UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #7. Benjamin Stahl. March 3, 2015

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1 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS A Homework #7 Benjamin Stahl March 3, 5 GRIFFITHS, 5.34 It will be shown that the magnetic field of a dipole can written in the following coordinate-free form: Assigning a coordinate system such that m = mẑ, this becomes: B dip = µ [3 m ˆr ˆr m]. 4πr 3 B dip = µ 4πr 3 [3mẑ ˆr ˆr mẑ] = µ m 4πr 3 [ 3cosθ ˆr cosθ ˆr sinθ ˆθ] = µ m 4πr 3 [ cosθ ˆr + sinθ ˆθ]. Where useful relationships on the back inside cover of Griffiths pertaining to ẑ and ˆr have been used. This leads to the exact result of Equation 5.88 of Griffiths, and thus the desired has been shown. GRIFFITHS, 5.35 It is given that a circular loop of wire, with radius R, lies in the x y plane centered at the origin and carries a current, I, running counterclockwise as viewed from the positive z axis. a The magnetic dipole moment will be calculated using Equation 5.86 from Griffiths: m I d a = I a = πr I ẑ. b Now, the approximate magnetic field at points far from the origin will be found by calculating the dipole field according to Equation 5.89 of Griffiths: B B dip = µ m [ µ cosθ ˆr + sinθ ˆθ] = πr 4πr 3 4πr 3 I [ cosθ ˆr + sinθ ˆθ ] = µ R I [ cosθ ˆr + sinθ ˆθ] 4r 3.

2 c Now this result is compared to that of Ex. 5.6 of Griffiths which is given as follows: Bz = µ I R R + z 3.3 Examining the limit where z >> R: µ I lim R z R z 3 R z + 3 = µ I Now, for the case of the z axis, θ = and r = z, thus the result found in part b becomes: B µ R I 4z 3 [] = µ I Thus, the result is consistent with that from Ex. 5.6 in when z >> r. R R z 3.4 z GRIFFITHS, 5.37 a It is given that a phonograph record of radius, R, carrying a uniform surface charge, σ, is rotating at constant angular velocity, ω. It s magnetic dipole moment will be found. In order to use the definition given in Equation 5.86 of Griffiths, the current, I, must be found. This will be done below, using the definition of surface current density: d I = K d I = K dr = σvdr = σr ωdr 3. dr Now, the magnetic dipole moment is calculated: dm = d I d a = d I πr ẑ = σr ωdr πr ẑ m = σωπ R r 3 dr ẑ = 4 σωπr4 ẑ 3. b The magnetic dipole moment of the spinning spherical shell in Ex. 5. will be found. The infinitesimal current in a ring cut parallel to the x y plane of the sphere will now be expressed: The infinitesimal area of such a ring will be: d I = σ ω πr sinθrdθ = σωr sinθrdθ 3.3 π da = πr sinθ 3.4 Thus the magnetic dipole moment can now be calculated by integrating over all the rings required to form the spherical shell: π dm = σωr sinθrdθπr sinθ m = πσωr 4 sin 3 θdθẑ = 4 3 πσωr4 ẑ 3.5 Using the result found above with Equation 5.85 of Griffiths allows for the dipole potential to be expressed: A dip r = µ m ˆr 4π r = µ 4πσωR 4 sinθ ˆφ 4π 3r = µ σωr 4 sinθ ˆφ 3r 3.6 The boxed result is exactly the same as Equation 5.69 of Griffiths for r > R and thus the desired has been shown.

