Fig. P3.14

Ampere’s 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).

We apply

(a)

(b)

(c) H = 0

P3.15: 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 + HL

For 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. P3.15

Fig. P3.16

Fig. P3.17a Fig. P3.17b

So 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 ≤ ≤ 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:

AP2: Ienc = I,

% MLP0317% generate plot for ACL problem

Fig. P3.18

a=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*a^2))*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 on

P3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J = 3 A/cm2 az (for in cm) plot the magnetic field intensity H versus radial distance from the z-axis over the range 0 ≤ ≤ 8 cm.

We’ll let a = 2 cm.

AP1 ( < a):

and

AP2 ( > a): Ienc = 2a3, so

The MATLAB plotting routine is as follows:% MLP0318% generate plot for ACL problem

a=2; %radius of solid wire (cm)N=40; %number of data points to plotrmax=8; %max radius for plot (cm)dr=rmax/N;

for i=1:round(a/dr) r(i)=i*dr; H(i)=r(i)^2;end

Fig. P3.19b

for i=round(a/dr)+1:N r(i)=i*dr; H(i)=a^3/r(i);endplot(r,H)xlabel('rho(cm)')ylabel('H (A/cm)')grid on

P3.19: 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 ≤ ≤ 12 cm.

(a)

(b) for < a, H = 0. For > 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)')grid on

Fig. P3.19a

Fig. P3.20bFig. P3.20a

P3.20: 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 ≤ ≤ 10 cm.

For each Amperian Path:

Now, for < a, Ienc = 0 so H = 0.

For a < < b,

% MLP0320% generate plot for ACL problem

a=4; %inner radius of pipe (cm)b=5; %outer radius of pipe(cm)N=120; %number of data points to plotI=3; %current (A)rmax=10; %max radius for plot (cm)dr=rmax/N;aoverdr=a/drboverdr=b/dr

Fig. P3.21a

for i=1:round(a/dr) r(i)=i*dr; H(i)=0;end

for i=round(a/dr)+1:round(b/dr) r(i)=i*dr; num(i)=I*(r(i)^2-a^2); den(i)=2*pi*(b^2-a^2)*r(i); H(i)=100*num(i)/den(i);end

for i=round(b/dr)+1:N r(i)=i*dr; H(i)=100*I/(2*pi*r(i));end

plot(r,H)xlabel('rho(cm)')ylabel('H (A/m)')grid on

P3.21: An infinite length line carries current I in the +az direction on the z-axis, and this is surrounded by an infinite length cylindrical shell (centered about the z-axis) of radius a carrying the return current I in the –az direction as a surface current. Find expressions for the magnetic field intensity everywhere. If the current is 1.0 A and the radius a is 2.0 cm, plot the magnitude of H versus radial distance from the z-axis from 0.1 cm to 4 cm.

Fig. P3.21b

The MATLAB routine used to generate Figure P3.21b is as follows:

% MLP0321% generate plot for ACL problemclcclear

a=2; %inner radius of cylinder(cm)N=80; %number of data points to plot

I=1; %current (A)rmax=4; %max radius for plot (cm)dr=rmax/N;

for i=1:40 r(i)=.1+(i-1)*dr; H(i)=100*I/(2*pi*r(i));end

for i=40:N r(i)=i*dr; H(i)=0;end

plot(r,H)xlabel('rho(cm)')ylabel('H (A/m)')grid on

P3.22: Consider a pair of collinear cylindrical shells centered on the z-axis. The inner shell has radius a and carries a sheet current totaling I amps in the +az direction while the outer shell of radius b carries the return current I in the –az direction. Find expressions for the magnetic field intensity everywhere. If a = 2cm, b = 4cm and I = 4A, plot the magnitude of H versus radial distance from the z-axis from 0 to 8 cm.

