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- Engineering Electromagnetics 8th Edition William H. Hayt Original
- Engineering Electromagnetics 8th Edition William H. Hayt Original

The McGraw-Hill Companies. Engineering Electromagnetics. Sixth Edition. William H. Hayt, Jr.. John A. Buck. Textbook Table of Contents. The Textbook Table. 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 - 7th Edition - William H. Hayt - Solution Manual. Arsh Khan. CHAPTER 1 Given the vectors M = −10ax + 4ay − 8az and N.

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Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. industry, Professor Hayt joined the faculty of Purdue University, where he served as. Hayt,Buck Engineering Electromagnetics 6e Pdf previous post Harrison Advanced Engineering Dynamics Pdf. next post Heat Exchanger. Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual. The BookReader requires JavaScript to be enabled. Please check that your browser.

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By expanding Eq.

Use Eq. The initial velocity in x is constant, and so no force is applied in that direction.

We integrate once: Integrating a second time yields the z coordinate: Have 1 1 K. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7. We begin by visualizing the problem. The positions are then found by integrating vx and vy over time: 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. This will be the vector sum of the forces on the four sides.

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. Uniform current sheets are located in free space as follows: Find the magnitude of the total force acting to split the outer cylinder apart along its length: 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. Find the force exerted on the: Compute the vector torque on the wire segment using: The rectangular loop of Prob. Find the vector torque on the loop, referred to an origin: They will add together to give, in the loop plane: 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.

We observe two things here: So we must use the given origin. Find H in a material where: 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: Compute for: Find a H everywhere: This result will depend on the current and not the materials, and is: The normal component of H1 will now be: HT 2 tangential component of H2 at the boundary: The core shown in Fig.

A coil of turns carrying 12 mA is placed around the central leg. Find B in the: We now have mmf The air gap reluctance adds to the total reluctance already calculated, where 0. 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. I will use the reluctance method here. The reluctance of each gap is now 0. We are not sure what to use for the permittivity of steel in this case, so we use the iterative approach. 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. In this case we use the magnetization curve, Fig.

A toroidal core has a square cross section, 2. The currents return on a spherical conducting surface of 0. We thus perform the line integral of H over a circle, centered on the z axis, and parallel to the xy plane: From Problem 9.

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. In this case the coil lies in the yz plane. The rings are coplanar and concentric.

We use the result of Problem 8. That solution is reproduced below: Determine this force: Now for the right hand side. The location of the sliding bar in Fig.

Find the voltmeter reading at: Have 0. The rails in Fig. In this case, there will be a contribution to the current from the right loop, which is now closed. Develop a function of time which expresses the ohmic power being delivered to the loop: This will be Id 0.

Then D 1. Now B 2. We set the given expression for Jd equal to the result of part c to obtain: Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance Sec.

The parallel plate transmission line shown in Fig. Note that B as stated is constant with position, and so will have zero curl. We use the expression for input impedance Eq. The Hard Way: Thus 1. Find L, C, R, and G for the line: A transmitter and receiver are connected using a cascaded pair of transmission lines. At the operating frequency, Line 1 has a measured loss of 0.

The link is composed of 40m of Line 1, joined to 25m of Line 2. At the joint, a splice loss of 2 dB is measured. If the transmitted power is mW, what is the received power? The total loss in the link in dB is 40 0. Suppose a receiver is rated as having a sensitivity of -5 dBm — indicating the minimum power that it must receive in order to adequately interpret the transmitted data. Consider a transmitter having an output of mW connected to this receiver through a length of transmission line whose loss is 0.

What is the maximum length of line that can be used? From this result, we subtract the maximum dB loss to obtain the receiver sensitivity: For this impedance to equal 50 ohms, the imaginary parts must cancel. If so what are they? At the input end of the line, a DC voltage source, V0 , is connected.

Here, the line just acts as a pair of lossless leads to the impedance. In a circuit in which a sinusoidal voltage source drives its internal impedance in series with a load impedance, it is known that maximum power transfer to the load occurs when the source and load impedances form a complex conjugate pair.

The condition of maximum power transfer will be met if the input impedance to the line is the conjugate of the internal impedance. What average power is delivered to each load resistor? First, we need the input impedance. The parallel resistors give a net load impedance of 20 ohms. For the transmission line represented in Fig. A ohm transmission line is 0. The line is operating in air with a wavelength of 0. Determine the average power absorbed by each resistor in Fig.

