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3-46Solutions for Chapter 3 Problems1.Magnetic Fields and Cross ProductsP3.1: Find AxB for the following:a. A = 2ax 3ay + 4az, B = 5ay - 1az b. A = ar + 2af + 4az, B = 2ar + 6az c. A = 2ar + 5aq + 1af, B = ar + 3af (a)(b)(c)P3.2: If a parallelogram has a short side a, a long side b, and an interior angle q (the smaller of the two interior angles), the area of the parallelogram is given byDetermine how you would use the cross product of a pair of vectors to find the area of a parallelogram defined by the points O(0,0,0), P(6,0,0), Q(8,12,0) and R(2,12,0). (Assume dimensions in meters)Fig. P3.2A = 6ax, B = 2ax + 12ayA x B = 72az,Area = 72 m2P3.3: Given the vertices of a triangle P(1,2,0), Q(2,5,0) and R(0,4,7), find (a) the interior angles, (b) a unit vector normal to the surface containing the triangle and (c) the area of the triangle.Fig. P3.3(a) (b) (c) 2.Biot-Savarts LawP3.4: A segment of conductor on the z-axis extends from z = 0 to z = h. If this segment conducts current I in the +az direction, find H(0,y,0). Compare your answer to that of Example 3.2.Fig. P3.4We use Eqn. (3.7) and change the limits:Note that if the line of current is semi-infinite (goes from z = 0 to z = ), wed have:Fig. P3.5aP3.5: An infinite length line with 2.0 A current in the +ax direction exists at y = -3.0 m, z = 4.0 m. A second infinite length line with 3.0 A current in the +az direction exists at x = 0, y = 3.0 m. Find H(0, 0, 0).This situation is shown in Figure P3.5a.Ho = H1 + H2Referring to the figure,R = 3ay 4az, R = 5,aR = 0.6ay 0.8azaf = ax x aR = 0.80ay + 0.60azFig. P3.5bHo = 159ax + 51ay +38az mA/mOr with 2 significant digitsHo = 160ax + 51ay +38az mA/mP3.6: A conductive loop in the shape of an equilateral triangle of side 8.0 cm is centered in the x-y plane. It carries 20.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 16cm).The situation is illustrated in Figure P3.6a. The sketch in Figure P3.6b is used to find the +x-axis intercept for the triangle. By simple trigonometry we have:Now for one segment we adapt Eqn. (3.7):with rar = RaR,H1 = -4.7ax 0.68az mA/mNow by symmetry the total H contains only the az component:Htot = -2.0 mA/m az.Fig. P3.6aFig. P3.6bP3.7: A square conductive loop of side 10.0 cm is centered in the x-y plane. It carries 10.0 mA current clockwise when viewed from the +az direction. Find H(0, 0, 10cm).Fig. P3.7We find H for one section of the squareby adopting Eqn. (3.7):With rar = RaR, we haveR = -5ax + 10az, |R| = 11.18x10-2 maR = -0.447ax + 0.894az,af = -ay x aR = -0.894 ax 0.447 azNow by symmetry the total H contains only the az component:HTOT = -10.4az mA/mP3.8: A conductive loop on the x-y plane is bounded by r = 2.0 cm, r = 6.0 cm, f = 0 and f = 90. 1.0 A of current flows in the loop, going in the af direction on the r = 2.0 cm arm. Determine H at the origin. Fig. P3.8By inspection of the figure, we seethat only the arc portions of the loop contribute to H.From a ring example we have:For the r = a segment of the loop:At r = b: ;So P3.9: MATLAB: How close do you have to be to the middle of a finite length of current-carrying line before it appears infinite in length? Consider Hf(0, a, 0) is the field for the finite line of length 2h centered on the z-axis, and that Hi(0, a, 0) is the field for an infinite length line of current on the z-axis. In both cases consider current I in the +az direction. Plot Hf/Hi vs h/a.Adapting Eqn. (3.7), for the finite length line we have:For the infinite length of line:The ratio we wish to plot is:The MATLAB routine follows.% M-File: MLP0309%Consider the field for a finite line of length 2h%oriented on z-axis with current I in +z direction.%The field is to be found a distance a away from %the current on the y axis (point (0,a,0).%We want to compare this field with that of an infinite%length line of current.