The MATLAB routine used to generate Figure P3.22b is as follows:% MLP0322% generate plot for ACL problem

Fig. P3.22a Fig. P3.22b

a=2; %inner radius of coax (cm)b=4; %outer radius of coax(cm)N=160; %number of data points to plot

I=4; %current (A)rmax=8; %max radius for plot (cm)dr=rmax/N;aoverdr=a/drboverdr=b/dr

for i=1:round(a/dr) r(i)=i*dr; H(i)=0;end

for i=round(a/dr)+1:round(b/dr) r(i)=i*dr; H(i)=100*I/(2*pi*r(i));end

for i=round(b/dr)+1:N r(i)=i*dr; H(i)=0;end

plot(r,H)xlabel('rho(cm)')ylabel('H (A/m)')grid on

P3.23: Consider the toroid in Figure 3.55 that is tightly wrapped with N turns of conductive wire. For an Amperian path with radius less than a, no current is enclosed and therefore the field is zero. Likewise, for radius greater than c, the net current

enclosed is zero and again the field is zero. Use Ampere’s Circuital Law to find an expression for the magnetic field at radius b, the center of the toroid.

Within the toroid, H = Ha, so

Then, Ienc by the Amperian path is: Ienc = NI.

4. Curl and the Point Form of Ampere’s Circuital LawP3.24: Find for the following fields:

a. A = 3xy2/z ax

b. A = sin2 a – 2 z cos a

c. A = r2sin ar + r/cos a

(a)

(b)

(c)

P3.25: Find J at (3m, 60, 4m) for H = (z/sin) a – (2/cos) az A/m.

Now find J by evaluating at the given point:

P3.26: Suppose H = y2ax + x2ay A/m. a. Calculate around the path , where A(2m,0,0),

B(2m,4m,0), C(0,4m,0) and D(0,0,0).b. Divide this by the area S (2m*4m = 8m2).

c. Evaluate at the center point.d. Comment on your results for (b) & (c).

(a) Referring to the figure, we evaluate

So we have

(b) dividing by S = 8m2, we have -2 C/m2

(c) Evaluating the curl of H:

, and at the center point (x = 1 and y = 2) we

have

(d) In this particular case, even though S is of appreciable size.

P3.27: For the coaxial cable example 3.8, we found:

Fig. P3.26

a. Evaluate the curl in all 4 regions.b. Calculate the current density in the conductive regions by dividing the current

by the area. Are these results the same as what you found in (a)?

(a)

(b)

Comment: is confirmed.

P3.28: Suppose you have the field H = r cos a A/m. Now consider the cone specified by = /4, with a height a as shown in Figure 3.56. The circular top of the cone has a radius a.

a. Evaluate the right side of Stoke’s theorem through the dS = dSa surface.b. Evaluate the left side of Stoke’s theorem by integrating around the loop.

(a)

ar derivative:

Fig. P3.28

a derivative:

So,

Now we must integrate this over the a surface:

(b)

Clearly in this case the circulation of H is the easiest approach.

5. Magnetic Flux DensityP3.29: An infinite length line of 3.0 A current in the +ay direction lies on the y-axis. Find the magnetic flux density at P(7.0m,0,0) in (a) Teslas, (b) Wb/m2, and (c) Gauss.

P3.30: Suppose an infinite extent sheet of current with K = 12ax A/m lies on the x-y plane at z = 0. Find B for any point above the sheet. Find the magnetic flux passing through a 2m2 area in the x-z plane for z > 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 < < 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 N

P3.35: 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.

Fig. P3.39

P3.38: A meter or so above the surface current of the previous problem there is an infinite length line conducting 1.0 A of current in the –ax direction. Determine the force per unit length acting on this line of current.

P3.39: Recall that the gravitational force on a mass m is where, at the earth’s surface, g = 9.8 m/s2 (-az). A line of 2.0 A current with 100. g mass per meter length is horizontal with the earth’s surface and is directed from west to east. What magnitude and direction of uniform magnetic flux density would be required to levitate this line?

By inspection, B = Bo(-ax)

The unit conversion to arrive at Newtons is as follows:

So we have Bo = 0.490 Wb/m2, and B = 0.490 Wb/m2 (-ax) (directed north)

Fig. P3.41

P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of 0.10 kg/m. One line sits on the ground and conducts 200. A in the +ax direction, and the other one, 1.0 cm above the first (and parallel), has sufficient current to levitate. Determine the current and its direction for line 2.