The next step is to determine the input impedance of the 2. The power dissipated by the ohm resistor is now 1 V 2 1 Find s on both sections 1 and 2: A lossless transmission line is 50 cm in length and operating at a frequency of MHz. Determine s on the transmission line of Fig. Note that the dielectric is air: With the length of the line at 2. To achieve this, the imaginary part of the total impedance of part c must be reduced to zero so we need an inductor.

Continuing, for this value of L, calculate the average power: The power delivered to the load will be the same as the power delivered to the input impedance. Using normalized impedances, Eq. A line drawn from the origin through this point intersects the outer chart boundary at the position 0. With a wavelength of 1. On the WTL scale, we add 0. A straight line is now drawn from the origin though the 0. A compass is then used to measure the distance between the origin and zin.

This is close to the value of the VSWR, as we found earlier. Problem This is close to the computed inverse of yL , which is 1. Now, the position of zL is read on the outer edge of the chart as 0.

The point is now transformed through the line length distance of 1. Drawing a line between this mark on the WTG scale and the chart center, and scribing the compass arc length on this line, yields the normalized input impedance. On the WTG scale, we read the zL location as 0. The distance is then 0. Using a compass, we set its radius at the distance between the origin and zL.

What is s on the remainder of the line? This will be just s for the line as it was before. This would return us to the original point, requiring a complete circle around the chart one- half wavelength distance. The distance from the resistor will therefore be: With the aid of the Smith chart, plot a curve of Zin vs.

Then, using a compass, draw a circle beginning at zL and progressing clockwise to the positive real axis. The intersections of the lines and the circle give a total of 11 zin values. The table below summarizes the results. A fairly good comparison is obtained. We mark this on the positive real axis of the chart see next page. The load position is now 0.

A line is drawn from the origin through this point on the chart, as shown. We then scribe this same distance along the line drawn through the. A line is drawn from the origin through this location on the chart.

Drawing a line from the chart center through this point yields its location at 0. Alternately, use the s scale at the bottom of the chart, setting the compass point at the center, and scribing the distance on the scale to the left. This distance is found by transforming the load impedance clockwise around the chart until the negative real axis is reached. This distance in wavelengths is just the load position on the WTL scale, since the starting point for this scale is the negative real axis.

So the distance is 0. Transforming the load through this distance toward the generator involves revolution once around the chart 0. A line is drawn between this point and the chart center. This is plotted on the Smith chart below.

We then set on the compass the distance between yL and the origin. The same distance is then scribed along the positive real axis, and the value of s is read as 2. First we draw a line from the origin through zL and note its intersection with the WTG scale on the chart outer boundary.

We note a reading on that scale of about 0. To this we add 0. A line drawn from the 0. This is at the zero position on the WTL scale. The load is at the approximate 0. The wavelength on a certain lossless line is 10cm. We begin by marking zin on the chart see below , and setting the compass at its distance from the origin.

First, use a straight edge to draw a line from the origin through zin , and through the outer scale. We read the input location as slightly more than 0. The line length of 12cm corresponds to 1. Thus, to transform to the load, we go counter-clockwise twice around the chart, plus 0. A line is drawn to the origin from that position, and the compass with its previous setting is scribed through the line. A standing wave ratio of 2.

Probe measurements locate a voltage minimum on the line whose location is marked by a small scratch on the line. When the load is replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm toward the source from the scratch.

Find ZL: This is a possible location for the scratch, which would otherwise occur at multiples of a half-wavelength farther away from that point, toward the generator. As a check, I will do the problem analytically. At the point X, indicated by the arrow in Fig.

With the short circuit removed, a voltage minimum is found 5cm to the left of X, and a voltage maximum is located that is 3 times voltage of the minimum. Use the Smith chart to determine: Again, no Smith chart is needed, since s is the ratio of the maximum to the minimum voltage amplitudes. Now we need the chart.

This point is then transformed, using the compass, to the negative real axis, which corresponds to the location of a voltage minimum. On the chart, we now move this distance from the Vmin location toward the load, using the WTL scale. A line is drawn from the origin through the 0. We begin with the general phasor voltage in the line: We use the same equation for V z , which in this case reads: With a short circuit replacing the load, a minimum is found at a point on the line marked by a small spot of puce paint.

