ECE 3209 Electromagnetic Fields Final Exam Example. University of Virginia Solutions
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1 ECE 3209 Electromagnetic Fields Final Exam Example University of Virginia Solutions (print name above) This exam is closed book and closed notes. Please perform all work on the exam sheets in a neat and legible manner. Support your answers with calculations and clearly show your reasoning. You may use a calculator and a table of vector differential operators is provided. Unjustified answers WILL NOT receive credit. If more space is needed, you may use the backs of pages or your own paper. On my honor as a student, I have neither given nor received unauthorized assistance on this exam. (sign name above) Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Total (100%) 1
2 1 Problem 1. A lossless transmission line of length l = 30 meters has a characteristic impedance of Z 0 = 50 Ω. A load with impedance Z L = 25 Ω is connected to the end of the line. The frequency of operation is 10 MHz and the phase velocity of the transmission line is v p = m/s. find (a) The VSWR on the transmission line, (b) The reflection coefficient seen looking into the input of the transmission line, and (c) The impedance seen looking into the input of the transmission line. The reflection coefficient at the load is given by, Thus the VSWR is given by, Γ L = Z L Z 0 Z L + Z 0 = = 1 3 VSWR = 1 + Γ L 1 Γ L = 1 + 1/3 1 1/3 = 2 The reflection coefficient seen looking into the input of the line is Γ in = Γ L e j2θ where θ is the electrical length of the transmission line. If the physical length of the line is l, Thus, θ = 2πl λ = 2π(30 [m]) ( [m/s]/ [Hz]) = 2π [radians]
3 2 Γ in = Γ L e j4π = Γ L = 1 3 The impedance seen looking into the input of the transmission line is Z in = Z Γ in 1 Γ in = 50 2/3 4/3 [Ω] = Γ L = 25 Ω
4 3 Problem 2. (a) The electric field inside a circular cylinder of radius a and height h (0 < x 2 + y 2 < a and 0 < z < h) is given by [ E = ẑ c h + where c and b are constants. b ] (3z 2 h 2 ) (a 2 x 2 y 2 ) 6ϵ 0 (i) Assuming the medium within the cylinder is empty space, find the total charge enclosed within the cylinder. We can find the charge density using Gauss s Law in differential form, ρ v = ϵ 0 E E x = ϵ0 x + E y y + E z z ρ v = bz(a 2 x 2 y 2 ) = bz(a 2 r 2 ) where used the relation for cylindrical coordinates, r 2 = x 2 + y 2. To find the total charge enclosed, we integrate the charge density over the volume of the cylinder, Q = V ρ v dv = 2πb a 0 h (a 2 r 2 ) rdr 0 [ a 2 r 2 z dz = 2πb 2 r4 4 ] a 0 [ z 2 2 ] h 0
5 4 Q = πba4 h 2 4 (ii) If the electric field outside the cylinder is zero, determine the surface charge residing on the side of the cylinder (at r = x 2 + y 2 = a) and on the ends of the cylinder (at z = 0 and z = h). To find the surface charge, we need to recall the boundary condition for the perpendicular (normal) component of the electric field, ( E 1 E 2 ) ˆn = ρ s ϵ where ˆn is the normal unit vector pointing from region 2 to region 1. Taking region 2 to be the volume of the cylinder and region 2 to be the medium surrounding the cylinder, ˆn will be the outward normal to the cylinder surface. On the side of the cylinder, ˆn = ˆr, and the normal to this surface is perpendicular to the field (which is in the z direction). Thus, there is no surface charge on the side of the cylinder. On the ends of the cylinder, ˆn = ±z (depending on which end) and we have for the left side (at z = 0, ˆn = ẑ), [ ] ρ s z=0 = ϵ 0 (ẑ E) c z=0 = h ϵ 0 + bh2 (a 2 x 2 y 2 ) 6 and on the right side (at z = h, ˆn = ẑ), ρ s z=h = ϵ 0 (ẑ E) z=h = [ c ] h ϵ 0 bh2 (a 2 x 2 y 2 ) 3
6 5 (b) The figure below shows a contour consisting of two semicircular sections of radius a and b (a < b) and two linear sections, each of length b a. The contour is in empty space and carries a total charge Q that is distributed uniformly along its length. Find the electric field (magnitude and direction) at the center point P. We can find the field at the center by applying Coulomb s Law and integrating over the charge distribution. Taking the charge per unit length on the contour to be ρ l, we have E = 1 4πϵ 0 S ρ l ( r r ) r r 3 dl By symmetry, the electric fields due to the lefthand and righthand straight sections will cancel and we only need worry about the semicircular sections. Again, by symmetry, the electric field will be pointing along the y axis, so we only need consider the y contributions, giving us ρ l E = ŷ 4πϵ 0 [ 0 π sin ϕ π b 2 bdϕ + 0 sin ϕ a 2 adϕ ] ρ l E = ŷ 4πϵ 0 [ cos ϕ b 0 π cos ϕ π] a 0 = ŷ ρ l 4πϵ 0 [ ] [ 2 b 2 = ŷ ρ l a 2πϵ 0 ] a b ab
