force reduces appropriately to the force exerted by one point charge on another. (20 points)

Transcription

1 Phsics III: Theor an Simulation Examination 3 December 4, 29 Answer All Questions Analtical Part: Due 5: p.m., M, 12/7/9 Name SOUTIONS 1. Two line charges A an B of the same length are parallel to each other an locate in the x plane. See figure.) Both have the same constant linear charge ensit i.e., charge per unit length) λ. i) Fin the total force that A exerts on B. ii) Show that for >>, this A B force reuces appropriatel to the force exerte b one point charge on another. 2 points) i) The smmetr about the common perpenicular bisector) inicates that the - component of total force exerte on B b A is zero. We shall fin the x component of the electric fiel ue to line charge A at the location of line charge B. The force exerte b A on a charge element of B is equal to the electric fiel at its position multiplie b the charge of the element. To fin the total force, we integrate over line charge B. The x-component of the electric fiel ue to a charge element of line charge A at a position on B that is at a height above the x axis is given b E x = E cos = 1 $ % r 2 r = $ % 4"#. ' 2 % & ) 2 3/ 2 ) Thus, E x = $ % $ ' & ) 4"# =, - ' 2 % & ) 2 3/ 2 ) 4"#, 2 & ) The x-component of the force on the charge element at a height on line charge B is then F x = qe x = $E x = $ 2 ',, F x = $ 2 ' &,, 2 & ) 2 = $ & 2"# ). & 2 & ) 2 ) = $ 2 ) -. Thus, Since there is onl an x-component, the total force is F = $ 2 ) ii) 2 2 = 2 2 ) 1/ 2 = 2 1 " 2 2 For, 1 2 ) 1/ 2 % Thus, ) & % 2 1/ 2 # = 1 2 $ 2 ) 1/ "# 2 2 & )î. 15) 2 2 & % Thus, F = $ 2"# ) % 2 & 2 ) 2"# ' 2 = )2 = q 2, which is the magnitue of the force 2 2 between two point charges of magnitue q. 5) r θ E E x x E

2 2. Consier an infinitel long, uniforml charge clinrical tube of inner raius a an outer raius b. The volume charge ensit i.e., charge per unit volume) is ρ. The regions r < a an r > b where r is the istance from the central axis, are empt. a) Use Gauss's law to fin the electric fiel in the regions i) r > b; ii) a < r < b; an iii) r < a. b) Fin the potential ifference between the inner an outer surfaces of the cliner. 2 points) Phsics III: Theor an Simulation, Examination 3 Shan 12/3/9) Page 2 of 7

3 3. Fin the magnetic fiel at an point on the axis of an infinitel long, thin half-cliner of raius R that carries a uniforml istribute current I along its length. See figure.) [Hint: Divie the half-cliner into infinitesimal strips.] 2 points) axis I Sie View strip 1 - B1 axis θ θ B 2 strip 2 Crosssectional View Phsics III: Theor an Simulation, Examination 3 Shan 12/3/9) Page 3 of 7

