electric field must be the same in the aluminum and steel regions. This comes from the requirement that E tangent to the boundary between two media must be continuous, and from the fact that when integrating E over the wire length, the applied voltage value must be obtained, regardless of the medium within which this integral is evaluated.
The total current passing radially outward through the medium between the cylinders is 3 A dc. Assume the cylinders are both of length l. So it works. A hollow cylindrical tube with a rectangular cross-section has external dimensions of 0. A current of A dc is flowing down the tube. Therefore the surface charge density at P is 0. The general procedure is to adjust the functions, f , such that the result for V is the same in all three integrations.
The concentration of both holes and electrons is 2. Electron and hole concentrations increase with temperature. Atomic hydrogen contains 5. In a certain region where the relative permittivity is 2. In Fig. The origin lies in region 1. Since the magnitude is negative, the normal component points into region 1 from the surface. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax , increase.
The voltage Vmax is limited by the field strength at which the dielectric breaks down, EBD. Trying this with the given materials yields the winner, which is barium titanate. A dielectric circular cylinder used between the plates of a capacitor has a thickness of 0. D will, as usual, be x-directed, originating at the top plate and terminating on the bottom plate.
The key here is that D will be constant over the distance between plates. This can be understood by considering the x-varying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x. The approximation becomes exact as the layer thicknesses approach zero. The width of the region containing R1 in Fig.
Having parallel capacitors, the capacitances will add, so R1 0 2. The region 0. Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1m. Then 6. These fields are plotted below. With reference to Fig. A 2 cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the potential of the cylinder be V and that of the plane be 0 V. Find the surface charge density on the: a cylinder at a point nearest the plane: The cylinder will image across the plane, producing an equivalent two-cylinder problem, with the second one at location 5 cm below the plane.
These dimensions are suitable for the drawing. The capacitance is thus. These dimensions are suitable for the actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from your sketch and from the exact formula. Symmetry allows us to plot the field lines and equipotentials over just the first quadrant, as is done in the sketch below shown to one-half scale.
Construct a curvilinear square map of the potential field between two parallel circular cylinders, one of 4-cm radius inside one of 8-cm radius. The two axes are displaced by 2. The drawing is shown below. Use of the drawing produces:. A solid conducting cylinder of 4-cm radius is centered within a rectangular conducting cylinder with a cm by cm cross-section.
Note that the five-sided region in the upper right corner has been partially subdivided dashed line in anticipation of how it would look when the next-level subdivision is done doubling the number of field lines and equipotentials.
NV is the number of squares between the circle and the rectangle, or 5. The inner conductor of the transmission line shown in Fig. The axes are displaced as shown. Some improvement is possible, depending on how much time one wishes to spend. Let the inner conductor of the transmission line shown in Fig. Construct a grid, 0.
Work to the nearest volt: The drawing is shown below, and we identify the requested voltage as 38 V. Use the iteration method to estimate the potentials at points x and y in the triangular trough of Fig. Work only to the nearest volt: The result is shown below. The mirror image of the values shown occur at the points on the other side of the line of symmetry dashed line.
Use iteration methods to estimate the potential at point x in the trough shown in Fig. Working to the nearest volt is sufficient. The result is shown below, where we identify the voltage at x to be 40 V. Note that the potentials in the gaps are 50 V.
Using the grid indicated in Fig. Half the figure is drawn since mirror images of all values occur across the line of symmetry dashed line. Conductors having boundaries that are curved or skewed usually do not permit every grid point to coincide with the actual boundary. Figure 6. The other two distances are found by writing equations for the circles: 0. The four distances and potentials are now substituted into the given equation:.
Consider the configuration of conductors and potentials shown in Fig. Use the computer to obtain values for a 0. Work to the nearest 0. Values along the vertical line of symmetry are included, and the original grid values are underlined.
Perfectly-conducting concentric spheres have radii of 2 and 6 cm. Find E and J everywhere: From symmetry, E and J will be radially-directed, and we note the fact that the current, I , must be constant at any cross-section; i. Since we know the voltage between spheres 1V , we can find the value of I through:.
This we find through. The cross-section of the transmission line shown in Fig. The result is independent of a, provided the proportions are maintained. The four radii are 1, 1. 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. Given the points M 0. The rotation direction is counter-clockwise when looking in the positive z direction. The rotation direction is clockwise when one is looking in the positive z direction. Two point charges of Q1 coulombs each are located at 0,0,1 and 0,0, 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. 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.
If the test charge is placed at the origin, the force on it is in the direction 0. So the two possible P coordinate sets are 0.
