Homework 8 Solutions - Electricity and Magnetism I | PHY 481, Assignments of Physics

Download Homework 8 Solutions - Electricity and Magnetism I | PHY 481 and more Assignments Physics in PDF only on Docsity! PHY481 - Outline of solutions to Homework 8 9.12. The free current is ~jf = j0î for −a ≤ z ≤ a. The magnetic susceptibility is χm = 0 inside the slab and χm finite outside the slab. Since there is a free current, the first step is to find the magnetic intensity using ∮ ~H · d~l = if . Since the current is uniform, we expect the field to be uniform. Using the right hand rule, the magnetic intensity is found to be directed along the negative y-axis. The Amperian loop to consider is a rectangle with long side l and short size 2z with normal along the x-axis, long sides parallel to the y-axis and centered on the x-axis. We have to consider two cases z < a, z > a. For z < a the enclosed free current is if = 2zlj0, while for z > a the enclosed free current is if = 2alj0. In both cases, the contour integral gives −2Hl, so we find that ~H = −zj0ĵ, −a < z < a; ~H = −aj0ĵ, z > a ~H = aj0ĵ, z < −a (1) The magnetic field is then found from ~B = µ ~H = µ0(1 + χm) ~H, so that, ~B = −µ0zj0ĵ, −a < z < a; ~H = −µ0(1 +χm)aj0ĵ, z > a ~H = µ0aj0ĵ, z < −a (2) The magnetization is given by, ~M = χm ~H, so that, ~M = 0, −a < z < a; ~M = −χmaj0ĵ, z > a ~M = χmaj0ĵ, z < −a (3) The surface bound currents are then found from ~Kb = − ~M ∧ k̂ at z = a, and ~Kb = ~M ∧ k̂ at z = −a. There are no bulk bound currents as ~∇∧ ~M = 0 in all regimes where ~M is unifom. The surface bound currents are then, ~Kb = χmaj0î, z = a; ~Kb = χmaj0î, z = −a (4) When the susceptibility is positive, the surface bound currents, Kb, are in the same direction as the free current leading to an enhancement of the field in the material. When the susceptibility is negative the surface bound currents are opposite the free current and reduce the field inside the material. 9.15. Two long thin concentric conducting cylinders of radii a and b are separated by a region of susceptibility χm. The inner cylinder carries current Ik̂, while the outer cylinder carries current −Ik̂. Using ∮ ~H · d~l = if = 2πrH for a circular contour, yields, ~H = 0, r < a; ~H = I 2πr φ̂ a < r < b; ~H = 0, r > b (5) 1 The magnetization is found from ~M = χm ~H, so that, ~M = 0, r < a; ~M = χmI 2πr φ̂ a < r < b; ~M = 0, r > b (6) The bound currents are finite only in the regime a ≤ r ≤ b. There are surface bound currents at r = a and at r = b, and we have to check for a bulk bound current in the regime a < r < b as the magnetization is not constant. The surface bound currents are found from ~Kb = ~M ∧ r̂ = χmI 2πb k̂, at r = b; and ~Kb = ~M ∧ r̂ = χmI 2πa (−k̂) at r = a (7) The bound currents are thus in opposite directions at a and b. However, when χm > 0 both of these bound currents produce a magnetic field inside the material that is larger than that in the absence of the material. Using the expression for the curl in cylindrical co-ordinates (see Table 2.3), the bulk bound currents are given by, ~Kb = ~∇∧ ~M = 1 r ∂(r I 2πr ) ∂r = 0 (8) Thus, even though the magnetization is not uniform the bulk bound current is zero. 9.17. a) Consider an infinite copper cylinder, with susceptibility χm = −9.6× 10−6, and with field ~B = B0~k, with B0 = 1T at the surface of the cylinder. Define H0 = B0/µ0. There are no free currents, so Hextt = H0 = H int t . We then have, ~Bint = µ0(1 + χm) ~H0; ~M = χmH0k̂; ~Kb = ~M ∧ r̂ = χmH0sinθφ̂ (9) b) Consider the cylinder with its axis along the z-direction and the field applied along the x- axis. Since there are no free currents, we may take ~H = −~∇φm. Gauss’s law for magnetism ~∇ · ~B = 0, along with the relation for a linear isotropic material ~B = µ ~H = µ0(1 + χm) ~H, implies that ~∇·(µ ~H) = 0. In regions of space or in parts of a material where µ is a constant, we have µ~∇ · ~H = 0. Using ~H = ~∇φm in this equation leads to ∇2φm = 0. We may thus solve the problem in a manner similar to that used for a dielectric sphere in a uniform electric field, though the boundary conditions are different. The applied field is B0, which corresponds to H0 = B0/µ0. Assume that the solutions to Laplace’s equations correspond to a uniform field in the interior of the cylinder and a uniform field and a dipole in the exterior. Note that this is a dipole in cylinderical co-ordinates (line dipole), so it has a different form than that for point dipoles, we have (the radius of the cylinder is a), φintm = −Ax = −Arcosφ, r < a; φextm = −B0rcosφ+ C a2 r cosφ r > a (10) 2 a flat edge parallel to a wire and distance b from the wire. The apex of the triangle is at distanc a+ b from the wire. The flux in the loop is then given by, φloop = 2 ∫ l 0 dx ∫ y 0 µ0idy ′ 2π(b+ y′) (26) where l = a/31/2, which is one half the length of a side and y = xtan60 = 31/2x. Evaluating give, φloop = Mi, with M = µ0b π31/2 [(1 + a b )ln(1 + a b )− a b ] (27) 10.27 Energy = volume ×B2/(2µ0). 10.28 The magnetic field of a dipole is, ~B = µ0 4π 3(~m · r̂)r̂ − ~m r3 (28) and the field energy density is given by, u = B2/2µ0, so the total magnetic energy is given by, U = ∫ ud3r = 2π ∫ π 0 ∫ ∞ RE r2sinθ 1 µ0 ( µ0 4πr3 )2(3(~m · r̂)r̂ − ~m) · 3(~m · r̂)r̂ − ~m (29) which gives, U = ∫ ud3r = 2π ∫ π 0 ∫ ∞ RE r2sinθ 1 µ0 ( µ0 4πr3 )2m2(3cos2θ + 1) (30) or U = ∫ ud3r = µ0m 2 8π2 ∫ π 0 ∫ ∞ RE 1 r4 sinθ(3cos2θ + 1) = µ0m 2 12πR3E (31) 11.2 a) Plates radius a, separation d with potential difference V (t). Using a circular contour and ∫ ~B · d~l = µ0i+ µ00 dφE dt (32) The LHS gives 2πB(r). The current enclosed is zero. For r < a, the electric flux enclosed in the loop is φE = V πr 2/d, while for r > a, the enclosed electric flux is V πa2/d. We then find that, B(r) = µ00r 2d dV dt ; r < a, B(r) = µ00a 2 2rd dV dt ; r > a (33) b) A wire carrying i has field µ0i/(2πr). The current is dQ/dt which is also CdV/dt, where C = 0A/d = 0πa 2/d. Combining these equations, gives the result quoted. 11.13 Maxwell’s equations in differential form, in vaccum, are given by, ~∇ · ~E = 0, (34) 5 ~∇ · ~B = 0 (35) ~∇∧ ~E = −∂ ~B ∂t (36) ~∇∧ ~B = µ00 ∂ ~E ∂t (37) To check that the solutions that are given satisfy these four equations, use Cartesian co- ordinates. The electric field is given by Ex = 0, Ey = 0, Ez = E0cos(πx/L)cos(πy/L)sin(ωt) (38) and the divergence of ~E is given by, ∂Ex ∂x + ∂Ey ∂y + ∂Ez ∂z = 0 (39) The magnetic field is Bx = −B0cos(πx/L)sin(πy/L)cos(ωt), By = B0sin(πx/L)cos(πy/L)cos(ωt) Bz = 0 (40) so the divergence of ~B is given by, ∂Bx ∂x + ∂By ∂y + 0 = B0π L sin(πx/L)sin(πy/L)cos(ωt)− B0π L sin(πx/L)sin(πy/L)cos(ωt) = 0 (41) The other parts are similar plug and chug 11.21 a) Intensity is the average energy per unit area per unit time. The energy per unit time per unit area or energy flux density is related to the electric and magnetic fields through the Poynting vector ~S = 1 µ0 ~E ∧ ~B (42) The electric and magnetic fields are perpendicular to each other and in phase so the cross product gives Peak energy flux density = 1 mu0 EpeakBpeak (43) The peak energy flux density is twice the intensity, and the peak magnetic and electric fields are related by, Epeak = cBpeak, so we have, 1300W/m2 = 1 2µ0 cB20 so that B0 = 2.3× 10−6T ; E0 = 700V/m (44) 6 Similar calculations for b) and c) d) n(t)hν = Intensity where n(t) is the number of photons arriving per unit area per unit time. 7