MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II November 9, 2014 Prof. Alan Guth PROBLEM SET 7 SOLUTIONS PROBLEM 1: J.J. THOMPSON AND THE CHARGE TO MASS RATIO OF THE ELECTRON (10 points) Griffiths Problem 5.3 (p. 216). ~ and the electric field E ~ (a) The beam is not deflected by the uniform magnetic field B when the forces exerted by these fields cancel each other: ~ + (~v × B) ~ =0. F~ = q E (1.1) ~ B ~ and the velocity ~v of the particles in the beam are perpendicular Since the fields E, to each other, both fields will exert a force along the same direction. This allows us to write Eq. (1.1) in terms of the magnitudes of the fields E and B and the magnitude of the velocity v: E = vB =⇒ v= E . B (1.2) (b) When the electric field is turned off, each particle will experience only the force from ~ This force will always be orthogonal to its velocity and thus the magnetic field B. will cause it to move along a circular orbit with the radius R. In this case, the force exerted by the magnetic field must supply the centripetal force for the orbit of radius R, which gives us v2 qvB = m R =⇒ q v = = m BR E . B2R (1.3) PROBLEM 2: EXAMPLES OF THE USE OF THE BIOT-SAVART LAW (15 points) Griffiths, Problem 5.8 (p. 228). (a) We can find the magnetic field in the center of a loop by summing the contributions of each of the four straight sides of the square loop. The figure on the right illustrates one convenient way to parametrize the calculation for a single side: we consider a straight piece of wire located at a distance s from P, the center of the loop, and break it into infinitesimal segments of wire, each lo- 8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014 p. 2 cated in the angular sector (θ; θ + dθ) with respect to P. Then the horizontal coordinate of each infinitesimal segment is `0 = s tan θ along the wire, and the length of the segment is d`0 = s d(tan θ) = s dθ/ cos2 θ. The distance from d~` to P is = s/ cos θ. The infinitesimal magnetic field at point P is pointed towards the reader and has magnitude s dθ cos2 θ µ0 I µ0 I |d~`0 × ˆ | µ0 I cos θ = = dθ cos θ. dB = 2 2 2 4π cos θ s 4πs 4π (2.1) Then the magnetic field produced by a piece of wire at a distance s in the angular sector (θ1 , θ2 ) with respect to P is equal θ2 Z B= dB(θ) = θ1 where s = R, θ2 = −θ1 = ~ tot B π 4 µ0 I (sin θ2 − sin θ1 ) , 4πs (2.2) for one side of the square. The total magnetic field is µ0 I =4 4πR √ √ ! 2 2 − − zˆ = 2 2 √ 2µ0 I zˆ , πR (2.3) where the z axis is perpendicular to the square, upwards in Fig. 5.22 of the problem set. (b) For an n-sided polygon, the angle between two lines drawn from the center to the two corners of one edge is 2π/n. Then we get θ2 = −θ1 = nπ . The total magnetic field is h i ~ tot = n µ0 I sin π − sin − π zˆ = B 4πR n n nµ0 I π sin zˆ . 2πR n (2.4) (c) When n → ∞, µ0 I sin x µ0 I sin nπ ~ zˆ = lim = Btot = lim π n→∞ 2R 2R x→0 x n µ0 I zˆ . 2R (2.5) This result agrees with our previous result for the magnetic field at the center of a circular current carrying wire. 8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014 p. 3 PROBLEM 3: MAGNETIC FIELD ON THE AXIS OF A TIGHTLY WOUND SOLENOID (15 points) (a) The wire forms a helix, but we can approximate each turn as a circle in the x-y plane. Since the number of turns per unit length is n, the current of a ring of length dz is given as dI = nIdz. Then the magnetic field at the origin produced by a ring at height z is given by Griffiths’ Eq. (5.41) (which was also derived in lecture) as dB = µ0 nIdz R2 , 2 (R2 + z 2 )3/2 (3.1) ~ = dB zˆ is the contribution from the ring at height z. The total magnetic where dB field at the origin is then Z ~ = µ0 nI zˆ B 2 z2 z1 R2 dz . (R2 + z 2 )3/2 (3.2) Using the change of variable z = R tan θ, with dz = R cosdθ2 θ , we find Z