%Plot Hf/Hi versus h/a. We expect that as h/a grows large,%the line will appear more infinite to an observation %point at (0,a,0).hova=0.01:.01:100;HfovHi=hova./sqrt(1+(hova).2);semilogx(hova,HfovHi)xlabel(h/a)ylabel(Hf/Hi)Fig. P3.9bFig. P3.9agrid onP3.10: MATLAB: For the ring of current described in MATLAB 3.2, find H at the following points (a) (0, 0, 1m), (b) (0, 2m, 0), and (c) (1m, 1m, 0).%M-File: MLP0310%Find the magnetic field intensity at any observation point%resulting from a ring of radius a and current I,%in the aphi direction centered in the x-y plane.df=1; %increment in degreesa=1; %ring radius in mI=1; %current in ARo=input(vector location of observation point: );for j=1:df:360; Fr=j*pi/180; Rs=a*cos(Fr) a*sin(Fr) 0; as=unitvector(Rs); dL=a*df*(pi/180)*cross(0 0 1,as); Rso=Ro-Rs; aso=unitvector(Rso); dH=I*cross(dL,aso)/(4*pi*(magvector(Rso)2); dHx(j)=dH(1); dHy(j)=dH(2); dHz(j)=dH(3);endH=sum(dHx) sum(dHy) sum(dHz)Now to run the program: MLP0310vector location of observation point: 0 0 1H = -0.0000 -0.0000 0.1768 MLP0310vector location of observation point: 0 2 0H = 0 0 -0.0431 MLP0310vector location of observation point: 1 1 0H = 0 0 -0.1907 MLP0310vector location of observation point: 0 1 1H = 0.0000 0.0910 0.0768So we see:(a) H = 0.18 az A/m(b) H = -0.043 az A/m(c) H = -0.19 az A/m(extra) H = 9.1ay + 7.7 az mA/mP3.11: A solenoid has 200 turns, is 10.0 cm long, and has a radius of 1.0 cm. Assuming 1.0 A of current, determine the magnetic field intensity at the very center of the solenoid. How does this compare with your solution if you make the assumption that 10 cm 1 cm?Eqn. (3.10): Or H = 1960 A/m azThe approximate solution, assuming 10cm 1cm, isP3.12: MATLAB: For the solenoid of the previous problem, plot the magnitude of the field versus position along the axis of the solenoid. Include the axis 2 cm beyond each end of the solenoid.% M-File: MLP0312% Plot H vs length thru center of a solenoid%clcclear% initialize variablesN=200; %number of turnsh=0.10; %height of solenoida=0.01; %radius of solenoidI=1; %currentdz=0.001; %step change in zz=-.02:dz:h+.02;zcm=z.*100;Fig. P3.12A1=(h-z)./sqrt(h-z).2+a2);A2=z./sqrt(z.2+a2);Atot=A1+A2;H=N*I.*Atot/(2*h);% generate plotplot(zcm,H)xlabel(z(cm)ylabel(H (A/m)grid onP3.13: A 4.0 cm wide ribbon of current is centered about the y-axis on the x-y plane and has a surface current density K = 2p ay A/m. Determine the magnetic field intensity at the point (a) P(0, 0, 2cm), (b) Q(2cm, 2cm, 2cm).Fig. P3.13a(a) Because of the symmetry (Figure P3.13a), we can use a modified Eqn. (3.14):(b) Referring to Figure P3.13b;Fig. P3.13bThis is separated into 3 integrals, eachone solved via numerical integration,resulting in:H=1.1083ax +0.3032az 1.1083az; orH = 1.1ax 0.80az A/m3.Amperes Circuit LawP3.14: A pair of infinite extent current sheets exists at z = -2.0 m and at z = +2.0 m. The top sheet has a uniform current density K = 3.0 ay A/m and the bottom one has K = -3.0 ay A/m. Find H at (a) (0,0,4m), (b) (0,0,0) and (c) (0,0,-4m).Fig. P3.14We apply (a) (b) (c) H = 0P3.15: An infinite extent current sheet with K = 6.0 ay A/m exists at z = 0. A conductive loop of radius 1.0 m, in the y-z plane centered at z = 2.0 m, has zero magnetic field intensity measured at its center. Determine the magnitude of the current in the loop and show its direction with a sketch.Htot = HS + HLFig. P3.15For the loop, we use Eqn. (3.10):where here(sign is chosen opposite HS).So, I/2 = 3 and I = 6A.P3.16: Given the field H = 3y2 ax, find the current passing through a square in the x-y plane that has one corner at the origin and the opposite corner at (2, 2, 0).Referring to Figure P3.6, we evaluate the circulation of H around the square path.Fig. P3.16So we have Ienc = 24 A. The negativeSign indicates current is going in the -az direction.P3.17: Given a 3.0 mm