Now, since the charge is at the origin, we expect to obtain only a radial component of EM. What volume charge density, appropriate for such time durations, should be assigned to that subregion? A uniform volume charge density of 0. Within what distance from the z axis does half the total charge lie?
What 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, the average density becomes 3. This leaves only the y component integrand, which has even parity. 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. What force per unit area does each sheet exert on the other?
For the charged disk of Problem 2. An electric dipole discussed in detail in Sec. 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 dime are glued to 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. Again, since the coins are insulated, they retain their original charges.
Answer: zero. First, from part b, the point charge will now lie inside. Find D and E everywhere. The enclosed charge is the result of part a. Does this indicate a continuous charge distribution?
A spherical surface of radius 3 mm is centered at P 4, 1, 5 in free space. Use the results of Sec. Show that div D is zero everywhere except at the origin. Using the formula for divergence in spherical coordinates see problem 3. The total surface charge should be equal and opposite to the total volume charge. Repeat Problem 3. We could just as well position the two points at the same z location and the problem would not change.
This means that when starting from either point, the initial force will be the same. This is also found by going through the same procedure as in part a, but with the direction roles of P and Q reversed. We therefore expect the same answer for all three paths. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.
How much work must be done to move one charge to a point equidistant from the other two and on the line joining them? A point charge Q is located at the origin. The result may be checked by conversion to spherical coordinates.
This will be just The electron therefore moves approximately along a streamline. Where does it leave the plate and in what direction is it moving at the time?
Considering the result of part a, we would expect the exit to occur along the bottom edge of the plate. Two point charges, 1 nC at 0, 0, 0. Under what conditions does the answer agree with Eq. We perform a line integral of Eq.
Using Eq. Find the total stored energy by applying a Eq. Four 0. The latter is just the potential at the square center arising from the original four charges, times the new charge value, or 4. Assume that a uniform electron beam of circular cross-section with radius of 0.
The requirement is that we have constant current throughout the beam path. Assuming that there is no transformation of mass to energy or vice-versa, it is possible to write a continuity equation for mass. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center. In all three cases mentioned in part a, the conductance is one-half the original value if all dimensions are reduced by one-half.
This is easily shown using the given formula for conductance. Thus 3. Show that the ratio of the current densities in the two materials is independent of a and b. The total current passing radially outward through the medium between the cylinders is 3 A dc.
A hollow cylindrical tube with a rectangular cross-section has external dimensions of 0. We therefore expect no z variation in E, and also note that the line integral of E between the bottom and top plates must always give V0. At a certain temperature, the electron and hole mobilities in intrinsic germanium are given as 0.
If the electron and hole concentrations are both 2. With the electron and hole charge magnitude of 1. Electron and hole concentrations increase with temperature.
A semiconductor sample has a rectangular cross-section 1. The material has electron and hole densities of 1. Atomic hydrogen contains 5. The origin lies in region 1. Since the magnitude is negative, the normal component points into region 1 from the surface. With the dielectric gone, re-calculate E, D, Q, and the energy stored in the capacitor. In the absence of friction in removing the dielectric, the charge and energy have returned to the battery that gave it.
Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax , increase. Trying this with the given materials yields the winner, which is barium titanate.
With the battery left connected, the plates are moved apart to a distance of 10d. Determine by what factor each of the following quantities changes: a V0 : Remains the same, since the battery is left connected.
Therefore, E has decreased by a factor of 0. This is also consistent with D having been reduced by 0. D will, as usual, be x-directed, originating at the top plate and terminating on the bottom plate. The key here is that D will be constant over the distance between plates. This can be understood by considering the x-varying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x.
The approximation becomes exact as the layer thicknesses approach zero. Repeat Problem 6. Therefore, V0 has increased by a factor of A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy plane, centered at the origin.
Potential V0 is on the top plate; the bottom plate is grounded. Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1m. With reference to Fig. Two 16 copper conductors 1.
A 2 cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the potential of the cylinder be V and that of the plane be 0 V. Find the surface charge density on the: a cylinder at a point nearest the plane: The cylinder will image across the plane, producing an equivalent two-cylinder problem, with the second one at location 5 cm below the plane. This is a quick one if we have already solved 6. These dimensions are suitable for the drawing. The capacitance is thus.
These dimensions are suitable for the actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from your sketch and from the exact formula.
The two axes are displaced by 2. The drawing is shown below. Use of the drawing produces:. A solid conducting cylinder of 4-cm radius is centered within a rectangular conducting cylinder with a cm by cm cross-section. NV is the number of squares between the circle and the rectangle, or 5.