Here we will use

So solving for I2:

P3.41: In Figure 3.57, a 2.0 A line of current is shown on the z-axis with the current in the +az direction. A current loop exists on the x-y plane (z = 0) that has 4 wires (labeled 1 through 4) and carries 1.0 mA as shown. Find the force on each arm and the total force acting on the loop from the field of the 2.0 A line.

So for B to C: F12 = -0.20 nN az

Likewise, from D to A: F12 = +0.20 nN az

P3.42: MATLAB: Modify MATLAB 3.4 to find the differential force acting from each individual differential segment on the loop. Plot this force against the phi location of the segment.

%MLP0342%modify ML0304 to find dF acting from the field% of each segment of current; plot vs phi

clearclcI=1; %current in Aa=1; %loop radius, in mmu=pi*(4e-7); %free space permeabilityaz=[0 0 1]; %unit vector in z directionDL1=a*2*(pi/180)*[0 1 0]; %Assume 2 degree increments%DL1 is the test element vector%F is the angle phi in radians%xi,yi is location of ith element on the loop%Ai & ai = vector and unit vector from origin% to xi,yi%DLi is the ith element vector%Ri1 & ri1 = vector and unit vector from ith% point to test point

for i=1:179 phi(i)=i*2; F=2*i*pi/180; xi=a*cos(F); yi=a*sin(F); Ai=[xi yi 0]; ai=unitvector(Ai); DLi=(pi*a/90)*cross(az,ai); Ri1=[a-xi -yi 0]; ri1=unitvector(Ri1); num=mu*I*cross(DLi,ri1); den=4*pi*(magvector(Ri1)^2); B=num/den; dFvect=I*cross(DL1,B); dF(i)=dFvect(1);endplot(phi,dF)xlabel('angle in degrees')

Fig. P3.42

ylabel('the differential force, N')

P3.43: MATLAB: Consider a circular conducting loop of radius 4.0 cm in the y-z plane centered at (0,6cm,0). The loop conducts 1.0 mA current clockwise as viewed from the +x-axis. An infinite length line on the z-axis conducts 10. A current in the +az direction. Find the net force on the loop.

The following MATLAB routine shows the force as a function of radial position around the loop. Notice that while there is a net force in the -y direction, the forces in the z-direction cancel.

% MLP0343% find total force and torque on a loop of% current next to a line of current

% variables% I1,I2 current in the line and loop (A)% yo center of loop on y axis (m)% uo free space permeability (H/m)% N number of segements on loop% a loop radius (m)% dalpha differential loop element% dL length of differential section% DL diff section vector% B1 I1's mag flux vector (Wb/m^2)% Rv vector from center of loop% to the diff segment% ar unit vector for Rv% y,z the location of the diff segment

clc

Fig. P3.43

clear

% initialize variablesI1=10;I2=1e-3;yo=.06;uo=pi*4e-7;N=180;a=0.04;dalpha=360/N;dL=a*dalpha*pi/180;ax=[1 0 0];

% perform calculationsfor i=1:N dalpha=360/N; alpha=(i-1)*dalpha; phi(i)=alpha; z=a*sin(alpha*pi/180); y=yo+a*cos(alpha*pi/180); B1=-(uo*I1/(2*pi*y))*[1 0 0]; Rv=[0 y-.06 z]; ar=unitvector(Rv); aL=cross(ar,ax); DL=dL*aL; dF=cross(I2*DL,B1); dFx(i)=dF(1); dFy(i)=dF(2); dFz(i)=dF(3);endplot(phi,dFy,phi,dFz,'--k')legend('dFy','dFz')Fnet=sum(dFy)

Running the program we get:

Fnet = -4.2932e-009>>So Fnet = -4.3 nN ay

P3.44: MATLAB: A square loop of 1.0 A current of side 4.0 cm is centered on the x-y plane. Assume 1 mm diameter wire, and estimate the force vector on one arm resulting from the field of the other 3 arms.