4 Phsics III: Theor an Simulation Examination 3 December 4, 29 Computational Part: Due 5: p.m., W, 12/9/9 Name SOUTIONS Note: Inclue all necessar graphs, numerical results, comments, an explanations in our solution to the computational problem. Where comparisons are mae, all the values that are being compare must be shown. Neeless to sa, points will be eucte if important ata are not presente. 1. a) Consier a stea current in a wire that forms an arc of a circle ling in the x plane with the center of curvature at the origin. See figure.) Write a MATAB R I program to calculate the magnetic fiel along the z-axis. Hint: You will nee to know the x an components of each current element l ) of the arc an the x α an components of its position. The current element ma be expresse in polar coorinates r, φ), an ou must integrate over φ.] a) Plot the magnitue of the α magnetic fiel an the z- component of the fiel as functions of position along the z-axis z 2. m) using the same graph axes. Use I = 5. A, R =.1 m, an α = π/3. Compare the two graphs an explain an similarities an ifferences. b) For α = π, plot the magnitue of the magnetic fiel an the z-component of the fiel as functions of position along the z-axis z 2. m) using the same graph axes. What o ou notice? Explain this result. c) For α = π, fin the magnetic fiel at the center of curvature. Compare our computational value with the analtical value. Is there agreement? 2 points) The position of the current element see figure to the right) is given b r el = Rcos)î Rsin) ĵ. The vector) length of the current element is given b see figure below) l = l sin"î l cos" ĵ = R" sin"î cos" ĵ). The Biot-Savart law is B = µ I l " z ˆk # r el ) 4 z ˆk # r, 3 el where z ˆk is the position vector of the fiel point along the z axis. a) Similarities: i) Both the magnitue of the magnetic fiel an its z-component fall off rapil with istance. This is expecte because the current istribution is finite an so the fall-off of both shoul be faster than 1/r infinitel long wire). ii) At z =, B = B z. This because for z =, both l an z ˆk r el are in the x- plane an so their cross prouct has onl a z component. Differences: i) Except for the fiel position z =, the magnetic fiel has both an x-component an a z-component. Thus, B B z. ii) B z falls off faster with istance than B. We can see wh b looking at the Biot-Savart law. B z falls off roughl as 1/r 3 because the component proucts in l z ˆk " r el ) that give rise to B z o not epen on z. Thus, we get ~z 3 in the enominator, i.e., a ~1/z 3 epenence. However, B falls off roughl as 1/r 2 because the component prouct in l z ˆk " r el ) that gives rise to B x is proportional to z. With the ~z 3 in the enominator of the Biot-Savart law, this gives an overall istance epenence of ~1/z 2. 6) b) For α = π, B = B z for all values of z. This is because we now have a complete circular loop an the higher smmetr eliminates both the x an components of the fiel along the z axis. 2) R φ α l φ l I x x Phsics III: Theor an Simulation, Examination 3 Shan 12/3/9) Page 4 of 7

5 c) For α = π, at z =, the computational value for the magnetic fiel is B = B z = T. For a circular loop, the fiel at the center is given b B center = µ I 2) R 4 " 1 #7 T $ A/m)5 A) = = " 1 #4 T. The agreement is excellent. 2.1m) % Phsics III: Theor an Simulation, Examination 3, Problem 4 29) help ph3ex3_p4_9 clear %% Declare constants I = 5; % Current in Amperes R =.1; % Raius of arc in meters alpha = pi/3; % Magnitue of maximum angle subtene b top an bottom of arc phi_top = alpha; % Angle subtene b top of arc phi_bot = -alpha; % Angle subtene b bottom of arc mu_ = 4pi1e-7; % Permeabilit of free space in SI units N = 2; % Number of subintervals for integral elta_phi = phi_top - phi_bot)/n; % ength of a subinterval %% Perform Biot-Savart Integral b mipoint metho count = 1; % Counter variable for number of fiel points use in graph for z = :.2:2. % Fiel point positions along z-axis B = [ ]; % Initialize magnetic fiel r_p = [ z]; % Position of fiel point for i=1:n phi = phi_bot i -.5)elta_phi; % Mipoint metho r_el = R[cosphi) sinphi) ]; % Position of current element l = Relta_phi[-sinphi) cosphi) ]; % Current element length vector r = r_p - r_el; % Position vector from element to fiel point B = crossl, r)/normr))^3; % Biot-Savart B = B B; en B = mu_i/4pi))b; % Final fiel value B_magcount) = normb); % Arra storing magnitues of B values B_zcount) = B3); % Arra for z-components of B z_pcount) = z; % Arra for positions of fiel point count = count 1; % Increment counter en %% Graphs plotz_p, B_mag, '-b', z_p, B_z, '.-r') xlabel'position of Fiel Point Along z-axis m)') label'magnetic fiel T)') title'magnetic Fiel vs. Position for Circular-Arc Current') legen' B ','B_z') gtext'\alpha = \pi/3') 1) Phsics III: Theor an Simulation, Examination 3 Shan 12/3/9) Page 5 of 7

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