The 1m distance is therefore 3. Therefore, with the actual load installed, the Vmin position as stated would be 3. This being the case, the normalized load impedance will lie on the positive real axis of the Smith chart, and will be equal to the standing wave ratio.

The Smith chart construction is shown on the next page. This point is to be transformed to a location at which the real part of the normalized admittance is unity. The stub is connected at either of these two points. The stub input admittance must cancel the imaginary part of the line admittance at that point. This point is marked on the outer circle and occurs at 0. The length of the stub is found by computing the distance between its input, found above, and the short-circuit position stub load end , marked as Psc.

The length of the main line between its load and the stub attachment point is found on the chart by measuring the distance between yL and yin2 , in moving clockwise toward generator.

In this case, everything is the same, except for the load- end position of the stub, which now occurs at the Poc point on the chart.

This occurs at 0. The attachment point is found by transforming yL to yin1 , where the former point is located at 0. The lossless line shown in Fig. For the line to be matched, it is required that the sum of the normalized input admittances of the shorted stub and the main line at the point where the stub is connected be unity. So the input susceptances of the two lines must cancel.

This line is one-quarter wavelength long, so the normalized load impedance is equal to the normalized input admittance. The Smith chart construction is shown below. To cancel the input normalized susceptance of We therefore write 2. The two-wire lines shown in Fig. In this case, we have a series combination of the loaded line section and the shorted stub, so we use impedances and the Smith chart as an impedance diagram. The requirement for matching is that the total normalized impedance at the junction consisting of the sum of the input impedances to the stub and main loaded section is unity.

In the transmission line of Fig. First, the load voltage is found by adding voltages along the right side of the voltage diagram at the indicated times. The load voltage as a function of time is found by accumulating voltage values as they are read moving up along the right hand boundary of the chart.

The pulse amplitudes are calculated as follows: In the charged line of Fig. This problem accompanies Example Plots of the voltage and current at the resistor are then found by accumulating values from the left sides of the two charts, producing the plots as shown.

Plot the load resistor voltage as a function of time: With the left half of the line charged to V0 , closing the switch initiates at the switch location two voltage waves: The results are summarized as follows: A simple frozen wave generator is shown in Fig. Determine and plot the load voltage as a function of time: Closing the switches sets up a total of four voltage waves as shown in the diagram below. With positive z travel, and with Es along positive x, Hs will lie along positive y.

We use the fact that each to component of Es , there will be an orthogonal Hs component, oriented such that the cross product of Es with Hs gives the propagation direction. First, we note that if E at a given instant points in the negative x direction, while the wave propagates in the forward y direction, then H at that same position and time must point in the positive z direction. We use the general formula, Eq.

Using With Es in the positive y direction at a given time and propagating in the positive x direction, we would have a positive z component of Hs , at the same time. Find the distance a uniform plane wave can propagate through the material before: In a non-magnetic material, we would have: Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: I will demonstrate: We apply our equation to the result of part a: Note that in Problem Given, a MHz uniform plane wave in a medium known to be a good dielectric.

Also, the specified distance in part f should be 10m, not 1km. We use the good dielectric approximations, Eqs. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. From part a, we have 4. This will be 4.

The external and internal regions are non-conducting. We use J 1. Use The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respec- tively. The dielectric is lossless and the operating frequency is MHz. Calculate the resistance per meter length of the: A hollow tubular conductor is constructed from a type of brass having a conductivity of 1.

The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter length at a frequency of a dc: In this case the current density is uniform over the entire tube cross-section. We write: Therefore we can approximate the resistance using the formula: Most microwave ovens operate at 2. Since the. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0. Assuming the conductor is non-magnetic, determine the frequency and the conductivity: The outer conductor thickness is 0.

Use information from Secs. The result is squared, terms collected, and the square root taken. Consider a left-circularly polarized wave in free space that propagates in the forward z direc- tion.

Since the wave propagates in the positive y direction and has equal x and z amplitudes, we identify the polarization as left circular. With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the out- put. Suppose that the length of the medium of Problem Describe the polarization of the output wave in this case: Given the general elliptically-polarized wave as per Eq.

What percentage of the incident power density is transmitted into the copper? We nevertheless proceed: Next we apply Eq. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2. There are two possible answers. Its intrinsic impedance is therefore approximated well by Eq. Solve for z to obtain ln 8. This is found using Eq. This is given by Eq.