% MLP0344 V2%% Square loop of current is centered on x-y plane. Viewed% from the +z axis, let current go clockwise. We want to% find the force on the arem at x = +2 cm resulting from the% current in arms at y = -2 cm, x = -2 cm and y = +2 cm.

% Wentworth, 12/3/03

% Variables% a side length (m)% b wire radius (m)% I current in loop (A)% uo free space permeability (H/m)% N number of segments for each arm% xi,yi location of test arm segment (at x = +2 cm)% xj,yj location of source arm segment (at y = -2 cm)% xk,yk location of source arm segment (at x = -2 cm)% xL,yL location of source arm segment (at y = +2 cm)% dLi differential test segment vector% dLj, dLk, dLL diff vectors on sources% Rji vector from source point j to test point i% aji unit vector of Rji

clc;clear;

a=0.04;b=.0005;I=1;uo=pi*4e-7;N=80;

for i=1:N xi=(a/2)+b; yi=-(a/2)+(i-0.5)*a/N; ypos(i)=yi; dLi=(a/N)*[0 -1 0]; for j=1:N xj=-(a/2)+(j-0.5)*a/N; yj=-a/2-b; dLj=(a/N)*[-1 0 0]; Rji=[xi-xj yi-yj 0];

aji=unitvector(Rji); num=I*CROSS(dLj,aji); den=4*pi*(magvector(Rji))^2; H=num/den; dHj(j)=H(3); end for k=1:N yk=(-a/2)+(k-0.5)*a/N; xk=-(a/2)-b; dLk=(a/N)*[0 1 0]; Rki=[xi-xk yi-yk 0]; aki=unitvector(Rki); num=I*CROSS(dLk,aki); den=4*pi*(magvector(Rki))^2; H=num/den; dHk(k)=H(3); end for L=1:N xL=(-a/2)+(L-0.5)*a/N; yL=(a/2)+b; dLL=(a/N)*[1 0 0]; RLi=[xi-xL yi-yL 0]; aLi=unitvector(RLi); num=I*CROSS(dLL,aLi); den=4*pi*(magvector(RLi))^2; H=num/den; dHL(L)=H(3); end H=sum(dHj)+sum(dHk)+sum(dHL); B=uo*H*[0 0 1]; F=I*CROSS(dLi,B); dF(i)=F(1);endFtot=sum(dF)plot(ypos,dF)

Running the program:

Ftot =

7.4448e-007

So Ftot = 740 nN Fig. P3.44

P3.45: A current sheet K = 100ax A/m exists at z = 2.0 cm. A 2.0 cm diameter loop centered in the x-y plane at z = 0 conducts 1.0 mA current in the +a direction. Find the torque on this loop.

P3.46: 10 turns of insulated wire in a 4.0 cm diameter coil are centered in the x-y plane. Each strand of the coil conducts 2.0 A of current in the a direction. (a) What is the magnetic dipole moment of this coil? Now suppose this coil is in a uniform magnetic field B = 6.0ax + 3.0ay + 6.0az Wb/m2, (b) what is the torque on the coil?

(a)

(b)

P3.47: A square conducting loop of side 2.0 cm is free to rotate about one side that is fixed on the z-axis. There is 1.0 A current in the loop, flowing in the –az direction on the fixed side. A uniform B-field exists such that when the loop is positioned at = 90, no torque acts on the loop, and when the loop is positioned at = 180 a maximum torque of 8.0 N-m az occurs. Determine the magnetic flux density.

At = 90°, . Also, since B is in direction of m, and therefore B = ±Boax.At = 180°, , and Therefore, B = -Boax and mBo = 8x10-6, so

7. Magnetic MaterialsP3.48: A solid nickel wire of diameter 2.0 mm evenly conducts 1.0 amp of current. Determine the magnitude of the magnetic flux density B as a function of radial distance from the center of the wire. Plot to a radius of 2 mm.