radius solid wire centered on the z-axis with an evenly distributed 2.0 amps of current in the +az direction, plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 r 9 mm.Figure P3.17 shows the situation along with the Amperian Paths. We have:This will be true for each Amperian path.AP1: So: Fig. P3.17aFig. P3.17bAP2: Ienc = I, % MLP0317% generate plot for ACL problema=3e-3; %radius of solid wire (m)I=2; %current (A)N=30; %number of data points to plotrmax=9e-3; %max radius for plot (m)dr=rmax/N;for i=1:round(a/dr) r(i)=i*dr; H(i)=(I/(2*pi*a2)*r(i);endfor i=round(a/dr)+1:N r(i)=i*dr; H(i)=I/(2*pi*r(i);endplot(r,H)xlabel(rho(m)ylabel(H (A/m)grid onP3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J = 3r A/cm2 az (for r in cm) plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 r 8 cm.Well let a = 2 cm.AP1 (r a): Ienc = 2pa3, so The MATLAB plotting routine is as follows:% MLP0318% generate plot for ACL problema=2; %radius of solid wire (cm)N=40; %number of data points to plotrmax=8; %max radius for plot (cm)dr=rmax/N;Fig. P3.18for i=1:round(a/dr) r(i)=i*dr; H(i)=r(i)2;endfor i=round(a/dr)+1:N r(i)=i*dr; H(i)=a3/r(i);endplot(r,H)xlabel(rho(cm)ylabel(H (A/cm)grid onP3.19: An infinitesimally thin metallic cylindrical shell of radius 4.0 cm is centered on the z-axis and carries an evenly distributed current of 10.0 mA in the +az direction. (a) Determine the value of the surface current density on the conductive shell and (b) plot H as a function of radial distance from the z-axis over the range 0 r 12 cm.(a) (b) for r a we have:The MATLAB routine to generate the plot is as follows:% MLP0319% generate plot for ACL problema=4; %radius of solid wire (cm)N=120; %number of data points to plotI=10e-3; %current (A)rmax=12; %max plot radius(cm)dr=rmax/N;for i=1:round(a/dr) r(i)=i*dr; H(i)=0;endfor i=round(a/dr)+1:N r(i)=i*dr; H(i)=100*I/(2*pi*r(i);endplot(r,H)xlabel(rho(cm)ylabel(H (A/m)Fig. P3.19bgrid onFig. P3.19aP3.20: A cylindrical pipe with a 1.0 cm wall thickness and an inner radius of 4.0 cm is centered on the z-axis and has an evenly distributed 3.0 amps of current in the +az direction. Plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 r 10 cm.For each Amperian Path:Now, for r a, Ienc = 0 so H = 0.For a r 0.This is valid at any point above the sheet.Now, P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of radius a carrying current I in the +az direction and outer shell of radius b carrying the return current. Find the magnetic flux passing through an area of length h along the z-axis bounded by radius between a and b.For a r b, 6.Magnetic ForcesP3.32: A 1.0 nC charge with velocity 100. m/sec in the y direction enters a region where the electric field intensity is 100. V/m az and the magnetic flux density is 5.0 Wb/m2 ax. Determine the force vector acting on the charge.P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m2 ay. Determine the force vector acting on the charge.P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it enters a magnetic field B = 1000. T ay (recall that a tesla T = Wb/m2). Calculate the force vector on the charge.The cross-product:Evaluating we find: F = -50ax + 30az mNP3.35: What electric field is required so that the velocity of the charged particle in the previous problem remains constant?P3.36: An electron (with rest mass Me= 9.11x10-31kg and charge q = -1.6 x 10-19 C) has a velocity of 1.0 km/sec as it enters a 1.0 nT magnetic field. The field is oriented normal to the velocity of the electron. Determine the magnitude of the acceleration on the electron caused by its encounter with the magnetic field.P3.37: Suppose you have a surface current K = 20. ax A/m along the z = 0 plane. About a meter or so above this plane, a 5.0 nC charged particle is moving along with velocity v = -10.ax m/sec. Determine the force vector on this particle.P3.38: A m