Pi Try measuring that. A MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown.

Determine the impedance of the unknown material: A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. The transmitted power fraction thus increases.

A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect con- ductor. This is a standing wave exhibiting circular polarization in time. Determine the standing wave ratio in front of the plate. Repeat Problem We can now evaluate the phase shift for the three given cases. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness.

Specify an arrangement of slabs and air spaces such that a the wave is totally transmitted through the stack: In this case, we look for a combination of half-wave sections. Let the inter-slab distances be d1 , d2 , and d3 from left to right. Two possibilities are i. Thus every thickness is one-quarter wavelength. The impedances transform as follows: The 50MHz plane wave of Problem Next we need the angle of refraction, which means that we need to know the refractive index of seawater at 50MHz.

Therefore, for s polarization,. The fraction transmitted is then 0. The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized.

A dielectric waveguide is shown in Fig. Find n0 in terms of n1 and n2: The prism of Fig. In the Brewster prism of Fig. The light is incident from air, and the returning beam also in air may be displaced sideways from the incident beam. More than one design is possible here. Using the result of Example For this to work, the Brewster angle must be greater than or equal to the critical angle.

Assume conductivity does not vary with frequency: In a good conductor: Over a small wavelength range, the refractive index of a certain material varies approximately. All frequency components arrive simultaneously.

Describe the pulse at the output of the second channel and give a physical explanation for what happened. In fact, we may write in general: The pulse, if originally transform-limited at input, will emerge, again transform-limited, at its original width.

Two aluminum-clad steel conductors are used to construct a two-wire transmission line. The radius of the steel wire is 0. The dielectric is air, and the center-to-center wire separation is 4 in. Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the bulk of the current resides in the aluminum, and we may neglect the steel. Each conductor of a two-wire transmission line has a radius of 0.

Pertinent dimensions for the transmission line shown in Fig. The conductors and the dielectric are non-magnetic. A transmission line constructed from perfect conductors and an air dielectric is to have a maximum dimension of 8mm for its cross-section. The line is to be used at high frequencies.

Specify its dimensions if it is: With the maximum dimension of 8mm, we have, using We use Eq. Line 1 is of 4mm width; line 2 unfortunately has been fabricated with a 5mm width. Determine the power loss in dB for waves transmitted through the junction. For line 1: How many modes propagate? Determine the maximum allowable plate separation, d: If the plate separation is 1 cm, determine the dielectric constant of the medium between plates: If the operating frequency is 32 GHz, which modes will propagate?

For the guide of Problem Assume a propagation distance of 10 cm: From Problem The So the answer is yes. In the guide of Figure To summarize, as frequency is lowered, the ray angle in guide 1 decreases, which leads to the incident angle at the interface increasing to eventually reach and surpass the critical angle. These will be: Two rectangular waveguides are joined end-to-end. This was already found in part a: Integrate the result of Problem First, the integration: This was considered in the discussion leading to Eq.

Using the result of Problem We use the condition expressed through The reasoning of part a applies to all modes, so the answer is the same, or 2. An asymmetric slab waveguide is shown in Fig. The minimum wave angle is thus determined by the greater of the two critical angles. Therefore, the percentage reduction required in the core radius will be 1. The answer to this can be found by inspecting Eq.

Using these values, along with our new equation, we write 0. In spherical coordinates, the components of E are given by and We next convert this to rectangular components: Prepare a curve, r vs. Substitute P directly to obtain: This will be just Assuming free space conditions, find: Since the field is constant at constant radius, we obtain the product: How much charge lies within the cylinder? Two point charges, 1 nC at 0, 0, 0. Use q. From the above field expression, the radial component magnitude is twice that of the theta component.

We use the general expression for the potential in the far field: This will be the surface of a cone, centered at the origin, along which E, in the ar direction, will exist. 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. Note first that current through the top and bottom surfaces will not exist, since J has no z component. We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume.

Find the total current I crossing the surface: Make the assumption that the electrons are emitted continuously as a beam with a 0. This is found using the equation of continuity: 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.

We may write the continuity equation for mass as follows, also invoking the divergence theorem: 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.

We make the estimate using the definition of divergence, but without taking the limit as the volume shrinks to zero: The continuity equation for mass reads: Therefore, the rate of change of density at the origin will be just the negative.