3 4 GRIFFITHS, 5.58 It is given that a uniform donut, carrying charge Q and mass M, rotates about its axis as shown in Figure 5.64 of Griffiths. a The ratio of its magnetic dipole moment to its angular momentum the gyromagnetic ratio will be found as follows: Q g = m L = I a π πr RMv = ω MωR = Q 4. M b The gyromagnetic ratio will now be calculated for a uniform spinning sphere. Since a sphere can be decomposed into an infinite number of infinitely thin donuts, and since the result for the gyromagnetic ratio above has no radial dependence, the above result is valid also for a sphere. Thus: g = Q M 4. c It is given that the angular momentum of a spinning electron is, the magnetic dipole moment will thus be calculated by using the derived relation for the gyromagnetic ratio: g = m di p L = Q M Where the fundamental physical constants have been evaluated in SI units. m di p = Q M L = e 4m e = Am GRIFFITHS, 6. The torque exerted on the square loop shown in Figure 6.6 of Griffiths, which is caused by the circular loop, will be calculated. Using the subscript of to denote the circular loop and to denote the square loop. The magnetic dipole moment of each loop can be defined as follows: m = m ẑ & m = m ˆr 5. Thus the torque on the square loop will be expressed using Equation 6. of Griffiths as: N = m B 5. Thus the magnetic field due to the circular loop must be calculated and will be done so as follows using Equation 5.89 of Griffiths: B = µ 4π r 3 [3 m ˆr ˆr m ] = µ 4π r 3 [3m ẑ ˆr ˆr m ẑ] = µ m 4π r ẑ 5.3 Now the torque can be calculated while noting from the diagram that ẑ and ˆr are orthogonal unit vectors which will be, by definition, mutually orthogonal to ˆx: N = m ˆr µ m 4π r ẑ = µ m m 4πr 3 ˆx 5.4 Computing the magnetic dipole moments as the product of current and area, the desired result is obtained: N = µ I πa Ib 4πr 3 ˆx = µ I ab 4r 3 ˆx 5.5 If the square loop is free to rotate, its equilibrium position will be where the net torque is zero. By inspection of the cross product that yielded the torque, the equilibrium will occur when the magnetic dipole moment of the square loop is pointing the same direction as the magnetic field created by the circular current loop. Such an orientation will occur when the square loop is in the ẑ direction. 3

4 6 GRIFFITHS, 6.4 In this problem the goal will be to derive the force law: F = m B. To do this, a scenario where a dipole is considered as an infinitesimal square of side length ɛ will be considered with current I running through the square loop oriented in the yz plane. The force along each of the four segments of wire will be calculated using d F = I d l B starting with the segment starting at the origin and moving counter-clockwise and then they will be summed: d F = I [ d y ŷ B, y, ] d F = I [ dzẑ B,ɛ, z ] d F 3 = I [ d y ŷ B, y,ɛ ] d F 4 = I [ dzẑ B,, z ] 6. d F = I [ d y ŷ B, y,ɛ B, y, dzẑ B,, z B,ɛ, z ] = I [ d y ŷ z + dzẑ y ] Thus the total force on the wire will be: F = I ɛ [ ŷ z + ẑ y ] = m [ ŷ z + ẑ y ] 6. Where the fact that the magnetic moment, m = I ɛ, has been used. Computing the cross products above leads to: [ By F = m ˆx y + B ] z Bx Bx + ŷ + ẑ 6.3 z y z Gauss law gives: B = B x x + B y y + B z z = B y y + B z z = B x x Substituting this into the above expression for the force and simplifying leads to the desired result: [ F = m ˆx B ] [ ] x Bx Bx Bx Bx Bx + ŷ + ẑ = m ˆx + ŷ + ẑ = m B x 6.5 x y z x y z Generalizing: 6.4 F = m B GRIFFITHS, 6.8 It is given that a long circular cylinder of radius R carries a magnetization M = ks ˆφ as shown in Fig. 6.3 of Griffiths. The magnetic field due to M will be found for points inside and outside the cylinder. First, the bound volume current is found: J b = M = sks ẑ = s s s 3ks ẑ = 3ksẑ 7. Next, the bound surface current is found: K b = M ˆn = ks ˆφ ŝ = kr ẑ 7. Now the magnetic field within the cylinder is found using Ampere s law with cylindrical symmetry: s s B d l = µ I enc B πs = µ J b da = µ 3ks πs ds = πµ ks 3 B = µ ks ˆφ 7.3 Now the magnetic field outside the cylinder is found using Ampere s law: [ R ] [ R ] B d l = µ I enc = µ J b da + K b dl = µ 3ks πs ds kr πr = B = 7.4 4