7 6 Note: if a = b, then the result goes to zero, as expected. Furthermore, if a < b, then the field points in the y direction, which is also expected since in this case the upper semicircle is closer to the center than the lower semicircle. Also note that the uniform linear charge density is given by, ρ l = Q π(a + b) + 2(b a)
8 7 Problem 3. Answer the following questions with a few short sentences and/or short calculations. Where appropriate, justify your answers. Unjustified answers will not receive credit. (a) Write down Ampére s Law in both differential and integral form for magnetostatic fields. For magnetostatic fields, there is no time-dependence. Thus, B d l = µ 0 J v ˆn da C S B = µ 0 Jv (b) Consider a straight conductor carrying steady current I (shown below) that passes through the center of a sphere of radius R. What is the total magnetic flux, through the spherical surface? Φ B = S B d a We know that the magnetic flux density, B, has no divergence. In other words, B ˆnda = 0, for any closed surface S including the spherical surface shown above!
9 8 (c) Consider the volume current density given below (k is a constant with appropriate units), [ ] J = k 2x 2 yˆx + 3y 2 zŷ (4xyz + 3yz 2 )ẑ Does this current density represent a steady current (that is, current flow in which there is no charge density accumulation over time)? If so, explain why. If not, give the rate at which the charge density is changing. From the continuity equation, J v = ρ v dt a steady current is one that has no divergence. Checking the divergence of the above current density, Thus, J = J x x + J y y + J z z = k(4xy + 6yz (4xy + 6yz)) = 0 The divergence is zero, so the current density above does represent a steady current!
10 Problem 4. 9 (a) Find the magnetic flux density B on both sides of a large thin sheet (in the z = 0 plane) carrying a radially ( r) directed surface current density of J s = ˆr I 2πr that flows uniformly outward from the center. Be sure to give both the magnitude and direction of B. Note that the coordinate system chosen for this problem has the z-axis pointing out of the page. (b) Demonstrate that the solution you found in part(a) satisfies the boundary conditions for magnetic fields in the z = 0 plane. The current in this problem spreads out radially from the origin in the xy-plane. Biot-Savart, From B = µ 0 4π S J s ˆ( r r ) r r 3 da As illustrated below, consider a point P above the x-axis. For each current element that contributes to field B-field at point P, there is a corresponding symmetric current element whose contribution to B at point P cancels the z-component, leaving only a field pointing
11 10 in the y-direction. Since the x-axis is arbitrary, we see the B field must thus in the ϕ (circumferential) direction for z > 0 and in the +ϕ direction for z < 0. As a result of the symmetry, we can find the B-field using Ampére s Law and applying it to a curved loop that symmetrically straddles the xy-plane and follows a r = constant contour parallel to the xy-plane as shown below: From Ampére s Law, C B d l = µ 0 I inc = B ϕ (2rΦ) = µ 0 Φ I 2πr rdϕ = µ 0I 2π Φ where Φ is the angle subtended by the arc of the path. Note that the sections of the path above and below the xy-plane both contribute to the path integral, but there is no contribution for the two vertical (z-directed) sections as the field is perpendicular to these paths. Consequently, the magnetic flux density is given by,
12 11 B = µ 0I 4πr { ˆϕ, z > 0 ˆϕ, z < 0 The boundary condition for B-fields can be expressed B 1 B 2 = µ 0 ( J s ˆn) ( ) where ˆn is the unit normal pointing from region 2 to region 1. Taking region 1 to be z > 0 and region 2 to be z < 0, and evaluating the difference in the B-fields at z = 0, we find B 1 B 2 = µ 0I 2πr ˆϕ Evaluating the right side of the boundary condition ( ) above, using the current in the xy-plane, µ 0 ( J s ˆn) = µ 0 I 2πr ˆr ẑ = µ 0I 2πr ˆϕ where we have used the relation ˆr ẑ = ˆϕ. Thus the field found in part (a) does indeed satisfy the boundary condition for B-fields.
