EEE161 Applied Electromagnetics Laboratory 4

Transcription

1 Dr. Milica Marković Applied Electromagnetics Laboratory page 1 EEE161 Applied Electromagnetics Laboratory 4 Instructor: Dr. Milica Marković Office: Riverside Hall milica@csus.edu Web: milica 1 Magneticc Field due to a Charge Distribution We will first find the magnetic field due to a loop carrying current I, positioned in X-Y plane, as shown in Figure??. To solve this problem, we will first divide the loop into small pieces, and label one of these pieces as dl. The magnetic field due to an infinitesimal current can be found using Biot-Savart s law. Magnetic field is labeled in Figure?? as db. The position of the infinitesimal current is defined by a position vector r 2. The position of point P, where the field will be calculated, is defined with the position vector r 1. The distance between the current and the observation point is labeled as r. The vector db is defined in Equation??. db l = µ I 4πr ( ˆdl ˆr) (1) 3 Figure 1: Loop of wire carrying current I. Magnetic field is shown due to a very small section (arc length) of the loop dl. The total magnetic field at a point P is then equal to the sum of all the fields due to the elemental currents as shown in Figure??. The equation for the total field is given in??.

2 Dr. Milica Marković Applied Electromagnetics Laboratory page 2 B = all inf. currents db (2) Figure 2: Loop of wire uniformly charged with line charge density ρ l. Electric field is shown due to several very small sections (arc length) of the loop dl. Each section is modeled by a point charge dq. The problem now is to represent all the variables in the Equation?? ( I, dl, ˆr and r) using appropriate coordinate system and given current distribution. As seen in Figure??, dl is an arc length in the direction of theta (blue arrow next to dl) dl = a dθa θ, where a is the radius of the loop. The vector r 2 is the position vector of the arc length dl and the vector r 1 is the position vector of the point be of the electric field calculation. Point P is an arbitrary point in the Cartesian coordinate system, P(x,y,z), therefore its vector is shown in Equation??. The vector r is the distance vector between the elemental current (the source) and the point at which we are calculating the electric field. r 1 = x a x + y a y + z a z (3) The vector r 2 can be written in Polar Coordinates as in Equation??,where a is the radius of the loop. The equation?? can be rewritten in Cartesian coordinate system as in Equation??. r 2 = a a r (4) a r = cosθ a x + sinθ a y (5) r 2 = a cosθ a x + a sinθ a y (6) The two vectors mark the beginning and the end of the distance vector r. The vector r is the sum of vectors r 2 and r 1.

3 Dr. Milica Marković Applied Electromagnetics Laboratory page 3 r = r 1 + ( r 2 ) (7) Therefore the vector s r magnitude and the unit vector are shown in Equations??-??. Vector r has the magnitude of: r = (x a cosθ) a x + (y a sinθ) a y + z a z (8) Unit vector in the direction of vector r is: r = (x a cosθ) 2 + (y a sinθ) 2 + z 2 (9) ˆr = ˆr = r r (10) r (x a cosθ)2 + (y a sinθ) 2 + z 2 (11) Cross product between the distance vector r and the vector of the direction of current Î is found in Equations??-??. dl = a dθ a θ (12) a θ = sinθ a x + cosθ a y (13) dl = a sinθ dθ a x + a cosθ dθ a y (14) dl r = ( a sinθ dθ a x + a cosθ dθ a y ) ((x a cosθ) a x + (y a sinθ) a y + z a z ) (15) (16) dl r = a x a y a z a sinθ dθ a cosθ dθ 0 (x a cosθ) (y a sinθ) z (17) dl r = (za cosθ dθ) a x + (a z sinθ dθ) a y + (a 2 a (y sinθ + x cosθ)) dθ a z (18) Replacing other variables in the Equations??-?? to the Equation??, we get the Equation?? for the electric field de at a point P. Components of the electric field are given in Equations??-??.