The inner conductor of the transmission line shown in Fig. The axes are displaced as shown. Some improvement is possible, depending on how much time one wishes to spend. For the coaxial capacitor of Problem 6. From Problem 6. A two-wire transmission line consists of two parallel perfectly-conducting cylinders, each hav- ing a radius of 0. A V battery is connected between the wires.
No, since the charge density is not zero. It is known that both Ex and V are zero at the origin. Both plates are held at ground potential. The problem did not provide information necessary to determine this. 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. Consider the parallel-plate capacitor of Problem 7.
Both plates are at ground potential. Repeat Problem 7. Therefore, the solutions to parts a and b are unchanged from Problem 7. The two conducting planes illustrated in Fig. The medium surrounding the planes is air. For region 1, 0. Then The capacitance will be Qnet What resistance is measured between the two perfect conductors? 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. The solution is found from Eq. Using thirteen terms,. The series converges rapidly enough so that terms after the sixth one produce no change in the third digit. 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. In Fig. Functions of this form are called circular harmonics.
Referring to Chapter 6, Fig. Construct a grid, 0. Work to the nearest volt: The drawing is shown below, and we identify the requested voltage as 38 V.
Use the iteration method to estimate the potentials at points x and y in the triangular trough of Fig. Work only to the nearest volt: The result is shown below. The mirror image of the values shown occur at the points on the other side of the line of symmetry dashed line. Use iteration methods to estimate the potential at point x in the trough shown in Fig.
The result is shown below, where we identify the voltage at x to be 40 V. Note that the potentials in the gaps are 50 V. Using the grid indicated in Fig. Conductors having boundaries that are curved or skewed usually do not permit every grid point to coincide with the actual boundary.
Figure 6. The other two distances are found by writing equations for the circles: 0. The four distances and potentials are now substituted into the given equation:. Using the method described in Problem 7.
Use a computer to obtain values for a 0. Work to the nearest 0. Values along the vertical line of symmetry are included, and the original grid values are underlined. The Biot-Savart method was used here for the sake of illustration. I will work this one from scratch, using the Biot-Savart law.
It is also possible to work this problem somewhat more easily by using Eq. Each carries a current I in the az direction. Determine the side length b in terms of a , such that H at the origin is the same magnitude as that of the circular loop of part a. Applying Eq. A disk of radius a lies in the xy plane, with the z axis through its center. Find H at any point on the z axis. Since the limits are symmetric, the integral of the z component over y is zero.
Find H in spherical coordinates a inside and b outside the sphere. The sketch below shows one of the slabs of thickness D oriented with the current coming out of the page. 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. We are now in a position to solve the problem. This point lies within the lower slab above its midpoint.
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. Thus H would not change with z. There would also be no change if the loop was simply moved along the z direction.
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. Inner and outer currents have the same magnitude.
We can now proceed with what is requested: a PA 1. We obtain 2. A balanced coaxial cable contains three coaxial conductors of negligible resistance. Assume a solid inner conductor of radius a, an intermediate conductor of inner radius bi , outer radius bo , and an outer conductor having inner and outer radii ci and co , respectively.
The intermediate conductor carries current I in the positive az direction and is at potential V0. A solid conductor of circular cross-section with a radius of 5 mm has a conductivity that varies with radius.
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. The x component of the curl is thus:.
To do this, we use the result of Problem 8. This leaves only the path segment that coindides with the axis, and that lying parallel to the axis, but outside. Their centers are at the origin. 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. Thus, using the result of Section 8.
The simplest form in this case is that involving the inverse hyperbolic sine. 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 circular orbit can be established if the magnetic force on the particle is balanced by the centripital force associated with the circular path. In either case, the centripital force must counteract the magnetic force. 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. 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. A planar transmission line consists of two conducting planes of width b separated d m in air, carrying equal and opposite currents of I A. Take the current in the top plate in the positive z direction, and so the bottom plate current is directed along negative z. The rectangular loop of Prob. Assume that an electron is describing a circular orbit of radius a about a positively-charged nucleus. 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 a H everywhere: This result will depend on the current and not the materials, and is: I 1. The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg. We now have mmf In Problem 9. Using this value of B and the magnetization curve for silicon. 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. A toroidal core has a square cross section, 2. The currents return on a spherical conducting surface of 0. Second method: Use the energy computation of Problem 9. The core material has a relative permeability of A coaxial cable has conductor dimensions of 1 and 5 mm. Find the inductance per meter length: The interfaces between media all occur along radial lines, normal to the direction of B and H in the coax line. B is therefore continuous and constant at constant radius around a circular loop centered on the z axis.
The rings are coplanar and concentric.
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