Using sin θ = R2 dz = (R2 + z 2 )3/2 √ z z 2 +R2 z2 →∞ find R2 cos3 θ Rdθ = R3 cos2 θ Z cos θdθ = sin θ . (3.3) and restoring the limits of integration, ~ = µ0 nI B 2 z2 (b) Since lim p Z z22 + R2 = 1 and " z2 p z22 + R2 lim z1 →−∞ −p z1 p z1 z12 + R2 z12 + R2 # zˆ . (3.4) = −1, for an infinite solenoid we ~ solenoid = µ0 nI zˆ . B (3.5) 8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014 p. 4 ~ FIELD (10 PROBLEM 4: VECTOR POTENTIAL FOR A UNIFORM B points) Griffiths Problem 5.25 (p. 248). ~ ·A ~ = 0: Let’s first check ∇ ~ ·A ~ = −1∇ ~ · (~r × B), ~ ∇ 2 1 ~ ~ ~ × B) ~ =0, =− B · (∇ × ~r) − ~r · (∇ 2 (4.1) ~ × ~r) = 0, where the first term vanishes because the curl of a radial vector field is zero, (∇ ~ is uniform. Then and the second term vanishes because the magnetic field B ~ ×A ~ = −1∇ ~ × (~r × B) ~ ∇ 2 1 ~ ~ ~ B ~ + ~r(∇ ~ · B) ~ − B( ~ ∇ ~ · ~r) . · ∇)~r − (~r · ∇) = − (B 2 (4.2) ~ B ~ = 0, The terms involving the partial derivative of the magnetic field give zero, (~r · ∇) ~ · B) ~ = 0. The first term in Eq. (4.2) is and ~r(∇ ~ · ∇)~ ~ r= (B ∂ ∂ ∂ + By + Bz (xˆ ex + yˆ ey + zˆ ez ) Bx ∂x ∂y ∂z ~ . = Bx eˆx + By eˆy + Bz eˆz = B (4.3) The last term in Eq (4.2) is, ∂x ∂y ∂z ~ ∇ ~ · ~r) = B ~ ~ . B( + + = 3B ∂x ∂y ∂z (4.4) ~ ×A ~ = − 1 (B ~ − 3B) ~ =B ~ . ∇ 2 (4.5) Then we get Is this result unique? In localized situations, where we can impose the boundary condition ~ → ~0 as |~r | → ∞, the conditions ∇ ~ ×A ~=B ~ and ∇ ~ ·A ~ have a unique solution. In that |A| ~ field extending to infinity, no such boundary condition is this case, however, with the B ~ is not unique. One possible change would be to add any constant vector. possible, so A ~ = B0 zˆ, then A ~ = B0 xˆ ~ = −B0 yˆ Bigger differences are possible, too. If B y or A x both work. 8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014 p. 5 PROBLEM 5: A CHARGED PARTICLE IN THE FIELD OF A MAGNETIC MONOPOLE (20 points) Griffiths Problem 5.45 (p. 258). (a) The force F~ on qe is, ~ = µ0 qe qm (~v × rˆ) . F~ = qe (~v × B) 4π r2 (5.1) Let the accelaration of qe be ~a, then F~ = m~a =⇒ ~a = µ0 qe qm (~v × ~r) . 4π mr3 (5.2) (b) Since the force on particle qe is found to be perpendicular to its velocity ~v , this force will not perform work on the particle. Therefore the kinetic energy of particle does not change and the speed v = |~v | is a constant of the motion. d 1 dEkin = m~v · ~v = m~a · ~v = 0 . dt dt 2 (5.3) (c) It is claimed that the vector quantity ~ ≡ m~r × ~v − µ0 qe qm rˆ Q 4π (5.4) ~ with respect to time, finding is conserved. To verify this, we differentiate Q ~ dQ µ0 qe qm d ~r = m(~v × ~v ) + m(~r × ~a) − , dt 4π dt r µ0 qe qm ~v µ0 qe qm ~r dr =0+ ~r × (~v × ~r) − − 2 . 4πr3 4π r r dt (5.5) The time derivative or r, dr ˆ, dr v · rˆ = 1r ~v · ~r. dt is the velocity component along r dt = ~ The triple product rule gives ~r × (~v ×~r) = r2~v −~r(~r ·~v ). Then putting these relations into Eq. (5.5), we get ~ ~v dQ µ0 qe qm 1 2 ~r (~v · ~r) = r ~v − ~r(~r · ~v ) − + 2 =0. dt 4π r3 r r r (5.6) 8.07 PROBLEM SET 7 SOLUTIONS, FALL 2014 p. 6 ~ points along z-axis, so Q ~ · rˆ = Q zˆ · rˆ = Q cos θ. (d) In the new coordinate system Q Then, from Eq. (5.4), we find µ0 qe qm ~ · rˆ = m(~r × ~v ) · rˆ − µ0 qe qm (ˆ Q r · rˆ) = , 4π 4π so Q cos θ = − µ0 qe qm . 4π (5.7) (5.8) ~ is a constant of the motion. Then Eq. (5.8) implies that In part (c) we found that Q the motion is confined to a fixed value of the polar angle θ, hence to a cone. The polar angle θ is cos θ = − µ0 qe qm . 4πQ (5.9) PROBLEM 6: THE MAGNETIC FIELD OF A SPINNING, UNIFORMLY CHARGED SPHERE (25 points) Griffiths Problem 5.60 (p. 264). This problem has been held over to Problem Set 8. PROBLEM 7: THE COMPLETE MAGNETIC FIELD OF A MAGNETIC DIPOLE: (10 points) This problem has been held over to Problem Set 8.