Ampere S Circuit Law

Ampere’s Circuit Law

P3.14: 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 = 3y2ax, 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.

So we have Ienc = 24 A. The negative

Sign 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

a=3e-3; %radius of solid wire (m)

I=2; %current (A)

N=30; %number of data points to plot

rmax=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);

end

for i=round(a/dr)+1:N

r(i)=i*dr;

H(i)=I/(2*pi*r(i));

end

plot(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/cm2az (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 = 2a3, 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 plot

rmax=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

for i=round(a/dr)+1:N

r(i)=i*dr;

H(i)=a^3/r(i);

end

plot(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 problem

a=4; %radius of solid wire (cm)

N=120; %number of data points to plot

I=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;

end

for i=round(a/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

P3.20: 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 plot

I=3; %current (A)

rmax=10; %max radius for plot (cm)

dr=rmax/N;

aoverdr=a/dr

boverdr=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;

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.

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

% MLP0321

% generate plot for ACL problem

clc

clear

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

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/dr

boverdr=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 = Ha, so

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

4.Curl and the Point Form of Ampere’s Circuital Law

P3.24: Find for the following fields:

a. A = 3xy2/zax

b. A = sin2a – 2z 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.

1. Calculate around the path , where A(2m,0,0), B(2m,4m,0), C(0,4m,0) and D(0,0,0).
2. Divide this by the area S (2m*4m = 8m2).
3. Evaluate at the center point.
4. 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:

1. Evaluate the curl in all 4 regions.
2. 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.

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

(a)

ar derivative:

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 Density

P3.29: 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 2m2area 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 Forces

P3.32: 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/m2ax. 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/m2ay. 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.

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 iswhere, 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)

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

clear

clc

I=1; %current in A

a=1; %loop radius, in m

mu=pi*(4e-7); %free space permeability

az=[0 0 1]; %unit vector in z direction

DL1=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);

end

plot(phi,dF)

xlabel('angle in degrees')

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

clear

% initialize variables

I1=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 calculations

for 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);

end

plot(phi,dFy,phi,dFz,'--k')

legend('dFy','dFz')

Fnet=sum(dFy)

Running the program we get:

Fnet = -4.2932e-009

So Fnet = -4.3 nNay

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);

end

Ftot=sum(dF)

plot(ypos,dF)

Running the program:

Ftot =

7.4448e-007

So Ftot = 740 nN

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 azoccurs. Determine the magnetic flux density.

At  = 90°, . Also, sinceB is in direction of m, and therefore B = ±Boax.

At  = 180°, , and

Therefore, B = -Boaxand mBo = 8x10-6, so

7.Magnetic Materials

P3.48: 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.

% 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 plot

rmax=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);

end

for i=round(a/dr)+1:N

r(i)=i*dr;

B(i)=uo*I/(2*pi*r(i));

End

rmm=r*1000;

plot(rmm,B)

xlabel('rho(cm)')

ylabel('B (Wb/m^2)')

grid on

8.Boundary Conditions

P3.49: 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 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.

%

clc

clear

% enter variables

disp('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 calculations

Hna=dot(Ha,a)*a;

Hta=Ha-Hna;

Htb=Hta;

Bna=ura*Hna; %ignores uo since it will factor out

Bnb=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: 6000

relative permeability in material 2: 3000

unit vector from mtrl 1 to mtrl 2: [0 0 1]

material where field is known (1 or 2): 1

known 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: 4

relative permeability in material 2: 1

unit vector from mtrl 1 to mtrl 2: [0 0 -1]

material where field is known (1 or 2): 1

known magnetic field intensity vector: [3 0 4]

ans =

The magnetic field in the other medium is:

Hb = 3 0 16

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)