UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #6. Benjamin Stahl. February 17, 2015

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1 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS A Homework #6 Benjamin Stahl February 7, 5 GRIFFITHS, 5.9 The magnetic field at a point, P, will be found for each of the steady current configurations shown in Fig. 5. using the Biot-Savart law: a) The two straight line segments make no contribution to the magnetic field at P. Thus the magnetic field contributions from the circular segments must be calculated. Using the Biot-Savart law 5.4 of Griffiths), an expression is derived for the magnetic field due to an arc of current: B = µ I d l ˆr r = µ I dl ˆk r = µ I Where s is the arc length: s =rθ. Thus the expression becomes: rs ˆk.) B = µ I θ r ˆk.) Now the desired field can be calculated: B = µ I π a π ) ˆk = µ I b 8 a ) out.) b b) The two line segments can be treated as one infinite line of charge. Using the expression given in Griffiths equation 5.9 for infinite lines of charge and the result found in the above part for an arc of charge. The desired field at P will be: B = µ I πr ) R π = µ I 4R π in.4) GRIFFITHS, 5. The magnetic field at a point, P, on the axis of a tightly wound solenoid consisting of n turns per unit length wrapped around a cylindrical tube of radius a and carrying current, I, will be found. Using Equation 5.4 of Griffiths and putting I nidz where the z axis is the one running through the center of the cylinder) and then

2 integrating: B = µ ni a a z ) dz.) Now the following trigonometric substitution is utilized: Implementing the substitution: z = a cotθ dz = a sin θ dθ & a z ) B = µ ni sin θ a a sin θ dθ = µ ni sinθdθ = µ ni In the case of an infinite solenoid, θ = and θ = π, so the field will be: B = µ ni = sin θ a.) [cosθ cosθ ].) [ )] = µ ni.4) GRIFFITHS, 5. Using the result of Ex. 5.6 of Griffiths, the magnetic field at the center of a uniformly charged spherical shell of radius, R, and total charge, Q, spinning at angular velocity, ω, will be found. Using the Biot-Savart law for surface currents as given in Equation 5.4 of Griffiths and setting K r ) = σ v = σr ω ˆφ: B r ) = µ K r ) ˆr r da = µ σrω Computing the remaining integral and putting σ = R da = µ σω R π π R sinθdφdθ = µ σωr Q yields the desired result: R π sinθdθ.) B = µ ωrq R [ cosπ) cos)] = µ ωq R.) 4 GRIFFITHS, 5.4 It is given that a steady current, I, flows down a long cylindrical wire of radius, a. The magnetic field both inside and outside the wire will be found for the following scenarios: a) The current is uniformly distributed over the outside surface of the wire. Using Ampere s Law: B d l = µ I enc B πs = µ I enc B = µ I enc πs There is no enclosed current within the wire in this scenario, thus the magnetic field will be: ˆφ 4.) B = {, s < a µ I πs ˆφ, s > a 4.) b) Now, the current is distributed such that J is proportional to s, that is J = ks, where k is a constant of proportionality. Then using Equation 5.45 of Griffiths: I tot = a J d a = a ksπs)ds = πka = I k = I πa 4.)

3 Having now found the constant of proportionality, the enclosed current can be expressed: I enc = s ks πs )ds = πks = πs I πa = s a I 4.4) Now using the result of Ampere s law found in the previous part the desired answer can be expressed: µ I s ˆφ, s < a πa B = µ I s > a πs ˆφ. 4.5) 5 GRIFFITHS, 5.5 It is given that a thick slab extending from z = a to z = a and infinite in the x and y directions) carries a uniform volume current, J = J ˆx. The magnetic field as a function of z both inside and outside the slab will be found. Using Ampere s law with an amperian loop in the xz plane with width l and height z: B d l = µ I enc Bl = µ lz J B = µ z J 5.) Using the right hand rule, the field within the slab will thus be: B = µ z J ŷ 5.) Outside the slab the field will be: B = { µ Jaŷ, z > a µ Jaŷ, z < a 5.) 6 GRIFFITHS, 5.4 The current density required to produce the vector potential, A = k ˆφ, will be found using Equation 5.64 of Griffiths: A = µ J J = µ A 6.) Using the vector Laplacian in cylindrical coordinates this becomes: J = [ A s A s µ s ) A φ s ŝ A φ A φ φ s ) ] A s ˆφ s A z ẑ φ 6.) Substituting the given vector potential allows for significant simplifications to be made, that yield the correct result: J = [ A ] φ µ s ˆφ = k µ s ˆφ 6.) 7 GRIFFITHS, 5.4 It is given that a current, I, flows to the right through a rectangular bar of conducting material in the presence of a uniform magnetic field, B, that points out of the page as shown in Fig of Griffiths).

