MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2013 Exam 3 Equation Sheet ! ! ! ! Force Law: F = q(E + v ! B ) Faraday’s law q ext q ext ! ! d ! E ! ds = # B ! nˆ da "" dt "" Force on Current Carrying Wire: closed open fixed path surface S ! ! ! F = Ids!"Bext # ! = Iind R wire Gauss’s Law for Magnetism: ! Magnetic Dipole B ! nˆ da = 0 ! """ µ = IAnˆ closed surface RHR Torque on a Magnetic Dipole Gauss’s Law: ! ! ! ! ! q ! = µ " B enc ext """ E ! dA = #o closed surface Force on a Magnetic Dipole ! ! ! ! F ( B) Electric Potential Difference = ! µ " B Electrostatics: ! z b Fz = µz ! ! !z !V = V "V # " E$ds b a % a Source Equations: ! ! ! ! E = !"V ! ! dq dq!(r " r!) E(r) = k rˆ = k e # 2 e # ! ! 3 source r source r " r! Potential Energy: ! !U = q!V ! ! µ Ids " rˆ B(r) = o 4 # 2 ! source r Capacitance: ! ! ! ! ! µ Ids$ " (r % r$) Q 1 Q2 1 B(r) = o C = U = = C!V 2 4 # ! ! 3 E ! source r % r$ !V 2 C 2 !B,Total N!B rˆ points from source to field point Inductance: L = = I I Ampere’s Law 1 2 ! back = "LdI / dt U = LI ! ! ! M 2 B ! d s = µ J ! nˆ da "" 0 "" closed S path Energy Density Stored in Fields: 1 2 1 2 uE = 2 ! 0 E ; uB = 2 B / µ0 1 Current Density and Current: Power Dissipated in Resistor: ! I J nˆ da P = I 2 R = !V 2 / R = "" ! Joule open surface Ohm’s Law: !V = I R ! ! Constants: J = ! cE where ! c is the conductivity ! ! $7 -1 µ0 ! 4" # 10 T % m % A E = !r J where !r is the resistivity ! " 1/ µ c2 8.85 # 10-12 C2 $ N-1 $ m-2 0 0 ! ! ! Power: P = F ! v k = 1/ 4!" 9.0 # 109 N $ m2 $C-2 e 0 ! Power from Voltage Source: P = I!V souce 2 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2013 Exam 3 Practice Problems Solutions Problem 1 Biot Savart A current loop, shown in the figure below, consists of two arc segments as shown, with a common center at P. One arc segment has an opening angle of 120 degrees and the other arc segment has an opening angle of 240 degrees. Two straight line segments join the arc segments. One arc segment has radius R and the other arc segment has radius R /2. A current I1 flows clockwise in the loop in the direction shown. a) What is the direction and magnitude of ! the magnetic field B at the point P ? Answer: The sides will not contribute because the cross product is zero there. ! ! ! µ I ds " rˆ µ I µ I 1 µ I B (r) = o 1 = $ zˆ o 1 ds = $zˆ o 1 2! R = $zˆ o 1 outer arc 4! # r 2 4 R2 # 4 R2 3( ) 6R ! ! ! ! ! µ I ds " rˆ µ I µ I 2 R 2µ I B (r) = o 1 = $ zˆ o 1 ds = $zˆ o 1 2! = $zˆ o 1 inner arc # 2 2 # 2 4! r 4! R / 2 4! R / 2 3 2 3R ( ) ( ) ! ! µ I 4µ I 5µ I Therefore the total field at P is given by B(r) = !zˆ o 1 ! zˆ o 1 = !zˆ o 1 6R 6R 6R Two fixed conducting rails are arranged as shown in the figure below. A metal bar of length s is placed at the origin, initially held in place, and a current I runs through the 2 rails and bar clockwise. The bar is then released. You may assume that the length of the bar is very short and that the magnetic field you calculated in part a) is uniform over the length of the bar. You may neglect the magnetic field due to the current through the rails. Solving 7-3 b) What is the direction and magnitude of the magnetic force acting on bar the instant it is released? Answer: The field at point P is inward from our calculations above, and the current in the bar is upward, so the total force on the bar is in the negative x direction, with c) ! 