5 8 GRIFFITHS, 6. An infinitely long cylinder of radius R is considered that carries a frozen-in magnetization, M = ksẑ. There is no free current anywhere. The magnetic field inside and outside the cylinder will be found by two different methods: a All of the bound currents will be found. Starting with the bound volume current: J b = M = s ks ˆφ = k ˆφ 8. Now, the bound surface current is found: K b = M ˆn = kr ˆφ 8. Now the magnetic field will be found inside the cylinder using Ampere s law: B d l = µ I enc Bl = µ [ J b da + ] K b dl = µ [ klrs + krl] = µ kls 8.3 Using the right hand rule, the result becomes: B = µ ksẑ 8.4 Now the field outside the cylinder is calculated, but there will be no current enclosed by the Amperian loop in such a case so the field must be: B = b Now the field will be found using the auxiliary H field. Using Ampere s law for this field for inside the cylinder: H d l = I fenc Hl = H = 8.5 Thus the magnetic field in the cylinder is found using Equation 6.8 of Griffiths: B = µ H + M = µ + ksẑ = µ ksẑ 8.6 Following the same arguments, the outside of the cylinder will be just B = because the magnetization outside the cylinder is. 9 GRIFFITHS, 6.7 It is given that a current I flows down a long straight wire of radius a. The magnetic field a distance, s, from the axis will be found for the case where the wire is made of a linear material with susceptibility, χ m, and where the current is distributed evenly. The auxiliary field is found for the region within the cylinder using Ampere s law: Hπs = I s a H = I πa s ˆφ 9. Where the fact that the current is distributed evenly has been used to express the enclosed free current. Thus the magnetic field can be expressed using B = µ H where µ = µ + χm : Now the auxiliary field outside the cylinder is found: B = µ + χm I πa s ˆφ 9. Hπs = I H = I πs ˆφ 9.3 5

6 Thus, the magnetic field will be: B = µ I πs ˆφ 9.4 Where the fact that µ = µ outside the material has been used. Now the bound volume current is found as follows: Now, the bound surface current is found: J b = M = χ m H = χ m H = χ m J m = χ m I πa ẑ 9.5 K b = M ˆn = χ m H ˆn = χ m I πa 9.6 Thus the net bound current flowing down the wire is: I b = χ m I πa πa χ m I πa = 9.7 GRIFFITHS, 6. a It will be shown that the energy of a magnetic dipole in a magnetic field is U = m B. First the torque is considered when the dipole makes an angle θ with the field: Thus the work done in rotating the dipole from one angle to another will be: N = mb sinθ. θ θ W = Ndθ = mb sinθdθ = mb [cosθ cosθ ]. θ θ Equating this work to the difference in potential energy in the two positions leads to the desired result: W = U θ U θ = mb [cosθ cosθ ] U θ = mb cosθ U = m B.3 b Now the interaction energy of two dipoles separated by a displacement, r, is found. Starting from the coordinate free magnetic field due to a dipole and evaluating for the first dipole field: Now the interaction energy can be found: B = µ 4πr 3 [3 m ˆr ˆr m ].4 U = m B = µ 4πr 3 [ m m 3 m ˆr m ˆr ].5 c If the angle between the two dipoles is θ θ, then the interaction energy can be expressed as: U = µ 4πr 3 [m m cosθ θ 3m cosθ m cosθ ].6 Using a trigonometric relation and simplifying, this becomes: U = µ 4πr 3 m m [sinθ sinθ cosθ cosθ ].7 6

7 For a stable position to be reached, the potential energy must take on a minimum value. Finding this first for θ : U θ = µ 4πr 3 m m [cosθ sinθ + sinθ cosθ ] = cosθ sinθ + sinθ cosθ =.8 Repeating for θ : U θ = µ 4πr 3 m m [sinθ cosθ + cosθ sinθ ] sinθ cosθ + cosθ sinθ =.9 Thus the equilibrium conditions are: θ = θ = or θ = π & θ = π. Evaluating these two conditions: U = µ 4πr 3 m m & U = µ 4πr 3 m m. Clearly, U is lower and thus the stable equilibrium will be when the dipoles are parallel and on the same line. d If numerous compass needles were arrange on pins in a line, they would all align so that the arrows pointed in the exact same direction because this is the stable equilibrium as was shown above. GRIFFITHS, 6.3A It is given that a toy consists of donut-shaped permanent magnets with magnetization parallel to the axis, which slide frictionlessly on a vertical rod as shown in Fig 6.3 of Griffiths. The magnets will be treated as dipoles, with mass m d and dipole moment, m. a If the two magnets are put back-to-back, the upper one will float. The magnetic force upward being balanced by the gravitational force downward. First the magnetic field of the first magnet is expressed: Thus, the magnetic force will be: F = m B = mu 4π B = µ m 4π z 3 ẑ. m z 3 = 3µ m π z 4 ẑ. Summing the gravitational and magnetic forces and requiring equilibrium allows for the desired result to be found: Fẑ = 3µ m π z 4 m 3µ d g = z = 4 m.3 πm d g GRIFFITHS, 7. It is given that two concentric metal spherical shells of radius a and b, respectively, are separated by a weakly conducting material of conductivity, σ, as shown in Figure 7.4a of Griffiths. 7