Using data tabulated in Appendix C, calculate the required diameter for a 2-m long nichrome wire that will dissipate an average power of W when V rms at 60 Hz is applied to it: Thus 3. Let the total current carried by this hybrid conductor be 80 A dc. 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. We first find J as a function of radius by dividing the current by the area of a sphere of radius r: 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. Converting all measurements to meters, the tube resistance over a 1 m length will be: The resistance of the filling will be: Find the magnitude of the electric field intensity in a conductor if: Find the equation of the conductor surface: Set the given potential function equal to 20, to find: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: The surface will be an equipotential, where the value of the potential is VP: I will work parts b and c in reverse order Find E at P: Such a unit normal can be construced from the result of part c: There are two reasons for this: Therefore the surface charge density at P is 0.

A more general method involves deriving the potential from the given field: The general procedure is to adjust the functions, f , such that the result for V is the same in all three integrations. Setting the given potential function equal to 0 and 60 and simplifying results in: First, find the electric field there: For each line charge, this will be: Vectors from all four charges to point A are: First, E at the origin is done as per the setup in part a, except the vectors are directed from the charges to the origin: We use the far-field potential for a z-directed dipole: The potential at any point is now: We just set the potential exression of part a equal to V to obtain: The concentration of both holes and electrons is 2.

Determine both the conductivity and the resistivity of this silicon sample: 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.

We need to find the components of E1 that are normal and tangent to the boundary, and then apply the appropriate boundary conditions. Since the magnitude is negative, the normal component points into region 1 from the surface. So in region 1: Again calculate E, D, Q, and the energy: With the source disconnected, the charge is constant, and thus so is D: 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. Which of these dielectrics will give the largest CVmax product for equal plate areas: 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.

R 0S 8. 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. Let w1 be the width of region 1. The above conditions enable us to write: Calculate the capacitance per square meter of surface area if: The region 0.

Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1m. Perfect dielectrics occupy the interior region: Then 6. These fields are plotted below. For a spherical capacitor, we know that: Again, find C: 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: 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 sketch is shown below. 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. As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression: 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. The result below could still be improved a little, but is nevertheless sufficient for a reasonable capacitance estimate.

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.

In this case NQ is the number of squares around the full perimeter of the circular conductor, or four times the number of squares shown in the drawing. 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.

The voltages at the grid points are shown below, where VA is found to be 19 V. 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: The four distances and potentials are now substituted into the given equation: Consider the configuration of conductors and potentials shown in Fig. Using the method described in Problem 10, write an expression for Vx not V0: Again, work to the nearest volt: The result is shown below, with values for the original grid points underlined: Use the computer to obtain values for a 0.

Work to the nearest 0. Values for the left half of the configuration are shown in the table below. 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. A few curvilinear squares are suggested: 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. If the potential of the inner conductor is V: We use the part a result, since the gravitational function must be the same as that for the electric potential.

We replace V0 by the maximum height, and multiply all dimensions by to obtain: The streamline is now specified by the equations: No, since the charge density is not zero. It is known that both Ex and V are zero at the origin. Our two equations are: We integrate the charge density found in part a over the specified volume: First, we find E: The problem did not provide information necessary to determine this.

Uniqueness specifies that there is only one potential that will satisfy all the given boundary conditions. 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. Find and sketch: We begin with the general solution of the one-dimensional Laplace equation in rectangular coordinates: The planes are parallel, and so we expect variation in potential in the direction normal to them. Using the two boundary condtions, our general potential function can be written: Working in volts and meters, we write the general one-dimensional solution to the Laplace equation in cylindrical coordinates, assuming radial variation: 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: From Eq. A narrow insulating strip separates them along the z axis. The boundary conditions are then substituted: Subtract the latter equation from the former to obtain: The two conducting planes illustrated in Fig. The medium surrounding the planes is air. For region 1, 0.

Use 1 dV 2. This will be. Then The capacitance will be Qnet Next, apply the boundary conditions: Find the capacitance between them: The region between the spheres is filled with a perfect dielectric. If the inner sphere is at V and the outer sphere at 0 V: 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. 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. These two equations can be unified to cover the entire range of x; the final expression for the electric field becomes: Use a development similar to that of Sec.

This situation is the same as that of Fig. The solution is found from Eq. We will use the first three terms to evaluate the potential at 3,4: 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.