4 Dr. Milica Marković Applied Electromagnetics Laboratory page 4 db z = db x = µi 4π (x a cosθ) 2 + (y a sinθ) 2 + z 23 (z a cosθ dθ) a x (19) db µi y = 4π (x a cosθ) 2 + (y a sinθ) 2 + z (a z sinθ dθ) a 23 y (20) µi 4π (x a cosθ) 2 + (y a sinθ) 2 + z (a2 a (y sinθ + x cosθ)) dθ a z 23 (21) Each component of the field can be integrated separately, as shown in Equations??-??. B x = B y = B z = 2π 0 2π 0 2π The magnetic field above is shown in Figure?? 0 µ I z a cosθ dθ 4 π (x a cosθ) 2 + (y a sinθ) 2 + z 23 a x (22) µ I a z sinθ dθ 4 π (x a cosθ) 2 + (y a sinθ) 2 + z 23 a y (23) µ I(a 2 a (y sinθ + x cosθ)) dθ 4 π (x a cosθ) 2 + (y a sinθ) 2 + z 23 a z (24) Figure 3: Magnetic Field of a Loop of Current Conclusion Solve all of the above examples by hand and compare the results with the ones you found during the lab. In addition, for each of the problems, write what you have learned. Do not write an essay, just in a few sentences write what you have done in each problem, and specifically what you have learned from the work done. Due Date

5 Dr. Milica Marković Applied Electromagnetics Laboratory page 5 Submit the printout of the lab work and the conclusion by NEXT FRIDAY. LAB IS DUE next Friday in the lab. No late labs. Appendices A Visualizing Scalar Fields in Matlab To visualize scalar fields in Matlab we can use the following functions: slice, contourslice, patch, isonormals, camlight and lightning. Please note that more detailed explanation about the these organization functions shown here can be found in Matlab help. A.1 slice Slice is a command that shows the magnitude of a scalar fields on a plane that slices the volume where the potential field is visualized. The format of this command is as shown below. slice(x,y,z,v,xslice,yslice,zslice) Where X, Y, and Z are coordinate of points where the scaler function is calculated, V is v the scalar function at those points, and the last three vectors xslice, yslice, and zslice are showing where will the volume will be sliced. An example of slice command is given below. in the example below there is an additional command colormap, that s collars the volume with a specific pallette. To see more about different color maps, see Matlab help. xslice has three points at which the x-axis will be slice. They are -1.2,.8, 2. This means that the volume will be slice with a plane that is perpendicular to x-axis and it crosses the x-axis at points -1.2,.8 and 2. clc clear all [x,y,z] = meshgrid(-2:.2:2,-2:.25:2,-2:.16:2); v = x.*exp(-x.^2-y.^2-z.^2); xslice = [-1.2,.8,2]; yslice = 1; zslice = [-2,0]; slice(x,y,z,v,xslice,yslice,zslice) colormap hsv A.2 contourslice Contourslice command will display equipotential lines on a plane being the volume where the potential field is visualized. An example of contourslice function is shown below. [x,y,z] = meshgrid(-2:.2:2,-2:.25:2,-2:.16:2); v = x.*exp(-x.^2-y.^2-z.^2); % Create volume data

6 Dr. Milica Marković Applied Electromagnetics Laboratory page 6 [xi,yi,zi] = sphere; % Plane to contour contourslice(x,y,z,v,xi,yi,zi) view(3) A.3 patch Patch command creates a patch of color. A.4 isonormals Command isonormals creates equipotential surfaces. A.5 camlight camlight( headlight ) creates a light at the camera position.camlight( right ) creates a ligh right and up from camera. camlight( left ) creates a light left and up from camera.camlight with no arguments is the same as camlight( right ). camlight(az,el) creates a light at the specified azimuth (az) and elevation (el) with respect to the camera position. The camera target is the center of rotation and az and el are in degrees. A.6 lighting lighting flat selects flat lighting. lighting gouraud selects gouraud lighting. lighting phong selects phong lighting. lighting none turns off lighting.