4 a) If the moving charges are positive, the direction that they will be deflected by the magnetic field will be determined. By the right hand rule and the Biot-Savart Law, the positive charges will be deflected down in this scenario, resulting in an accumulation of positive charge on the lower surface. b) The resulting potential difference between the top and bottom of the bar will be found in terms of B, v, and the relevant dimension of the bar. As given there is an equilibrium that is reached, thus the electric and magnetic forces can be equated to find the desired result: qvb = qe E = vb = V t V = vbt 7.) Note that the bottom will be at the higher potential because that is where the positive charges are accumulated. c) If the moving charges were negative and moving to the left, then the results would be: The charges would be deflected down and thus negative charge would be deposited on the bottom. The magnitude of the potential difference would be unchanged, but the higher potential would be at the top. 8 GRIFFITHS, 5.4 It is given that a plane wire loop of irregular shape is situated so that part of it is in a uniform magnetic field, B, as shown in Fig of Griffiths. The field occupies the shaded region in the figure and points perpendicular to the plan of the loop. The loop carries a current, I. It will be shown that the net magnetic force on the loop is F = I Bw, where w is the chord subtended. Starting with Equation 5.6 of Griffiths: F = I d l B ) = I [ ] d l B Where the fact that the magnetic field is constant has been used to pull it out of the integration. Now the remaining integral is just just the vector displacement from the point where the wire enters the field first to where it leaves, which is just wa. Recognizing that w and B are orthogonal leads to the desired result: Where the force is orthogonal to w 8.) F = I [ w B ] F = I Bw 8.) 9 GRIFFITHS, 5.47 It is given that the magnetic field on the axis of a circular current loop falls of sharply with increasing z. However a more uniform field can be created by using two such loops a a distance d apart as shown in Fig of Griffiths. a) The magnetic field as a function of z will be found, and it will then be shown that B z = at z =. Starting from Equation 5.4 of Griffiths and using the principle of superposition, the desired field can be found: B = µ I R [ ] R z d [ ] R z d 9.) Differentiating the result with respect to z: B z = µ I R z d R z d z d R z d 9.) 4

5 Simplifying and evaluating for z = : As expected, this result in zero. B z = µ I R z= d [ R d ) ] 5 d R d = 9.) b) Now d will be found such that B = at z =. Starting from the simplified expression for the first derivative: z B z = µ I R z d R z d z d R z d 9.4) Now differentiating again and simplifying: B z = µ I R R z d [ ] z d 5 R z d z d [ ] 5 R z d R z d [ ] z d 5 R z d z d [ ] 5 R z d = µ I R R z d [ ] 5 z d R z d [ ] 5 R z d R z d [ ] 5 z d R z d [ ] 5 R z d 9.5) Now putting z = and equating the expression to allows for d to be determined: R [ ] d 5 d R d = µ I R [ ] 5 R d R d [ ] 5 d R d [ ] 5 R d 9.6) [ d ] 5 = R d [ d ] 5 R [ = R d ] [[ d R ) ] 5 d ] The solution corresponding to the first factored term is discarded because it offers non-physical solutions. Now focusing only on the second factor: Selecting only the useful solution: d d d d R 5 = R = 4 R = ± 9.7) d = R 9.8) 5

6 Now the resulting magnetic field at the center will be found by evaluating for d = R and z = : B = µ I R [ R R ) ] [ R R ) ] = µ I R 5 R = 8µ I 5 5R 9.9) GRIFFITHS, 5.5 A plane loop of wire that carries a steady current I will be considered. The magnetic field at a point the origin) in the plane will be calculated. The shape of the wire is given, in polar coordinates, by a specified function, r θ) and the scenario is shown in Fig. 5.6 of Griffiths. a) The magnitude of the magnetic field will be found starting from the Biot-Savart Law: B = µ I d l r r = µ I d l r r.) Where the hint given in the problem statement r = r ) has been utilized. Now using the definition of the cross product, and subsequently the definition of arc length, in conjunction with the given figure this becomes: B = µ I dlsinφ r = µ I rdθ r = µ I dθ r.) This was the desired result. b) The result found in the above part will be tested by calculating the field at the center of a circular loop of radius, R: B = µ I dθ R = µ I π dθ = µ I.) R R This matches exactly the result from using Equation 5.4 of Griffiths with z = : B = µ I R R z ) = µ I R R ) = µ I R.4) c) Now, the lituus spiral will be considered. It is defined by: r θ) = θ, < θ π).5) The lituus spiral is plotted in the following figure: 6

7 π π π Figure.: The lituus sprial plotted for with a =. Now the magnetic field at the origin will be found by completing the loop with a straight line segment along the x axis which will not contribute to the magnetic field) using the result found in part a: B = µ I θdθ = µ I a 6πa θ d) Now, it is given that for a conic section with focus at the origin: π p r θ) = e cosθ = µ I π.6) a.7) Where p is the semilatus rectum y intercept) and e is the eccentricity. The field will be found using the result found in part a: B = µ I e cosθ)dθ = µ I π [θ e sinθ] p p = µ I.8) p This is the expected result. GRIFFITHS, 5.5 a) It is given that a way of filling in the missing link in Fig of Griffiths is to exploit the analogy between the defining equation for A and Maxwell s equation for B. Using the pertinent relations, A will be expressed in terms of B. The following relations are known for the magnetic field: B = µ J & B = & B = µ It is also known that A = B and A =. Thus, the vector potential can be expressed as: A = b) It is given that the electrical analog of the above result is: V r ) = J ˆr r dτ.) B ˆr r dτ.) Er ) ˆr r dτ.) 7

8 Taking the gradient in non-primed coordinates of the above result: V = Er ) ˆr ) r dτ = This result is true, and thus the given assertion must also be true. Er )δ r r )dτ = Er ).4) 8