Since we want V at the center of a square trough, it no longer matters on what boundary each of the given potentials is, and we can simply write: Solve for the potential at the center of the trough: The series solution will be of the form: From part a, the radial equation is: Functions of this form are called circular harmonics.

Find H in cartesian components at P 2, 3, 4 if there is a current filament on the z axis carrying 8 mA in the az direction: Find H if both filaments are present: 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. Find H at the origin: We use the Biot-Savart law, which in this case becomes: Again, find H at the origin: The parallel filamentary conductors shown in Fig.

We need an expression for H in cartesian coordinates. We can start with the known H in cylindrical for an infinite filament along the z axis: The results of part a should help: For the finite-length current element on the z axis, as shown in Fig. The Biot-Savart law reads: Calculate H at P 0, 0, 3: 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 Biot-Savart law is applied to the two wire segments using the following setup: Three uniform cylindrical current sheets are also present: 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. Find H at: This point lies within the lower slab above its midpoint. Thus the field will be oriented in the negative x direction. Referring to Fig. The total field will be this field plus the contribution from the upper slab current: Here the fields from both slabs will add constructively in the negative x direction: Since 0.

The field from the lower slab will be negative x-directed as well, leading to: This point lies above both slabs, where again fields cancel completely: Consider this situation as illustrated in Fig. 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.

Suppose the rectangular loop was drawn such that the outside z-directed segment is moved further and further away from the cylinder. 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. With our rectangular path set up as in part a, we have no path integral contributions from the two radial segments, and no contribution from the outside z-directed segment.

Find H everywhere: Between the cylinders, we are outside the inner one, so its field will not contribute. A toroid having a cross section of rectangular shape is defined by the following surfaces: The construction is similar to that of the toroid of round cross section as done on p. Inner and outer currents have the same magnitude. We can now proceed with what is requested: 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. Calculate H at: Here, we are either just inside or just outside the first current sheet, so both we will calculate H for both cases.

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. Current density is distributed as follows: 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. We can use the results for regions 1 and 2 to construct the field everywhere: The value of H at each point is given.

We use the approximation: 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. The x component of the curl is thus: If so, what is its value?

So, YES. The integral is: Let the surface have the ar direction: 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. When evaluating the curl of G using the formula in spherical coordinates, only one of the six terms survives: 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.

We first find the current density through the curl of the magnetic field, where three of the six terms in the spherical coordinate formula survive: A long straight non-magnetic conductor of 0. Here we use H outside the conductor and write: A solid nonmagnetic conductor of circular cross-section has a radius of 2mm. If the conductor is 1m in length and has a voltage of 1mV between its ends, find: We first need to find J, H, and B: The current density will be: This is part a over again, except we change the upper limit of the radial integration: This is entirely outside the current distribution, so we need B there: All surfaces must carry equal currents.

With this requirement, we find: Use an expansion in cartesian coordinates to show that the curl of the gradient of any scalar field G is identically equal to zero. We proceed as follows: Thus, using the result of Section 8. We can then write: The solenoid shown in Fig. Since H is only in the z direction, Vm should vary with z only.

Since we have parallel current sheets carrying equal and opposite currents, we use Eq. Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Fig. Select the proper zero reference and sketch the results on the figure: By expanding Eq.

Use Eq. The initial velocity in x is constant, and so no force is applied in that direction. We integrate once: Integrating a second time yields the z coordinate: Have 1 1 K. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7. We begin by visualizing the problem. We can construct the differential equations for the forces in x and in y as follows: The positions are then found by integrating vx and vy over time: It establishes a negative x-directed velocity as it leaves the field, given by the acceleration times the transit time, tt: 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. The field from the long wire now varies with position along the loop segment. This will be the vector sum of the forces on the four sides. 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. Uniform current sheets are located in free space as follows: Find the vector force per meter length exerted on a current filament carrying 7 mA in the aL direction if the filament is located at: Find the vector force on a 1-m length of the 1-mA filament and plot F versus k: The total B field arising from the two 25A filaments evaluated at the location of the 1-mA filament is, in cartesian components: Find the magnitude of the total force acting to split the outer cylinder apart along its length: 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: 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. Compute the vector torque on the wire segment using: The rectangular loop of Prob. Find the vector torque on the loop, referred to an origin: The fields from both current sheets, at the loop location, will be negative x-directed.