Fig. P3.48

% MLP0348% generate plot for ACL problem

a=2e-3; %radius of solid wire (m)I=1; %current (A)N=30; %number of data points to plotrmax=4e-3; %max radius for plot (m)dr=rmax/N;uo=pi*4e-7;ur=600;

for i=1:round(a/dr) r(i)=i*dr; B(i)=(ur*uo*I/(2*pi*a^2))*r(i);endfor i=round(a/dr)+1:N r(i)=i*dr; B(i)=uo*I/(2*pi*r(i));Endrmm=r*1000;plot(rmm,B)xlabel('rho(cm)')ylabel('B (Wb/m^2)')grid on

8. Boundary ConditionsP3.49: A planar interface separates two magnetic media. The magnetic field in media 1 (with r1) makes an angle 1 with a normal to the interface. (a) Find an equation for 2, the angle the field in media 2 (that has r2) makes with a normal to the interface, in terms of 1 and the relative permeabilities in the two media. (b) Suppose media 1 is

Fig. P3.49

nickel and media 2 is air, and that the magnetic field in the nickel makes an 89 angle with a normal to the surface. Find 2.

P3.50: MATLAB: Suppose the z = 0 plane separates two magnetic media, and that no surface current exists at the interface. Construct a program that prompts the user for r1 (for z < 0), r2 (for z > 0), and one of the fields, either H1 or H2. The program is to calculate the unknown H. Verify the program using Example 3.11.

% M-File: MLP0350%% Given H1 at boundary between a pair of% materials with no surface current at boundary,% calculate H2.%clcclear% enter variablesdisp('enter vectors quantities in brackets,')disp('for example: [1 2 3]')ur1=input('relative permeability in material 1: ');ur2=input('relative permeability in material 2: ');a12=input('unit vector from mtrl 1 to mtrl 2: ');F=input('material where field is known (1 or 2): ');Ha=input('known magnetic field intensity vector: ');

if F==1 ura=ur1; urb=ur2; a=a12;

else ura=ur2; urb=ur1; a=-a12;end

% perform calculationsHna=dot(Ha,a)*a;Hta=Ha-Hna;Htb=Hta;Bna=ura*Hna; %ignores uo since it will factor outBnb=Bna;Hnb=Bnb/urb;display('The magnetic field in the other medium is: ')Hb=Htb+Hnb

Now run the program (for Example 3.11):enter vectors quantities in brackets,for example: [1 2 3]relative permeability in material 1: 6000relative permeability in material 2: 3000unit vector from mtrl 1 to mtrl 2: [0 0 1]material where field is known (1 or 2): 1known magnetic field intensity vector: [6 2 3]

ans =The magnetic field in the other medium is: Hb = 6 2 6

For a second test, run the program for problem P3.52(a).

enter vectors quantities in brackets,for example: [1 2 3]relative permeability in material 1: 4relative permeability in material 2: 1unit vector from mtrl 1 to mtrl 2: [0 0 -1]material where field is known (1 or 2): 1known magnetic field intensity vector: [3 0 4]

ans =The magnetic field in the other medium is: Hb = 3 0 16

Fig. P3.51

P3.51: The plane y = 0 separates two magnetic media. Media 1 (y < 0) has r1 = 3.0 and media 2 (y > 0) has r2 = 9.0. A sheet current K = (1/o) ax A/m exists at the interface, and B1 = 4.0ay + 6.0az Wb/m2. (a) Find B2. (b) What angles do B1 and B2

make with a normal to the surface?

(a)

(b)

(c)

(d)

(e)

(f)

(g)

Now for step (e):

Angles:

P3.52: Above the x-y plane (z > 0), there exists a magnetic material with r1 = 4.0 and a field H1 = 3.0ax + 4.0az A/m. Below the plane (z < 0) is free space. (a) Find H2, assuming the boundary is free of surface current. What angle does H2 make with a normal to the surface? (b) Find H2, assuming the boundary has a surface current K = 5.0 ax A/m.