8 a It is given that a potential difference, V, is maintained between the shells. The current will be found that flows between the shells. First the electric field between the two shells will be found using Gauss law and assuming that the charge on the inner shell is Q: E d a = Q enc E 4πr = Q E = Q ˆr. ɛ ɛ 4πɛ r Thus, the potential difference can now be expressed as follows: V = V a V b = E d l = a Q 4πɛ r dr = Q 4πɛ a b Now the current can be calculated: I = J d a = σ b E d a = σ Q 4πɛ r 4πr = σ Q = 4πσV ɛ a b Q = 4πɛ V a b..3 b Now the resistance between the shells is calculated as follows: V = I R R = V I = 4πσ a b.4 c For values of b where b >> a, it is easily observed from the forms above that the resistance will go to 4πσa. Thus b is irrelevant. This is explained by realizing that the majority of the resistance lies in the region immediately surrounding the sphere and decreases as distance is added. For two submerged spheres, the resistance will be: R = = 4πσa πσa Therefore the current will be:.5 I = V = πσav.6 R 3 GRIFFITHS, 7. It is given that a capacitor, C, has been charged up to potential V. At time t =, it is connected to a resistor, R, and begins to discharge as shown in Fig. 7.5a of Griffiths. a The charge on the capacitor will be found as a function of time. First the voltage across the capacitor is expressed in the following ways: V = Q C = I R I = Q RC dq dt = I dq dt = Q RC 3. Where the definition of current has been used to yield the above differential equation. Note the negative sign to denote that the charge on the capacitor is decreasing. Solving the differential equation yields: dq Q = dt lnq = RC RC t + A Q = Ae RC t 3. Applying the given condition that Q = CV gives the desired result: Q = A = CV Qt = CV e RC t 3.3 Differentiating this result and multiplying by for the same reason as above gives the current as a function of time: I t = dq dt = CV e RC t = V RC R e RC t 3.4 8

9 b The original energy stored in the capacitor is found by using Equation.55 of Griffiths: W = CV = CV 3.5 Now it will be confirmed by integrating Equation 7.7 of Griffiths that the heat delivered to the resistor is equal to the energy lost by the capacitor: W = Pdt = I Rdt = R V R e RC t dt = V R e RC t dt = RC V R e RC t = CV 3.6 c This and subsequent parts of the problem will be considering the case where the capacitor and resistor are connected to a battery of voltage V at t = as shown in Fig. 7.5b. The charge on the capacitor as a function of time will again be found. Expressing the voltage of this series circuit: V = Q C + I R I = V R Q RC I = dq dt = RC CV Q 3.7 Solving the differential equation: dq CV Q = RC dt lncv Q = RC t + A Q = CV Ae RC t 3.8 Now using the given condition that the capacitor is uncharged at t = gives the desired result: Q = CV A = A = CV Qt = Q = CV e RC t 3.9 Thus the current will be: I = dq dt = CV RC e RC t = V R e RC t 3. d Now the total energy output of the batter will be found: V Idt = V R e RC t dt = RC V R e RC t = CV 3. Since the current is the same as that in the first scenario, the energy delivered to the resistor is still CV. Thus the final energy stored in the capacitor will be CV which is half of the energy from the battery goes to the capacitor and half to the resistor. 4 GRIFFITHS, 7.4 In this problem, the conductivity of the material separating the cylinders in Ex. 7. of Griffiths is non-uniform, specifically σs = k s. The resistance between the cylinders will be found. For the geometry of this problem, the surface current density can be defined as: J = I A = I 4. πsl However, it is also defined as J = Eσ. Substituting leads to the following: Eσ = I πsl E k I s πsl E = I πkl 4. 9

10 Now the potential difference can be found: V = a Now the resistance can be calculated as follows: b E d l = I a b = I b a 4.3 πkl πkl R = V I = b a 4.4 πkl