13 12 Problem 5. Consider the situation below depicting two very long, tightly-wound solenoids, of radii a and b (with b > a), that are nested (i.e, they share the same axis and the solenoid of smaller radius lies within the solenoid of large radius). The same current, I, flows along each solenoid, but in opposite directions as shown in the figure (a) If the radius of the larger solenoid is twice that of the smaller solenoid (b = 2a) and the windings per unit length for the smaller solenoid (n 1 ) is four times that of the larger solenoid (n 2 = 0.25 n 1 ), then find the magnetic flux density ( B) in each of the three regions, (i) r < a, (ii) a < r < b, and (iii) r > b. (b) Determine the mutual inductance per unit length between the two solenoids. (c) Suppose a loop of wire, with resistor R and radius r > b, encircles the axis of the solenoids as shown in the cross-sectional diagram below, If the current I flowing in the solenoids is sinusoidal, I(t) = I 0 cos (ωt) what current flows around the circuit loop?
14 13 The magnetic flux density for a solenoid of n turns per unit length and carrying current I is (found from Ampe re s Law), B = ẑ µ 0 ni, inside and B = 0, outside Thus, for the nested solenoids in this problem, we have (using superposition) B = ẑµ 0 I(n 1 n 2 ) = 0.75µ 0 n 1 ẑ, r < a B = ẑµ 0 In 2 = 0.25µ 0 n 1 ẑ, a < r < b B = 0, r > b The mutual inductance (per unit length) between the solenoids can be found from the flux through one of the solenoids that is produced by current around the other solenoid. Consider the flux through the inner solenoid due to the outer solenoid, Φ inner = µ 0 In 2 (n 1 πa 2 ) = MI M = 0.25n 2 1πa 2 µ 0 This, of course, is the same mutual inductance we would get by finding the flux through the outer solenoid due to current flowing through the inner solenoid, Φ outer = µ 0 In 1 (n 2 πa 2 ) = MI M = 0.25n 2 1πa 2 µ 0 The net magnetic flux through the loop with resistor R is given by Φ total = µ 0 I(n 1 n 2 )πa 2 µ 0 In 2 π(b 2 a 2 ) = 0.75n 1 πa 2 µ 0 I 0.25n 1 π(4a 2 a 2 )µ 0 = 0.75n 1 πa 2 µ 0 I 0.75n 1 πa 2 µ 0 I = 0 Thus, the induced emf around the circuit is
15 14 emf = dφ total dt = 0
16 15 Problem 6. (a) Write Maxwell s equations in differential form (5%). E = ρ v ϵ 0, B = 0 E = B t B = µ 0 Jv + µ 0 ϵ 0 E t (b) Consider the circular parallel-plate capacitor filled with dielectric material, ϵ, shown below. A dc current is switched on at t = 0. Find the fields ( E, H) and the Poynting vector, S, as functions of position and time inside the capacitor. Assume that a d so that the fringing fields are negligible, all fields are completely confined within the capacitor, and charge distributes uniformly on the two conducting plates. Also, find the electromagnetic energy density, u, stored in the capacitor and show that, S + t U = 0 Applying Gauss s Law to the capacitor (assuming a uniform charge distribution on the plates, which is OK if we neglect the edge effects) gives us, S E d a = EA = q πa 2 ϵ A where a is the radius of the cylindrical capacitor. Noting that q = t 0 i(t)dt = I 0 t
17 16 for a constant current I 0 charging the capacitor, we have E = ẑ I 0t ϵπa 2, inside the capacitor Now, applying Ampére s Law to find the magnetic field, ) 2 C H d l = ϵ d dt S E d a = I 0 ( r a where we have chosen our surface and contour C to lie on a circle inside the capacitor of radius r. Thus, H = ˆϕ I 0r 2πa 2, inside the capacitor The Poynting vector (instantaneous form) is thus, S(t) = E H = I2 0 tr 2ϵπ 2 a 4 ˆr Note that S is directed radially inward. The energy density stored in the capacitor is, W = 1 2 ϵe µh2 = 1 2 ( I 2 0 t 2 ϵπ 2 a 4 + µi2 0 r 2 4π 2 a 4 ) Now, let s check Poynting s Theorem. The relevant terms are: and S = 1 r ) (rs r = I2 0 t ( 2r ) r 2ϵπ 2 a 4 = I2 0 t r ϵπ 2 a 4 W t = I2 0 t ϵπ 2 a 4 Thus, we see that
18 17 S + W t = 0 as required by Poynting s Theorem.
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