(a)

(1) , (2) , (3) ,(4)

(5) , (6) , (7)

(b)Now step (6) becomes , shere a21 = az.

Let’s let , then

Solve for A and B:

; so A = 3 and B = 5

Finally, .

P3.53: The x-z plane separates magnetic material with r1 = 2.0 (for y < 0) from magnetic material with r2 = 4.0 (for y > 0). In medium 1, there is a field H1 = 2.0ax + 4.0ay + 6.0az A/m. Find H2 assuming the boundary has a surface current K = 2.0ax – 2.0az A/m.

P3.54: An infinite length line of 2 A current in the +az direction exists on the z-axis. This is surrounded by air for ≤ 50 cm, at which point the magnetic media has r2 = 9.0 for > 50 cm. If the field in media 2 at = 1.0 m is H = 5.0a A/m, find the sheet current density vector at = 50. cm, if any.

Method 1:

From just the line of current we would have

Now, since is the contribution from the sheet current.

Method 2:From I1 at boundary we have

since H

varies as 1/. So

Faraday’s law and EMF

Fig. P4.9

P4.9: The magnetic flux density increases at the rate of 10 (Wb/m2)/sec in the z direction. A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch) and magnitude of the induced current in the conducting loop.

I=

I=10mA clockwise (when viewedFrom +z)

P4.10: A bar magnet is dropped through a conductive ring. Indicate in a sketch the direction of the induced current when the falling magnet is just above the plane of the ring and when it is just below the plane of the ring, as shown in Figure 4.22.

Refer to Figure P4.10.When the north pole first goes through the loop, flux is increasing and the current induced to oppose this change in flux is as shown.

When the south pole is exiting the loop, flux is decreasing and the current induced acts to oppose this change in flux.

P4.11: Considering Figure 4.7, suppose the area of a single loop of the pair is 100 cm2, and the magnetic flux density is constant over the area of the loops but changes

Fig. P4.10

with time as where Bo = 4.0 mWb/m2 and = 0.30 Np/sec. Determine VR at 1, 10, and 100 seconds.

at t = 1 sec, VR = 17.8 Vat t = 10 sec, VR = 1.20 Vat t = 100 sec, VR = 2.25x10-18 V

P4.12: Sometimes a transformer is used as an impedance converter, where impedance is given by v/i. Find an expression for the impedance Z1 seen by the primary side of the transformer in Figure 4.11 that has a load impedance Z2 terminating the secondary.

We have and

P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It is placed in a plane normal to a magnetic field increasing with time as B = 1.0 t Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center of the loop resulting from the induced current, and compare this with the original magnetic flux density that generated the induced current at t = 1.0 sec.

We find the distributed resistance of the loop and work the problem assuming this resistance is lumped in one spot as shown in the figure.(a) The induced current is Vemf divided by the distributed resistance of the wire loop.

(note that this answer has no time dependence)

(b) The field at the center of the loop from a single arm of the loop is found from Eqn. (3.7):

So

P4.16: Referring to Figure 4.23, suppose a conductive bar of length h = 2.0 cm moves with velocity u = -1.0 m/s a towards an infinite length line of current I = 4.0 A. Find an expression for the voltage from one end of the bar to the other when reaches 10 cm and indicate which end is positive.In Figure P4.16, an imaginary circuit has been chosen. For the chosen circulation direction, we have the sign for Vemf as shown. Then,

,

Therefore, the bottom of the bar is positive.

Fig. P4.13

Fig. P4.16

Fig. P4.19bFig. P4.19a

P4.17: Suppose we have a conductive bar moving along a pair of conductive rails as in Figure 4.12, only now the magnetic flux density is B = 4.0ax + 3.0az Wb/m2. If R = 10. , w = 20. cm, and uy = 3.0 m/s, calculate the current induced and indicate its direction.

(clockwise when viewed from the +z axis)

P4.18: The radius r of a perfectly conducting metal loop in free space, situated in the x-y plane, increases at the rate of (r)-1 m/sec. A break in the loop has a small 2.0 ohm resistor across it. Meanwhile, there exists a magnetic field B = 1.0 az T. Determine the current induced in the loop, and show in a sketch the direction of flow.