They will add together to give, in the loop plane: 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. Obtain an expression for the torque exerted on the finite conductor about an origin located at 0, 2, 0: 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. The field is uniform and so does not produce any translation of the loop. We observe two things here: So we must use the given origin.

Find H in a material where: 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: Three current sheets are located as follows: Find B everywhere: With this in mind, we can construct the following expressions for the B field in all four regions: Compute for: 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: Point P 2, 3, 1 lies on the planar boundary boundary separating region 1 from region 2. Find H2: The normal component of H1 will now be: The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg. Find B in the: We now have mmf The air gap reluctance adds to the total reluctance already calculated, where 0.

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.

I will use the reluctance method here. The reluctance of each gap is now 0. We are not sure what to use for the permittivity of steel in this case, so we use the iterative approach. 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. Calculate the total flux in the toroid if: This is d 0. In this case we use the magnetization curve, Fig.

We can begin with the value of 6 found in part a, assuming infinite permeability: I will leave the answer at that, considering the lack of fine resolution in Fig.

Determine the total energy stored in a spherical region 1cm in radius, centered at the origin in free space, in the uniform field: First we find the energy density: This field differs from H2 only by the negative x component, which is a non-issue since the component is squared when finding the energy density. A toroidal core has a square cross section, 2. First, the magnetic field strength will be the same as in part a, since the calculation is material-independent.

Three planar current sheets are located in free space as follows: The currents return on a spherical conducting surface of 0. We thus perform the line integral of H over a circle, centered on the z axis, and parallel to the xy plane: Find the inductance of the cone-sphere configuration described in Problem 9.

The inductance is that offered at the origin between the vertices of the cone: From Problem 9. Second method: Use the energy computation of Problem 9. The core material has a relative permeability of If the core is wound with a coil containing turns of wire, find its inductance: The stored energy within the specified volume will be: 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. A rectangular coil is composed of turns of a filamentary conductor. Find the mutual inductance in free space between this coil and an infinite straight filament on the z axis if the four corners of the coil are located at a 0,1,0 , 0,3,0 , 0,3,1 , and 0,1,1: In this case the coil lies in the yz plane.

Find the mutual inductance of this conductor system in free space: We first find the magnetic field inside the conductor, then calculate the energy stored there. It may be assumed that the magnetic field produced by I t is negligible. The magnetic flux will be: The location of the sliding bar in Fig.

Find the voltmeter reading at: Have 0.

First the flux through the loop is evaluated, where the unit normal to the loop is az. The rails in Fig. In this case, there will be a contribution to the current from the right loop, which is now closed. Develop a function of time which expresses the ohmic power being delivered to the loop: First, since the field does not vary with y, the loop motion in the y direction does not produce any time-varying flux, and so this motion is immaterial.

We can evaluate the flux at the original loop position to obtain: This will be Id 0. Then D 1. We set the given expression for Jd equal to the result of part c to obtain: Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance Sec. This we already found during the development in part a: We have.

The parallel plate transmission line shown in Fig. Neglect fields outside the dielectric. We evaluate the flux integral of Jd over the given cross section: Note that B as stated is constant with position, and so will have zero curl.

The equation is thus not valid with these fields. Nevertheless, we press on: We use the result of part a: The procedure here is similar to the development that leads to Eq. Begin by taking the curl of both sides of the Faraday law equation: In free space, the magnetic field of the uniform plane wave can be easily found using the intrinsic impedance: First, we note that if E at a given instant points in the negative x direction, while the wave propagates in the forward y direction, then H at that same position and time must point in the positive z direction.

We use the general formula, Eq. Using With Es in the positive y direction at a given time and propagating in the positive x direction, we would have a positive z component of Hs , at the same time. Find the distance a uniform plane wave can propagate through the material before: A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region.

In a non-magnetic material, we would have: Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: Then the power factor is P. Assume x-polarization for the electric field. At this point a flaw becomes evident in the problem statement, since solving this part in two different ways gives results that are not the same.

I will demonstrate: We apply our equation to the result of part a: Note that in Problem A positive y component of E requires a posi- tive z component of H for propagation in the forward x direction. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. From part a, we have 4. This will be 4. The external and internal regions are non-conducting.

We use J 1. Use The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respectively. The dielectric is lossless and the operating frequency is MHz. Calculate the resistance per meter length of the: Again, 70 applies but with a different conductor radius.