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Massachusetts Institute of Technology Department of Physics Physics 8.07 Fall 2005 Problem Set 5 Solutions Problem 1: Fields in a non-uniform material a) Griffths Problems 4.33 (p. 198) ~ t ] = 0. With the Boundary conditions at the interface are [Dn ] = 0 and [E geometry of Griffiths Figure 4.34, the normal components are given by a ~ = E, ~ factor cos θ and the tangential components by a factor sin θ. With D this gives Dn = 1 E1 cos θ1 = 2 E2 cos θ2 , Et = E1 sin θ1 = E2 sin θ2 . Combining these gives tan θ2 2 = . tan θ1 1 Because the sense of deflection is similar to Snell’s law, a convex dielectric “lens” will “focus” the electric field. It’s fascinating that the amount of focusing is different for static field lines than for electromagnetic waves – EM waves don’t necessarily travel along the static field lines! ~ ·D ~ = 0 or b) We want to find Ex (z) and Ez (z). Instead of [Dn ] = 0 we have ∇ d (Ez ) = 0 . dz ~ t ] = 0 we have The solution of this is (z)E(z) = constant. Instead of [E ~ ×E ~ = 0 or ∇ dEx =0. dz The solution is Ex = constant. With the values given at z = 0, we get 0 Ex = E0 , Ez = E0 . (z) These imply Dx = (z)E0 , Dz = 0 E0 , and, since ρ = ρf + ρb = ρb , ~ ·E ~ = −E0 0 ρb = 0 ∇ 1 2 d . dz Problem 2: Griffiths Problem 5.3 (p. 208) ~ + ~v × B ~ ) = 0. With ~v , E ~ and B ~ mutually perpena) No deflection means q(E dicular, this implies E = vB or v = E/B. ~ = m~a. Charges will travel in circular b) With only a magnetic field, F~ = q~v × B ~ and centripetal acceleration orbits of radius R with ~v perpendicular to B v 2 /R, implying mv 2 /R = qvB or q/m = v/(RB) hence q E = 2 . m B R Problem 3: Griffiths Problem 5.10 (p. 220) This problem requires the magnetic field of a long, straight wire carrying steady current ~ = µ0 I/(2πs)~eφ where s is the distance from the wire and ~eφ is the azimuthal unit I: B vector whose direction is given by the right-hand rule. Also, the force on a current I ~ In this problem, we define the Cartesian coordinate flowing along d~l is dF~ = Id~l × B. system so that the loop lies in the x-y plane, with the long wire pointing along ~ex . The magnetic field on the loop points in the direction ~ez . a) The forces on the two segments perpendicular to the long wire are equal and opposite and therefore cancel. The other two segments have d~l = ±~ex dx giving forces in the direction ±~ex × ~ez = ∓~ey . Putting in the distances y = s and y = s + a, and integrating over x from 0 to a, we get " a2 1 µ 0 I 2 a2 µ0 I 2 a 1 − 1 + O ~ey = F~ = 2π s s+a 2πs2 s2 !# ~ey . For a s, the magnetic field at the center of the loop is ~ = B µ0 I µ0 I ~ez ≈ ~ez . 2π(s + a/2) 2πs For a s, the force on the loop may therefore be written ~ m ~), m F~ = ∇( ~ ·B ~ = −Ia2~ez . b) The geometry of the triangular loop is a bit more complicated. Label the three sides 1 (left), 2 (right), and 3 (bottom, parallel to the long wire). Along segments 1 and 2, √ ! 1 3 d~l = dl ~ex ± ~ey . 2 2 2 The forces arising from the y-component of d~l cancel in the sum. For a point that is a distance ` from the bottom corner √ of size 1 or 2, the y-value (needed to compute the magnetic field) is s + ` 3/2. The net force contribution from sides 1 and 2 is therefore √ ! 2 Z a 2 µ I 3 (d`/2) (~ e × ~ e ) µ I a 0 x z 0 √ F~1+2 = 2 = − √ ln 1 + ~ey . 2π 2s 0 s + ` 3/2 π 3 Adding to this the force on the horizontal segment gives the total force: √ !# " µ0 I 2 a 3a 2 ~ F = − √ ln 1 + ~ey . 2π s 2 s 3 For a small loop, a s, we can expand the logarithm using ln(1 + ) = − 2 /2 + · · ·, yielding √ √ √ ! 3 µ 0 I 2 a2 3 µ 0 I 2 a2 3 2 ~ ~ ~ ~ F = = ∇(m ~ ·B) , m ~ =− ~ey = ∇ − Ia ~ez . 2 8π s 8π s 4 √ The area of the equilateral triangle is ( 3/4)a2 , so that m = I ∗ Area, with direction perpendicular to the loop chosen by the right-hand rule. The vector m ~ is called the magnetic moment and is defined in Griffiths equation (5.84). Problem 4: Griffiths Problem 5.38 (p. 247) The current density is J~ = ρ−~v where ρ− < 0. Assuming J~ is uniform in the wire, the magnetic field is circumferential with magnitude given by Ampère’s law, 1 1 B = µ0 Js = µ0 ρ− vs . 2 2 Now, in order that the negative charges be in a steady state, the net force on them must ~ + ~v × B ~ = 0 yielding vanish, implying E ~ = −|v− B|~es = 1 µ0 ρ− v 2~s . E 2 (Note the sign choice is needed because with ρ− < 0 the field points radially toward the axis, so that negative charges feel an outwards electric force which cancels the inward magnetic force.) This field is generated by a net charge density ~ ·E ~ = 0 µ 0 ρ− v 2 , ρ = ρ− + ρ+ = 0 ∇ giving ρ+ = −(1 − 0 µ0 v 2 )ρ− or, with c2 = 1/(0 µ0 ), ρ− = −γ 2 ρ+ , γ = (1 − v 2 /c2 )−1/2 . The negative charges do not pinch because of the electric repulsion away from the axis. The compensating positive charge resides on the outside of the wire. 3