Here we’ve assumed dS = -dSaz to get iind and Vemf as shown. Our approach will be to find , then Vemf = -d/dt.

P4.19: Rederive Vemf for the rectangular loop of Figure 4.16 if the magnetic field is now B = Boaz.

We see in Figure P4.19a that for the line sections.For the section we have:

Fig. P4.18

Fig. P4.21

, so for the section, the contributions to

Vemf cancel. This will also be the case for the section, and therefore Vemf = 0; no current is induced.

P4.20: In Figure 4.16, replace the rectangular loop with a circular one of radius a and rederive Vemf.

P4.21: A conductive rod, of length 6.0 cm, has one end fixed on a grounded origin and is free to rotate in the x-y plane. It rotates at 60 revolutions per second in a magnetic field B = 100. mT az. Find the voltage at the end of the bar.

We can confirm the sign by observing that a positive charge placed in the middle of the bar would move to the ungrounded end by the Lorentz force equation.

P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make conductive contact with a ring to make one end of the electrical contact to R; the other contact is made to the center of the bar. Given Bo = 100. mWb/m2, a = 6.0 cm, and R = 50. , determine I if the bar rotates at 1.0 revolution per second.

Fig. P4.22

Figure P4.22 indicates one of the paths for the circulation integral.

P4.23: A Faraday Disk Generator is similar to the rotating conductor of P4.22, only now the rotating element is a disk instead of a bar. Derive an expression of the V emf

produced by a Faraday Disk Generator, and using the parameters given in problem 4.22, find I.

Worked exactly as P4.22.

P4.24: Consider a sliding rail problem where the conductive rails expand as they progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y direction, and uy = 2.0 m/sec, find the Vemf across a 100. resistor at the instant when y = 10. cm if the field is Bo = 100. mT.

First we modify the figure so that the top rail is horizontal and all the spreading occurs via the bottom rail. As before, our approach will be to find and then d /dt. We have:

Now, notice that x and y are not independent and are in fact related: x=y+wSo we have

Fig. P4.24

Alternate Method:

Uniform Plane Wave and Power Transmission

1. A wave with = 6.0 cm in air is incident on a nonmagnetic, lossless liquid media. In the liquid, the wavelength is measured as 1.0 cm. What is the wave’s frequency (a) in air? (b) in the liquid? (c) What is the liquid’s relative permittivity?

(a)

(b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz(c)

_____________________________________________________________________2. Given = 1.0x10-5 S/m , r = 2.0, r = 50., and f = 10. MHz, find , , , and .

Inserting these into the expressions for and ,

_____________________________________________________________________3. Suppose in free space, H(x,t) = 100.cos(2x107t – x + /4) az mA/m. Find E(x,t).

Since free space is stated,

and then

_____________________________________________________________________4. A 100 MHz wave in free space propagates in the y direction with an amplitude of 1 V/m. If the electric field vector for this wave has only an az component, find the instantaneous expression for the electric and magnetic fields.

From the given information we have and

or .

Now to find H.

So

or

_____________________________________________________________________5. In a lossless, nonmagnetic material with r = 16, H = 100 cos(t – 10y) az mA/m. Determine the propagation velocity, the angular frequency, and the instantaneous expression for the electric field intensity.

_____________________________________________________________________6. In a media with properties = 0.00964 S/m , r = 1.0, r = 100., and f = 100. MHz, a 1.0 mA/m amplitude magnetic field travels in the +x direction with its field vector in the z direction. Find the instantaneous form of the related electric field intensity.

Finally,

_____________________________________________________________________7. In seawater, a propagating electric field is given by E(z,t) = 20.e-z cos(xt – z + 0.5) ay V/m. Assuming ’’=0, find (a) and , and (b) the instantaneous form of H.

For seawater we have r = 72, = 5, and r = 1.So:

or with appropriate significant digits:

_____________________________________________________________________8. For Nickel ( = 1.45 x 107, r = 600), make a table of , , , up, and for 1Hz, 1kHz, 1MHz, and 1 GHz.

For Ni we have = 1.45x107S/m, r = 600

= 1/

Table f(Hz)= 1 103 106 109

(Np/m) 185 5860 185x103 5.9x106

(rad/m) 185 5860 185x103 5.9x106

18ej45º 570ej45º 18ej45ºm 0.57ej45º 5.4mm 170m 5.3m 170nm

up(m/s) 12x106 12x106 12x106 12x106

9. A semi-infinite slab exists for z > 0 with = 300 S/m, r = 10.2, and r = 1.0. At the surface (z = 0),

E(0,t) = 1.0 cos( x 106t) Find the instantaneous expressions for E and H anywhere in the slab.

The general expression for E is:

Here, (i.e. it is a good conductor), so

So now we have

To find B we’ll work in phasors.

_____________________________________________________________________10. A 600 MHz uniform plane wave incident in the z direction on a thick slab of Teflon (r = 2.1, r = 1.0) imparts a 1.0 V/m amplitude y-polarized electric field intensity at the surface. Assuming = 0 for Teflon, find in the Teflon (a) E(z,t), (b) H(z,t) and (c) Pav.

Teflon: = 0 so = 0,

and

(a)

(b)

(c)

_____________________________________________________________________11. A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0 mV/m amplitude at the surface. How much power passes through a square meter at the surface? How much power passes through a square meter area 10. m beneath the surface?

Cu:

Now at 10 m beneath the surface, we have

_____________________________________________________________________12. Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ay V/m, find the polarization and handedness.

The field can be rewritten as E(z,t) = 10.cos(t-z)ax + 20.cos(t-z-45-180°)ay

or E(z,t) = 10.cos(t-z)ax + 20.cos(t-z+135°)ay

_____________________________________________________________________13. Given

we say that Ey leads Ex for 0 < < 180, and that Ey lags Ex when –180 < < 0. Determine the handedness for each of these two cases.

For 0 < < 180°, we have LHP

For 180° < < 360°, we have RHP_____________________________________________________________________14. Suppose a UPW in air carrying an average power density of 100 mW/m 2 is normally incident on a nonmagnetic material with r = 11. What is the time-averaged power density of the reflected and transmitted waves?

_____________________________________________________________________15. A UPW in a lossless nonmagnetic r = 16 media (for z < 0) is given by

E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m.This is incident on a lossless media characterized by r = 12, r = 6.0 (for z > 0). Find the instantaneous expressions for the reflected and transmitted electric field intensities.

or , so

.

_____________________________________________________________________16. The wave Ei = 10.cos(2x 108t - 1z) ax V/m is incident from air onto a copper conductor. Find Er, Et and the time-averaged power density transmitted at the surface.

For copper we have

so

We find

So Er = -10.cos(2x 108t + 1z) ax V/m

and

17. A wave specified by Ei = 100.cos(x107t-1z)ax V/m is incident from air (at z < 0) to a nonmagnetic media (z > 0, = 0.050 S/m, r = 9.0). Find Er, Et and SWR. Also find the average power densities for the incident, reflected and transmitted waves.

In this problem we find in medium 2 (z > 0) that = 0.0025 and = 0.05. These values are too close to allow for simplifying assumptions. Using (5.13) and (5.31), we calculate:

.

Then,

,

so

,

so

(check: 13.3 W/m2 = 10.7 W/m2 + 2.6 W/m2)_____________________________________________________________________18. A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incident from air (z < 0) onto a slab of lossless, nonmagnetic material with r = 25 (z > 0). The angle of incidence is 40. Calculate (a) the angle of transmission, (b) the reflection and transmission coefficients, and (c) the incident, reflected and transmitted fields.

(a) The material parameters in this problem are the same as for P5.48. So, once again we have t = 7.4°. Also, 1 = 2.09 rad/m and 2 = 10.45 rad/m.(b)

(c) Incident:

Reflected:

transmitted:

_____________________________________________________________________