5µo I1 F = !xˆ I2s 6R Solving 7-4 Problem 2: Current Ring A circular ring of radius R has a current I in the counterclockwise direction as seen from above. c) Calculate the magnetic field due to the current at an arbitrary point a distance z along the z -axis passing through the center of the ring perpendicular to the plane of the ring. z =+# ! ! d) Calculate the line integral $ B ! ds along the z-axis from z = !" to z = +! . z="# e) The path is not a closed and yet it satisfies Ampere’s Law. Why? Solution: We shall apply the Biot-Savart law to calculate the magnetic field at the point P a distance z along the positive z-axis # =2! ! # =2! ! ! ! ! µ I d s " rˆ µ I d s " (r - r%) B(P) = 0 = 0 $ 4 r 2 $ 4 ! ! 3 # =0 ! # =0 ! r - r% Choose a coordinate system and unit vectors as shown in the figure below. Solving 7-5 ! ! ! Then d s = Rd ˆ , r = z kˆ , and r = R rˆ . Then the Biot Savart Law becomes ! ! ! ! " =2! µ I (Rd" "ˆ) # (z kˆ - R rˆ) B(z) = 0 . (1.1) $ 4 3 " =0 ! z kˆ - R rˆ ˆ ˆ ˆ ˆ We now explicitly calculate the cross products noting that ! " k = rˆ and ! " (#rˆ) = k and Eq. (1.1) becomes ! µ IzR " =2! µ IR2 " =2! B(z) = 0 d"rˆ + 0 d" kˆ (1.2) 4!(z2 + R2 )3/ 2 # 4!(z2 + R2 )3/ 2 # " =0 " =0 The first integral vanishes because as we integrate around the circle the unit vector rˆ always points radially outward and hence sums to zero. Equivalently if we choose ˆ ˆ coordinates such that rˆ = cos!i + sin!j then ! µ IzR " =2! µ IzR " =2! B(z) = 0 d"rˆ = 0 d"(cos"ˆi + sin"ˆj) 2 2 3/ 2 # 2 2 3/ 2 # 4!(z + R ) " =0 4!(z + R ) " =0 µ IzR " =2! " =2! ! = 0 (sin" ˆi + $cos" ˆj) = 0 2 2 3/ 2 " =0 " =0 4!(z + R ) The second integral in Eq. (1.2) is straightforward and so the magnetic field along the z- axis is given by 2 2 ! µ IR " =2! µ IR B(z) = 0 " kˆ = 0 kˆ . (1.3) 4!(z2 + R2 )3/ 2 " =0 2(z2 + R2 )3/ 2 z =+# ! ! We now calculate the line integral $ B ! ds along the z-axis from z = !" to z = +! . z="# z z 2 z 2 =+# ! ! =+# µ IR =+# µ IR B ! ds = 0 kˆ ! dzkˆ = 0 dz (1.4) $ $ 2(z2 + R2 )3/ 2 $ 2(z2 + R2 )3/ 2 z="# z="# z="# This integral is equal to z z 2 z=+# =+# ! ! =+# µ IR µ Iz µ I µ I B ! ds = 0 dz = 0 = 0 " " 0 = µ I (1.5) $ $ 2(z2 + R2 )3/ 2 2(z2 + R2 )1/ 2 2 2 0 z="# z="# z="# Solving 7-6 where we have used the fact that µ I(z) µ I µ I(z) µ I lim 0 = 0 and lim 0 = " 0 . z 2 2 1/ 2 z 2 2 1/ 2 !" 2((z) + R ) 2 !"# 2((z) + R ) 2 ! ! Recall that Ampere’s Law states that B ! ds = µ I however the integral above is not a "" 0 enc closed line integral. ! However if we integrate B along a semi-circular path at infinity from z = +! to z = !" and add our two line integrals together we get a closed line integral z =+# ! ! ! ! ! ! $ B ! ds + $ B ! ds = "$ B ! ds (1.6) z="# semi-circle atinfinity ! Although we have only calculated B(z) along the z-axis, we see that ! µ IR2 ! limB(z) = lim 0 kˆ = 0 . This holds everywhere on the semi-circle at infinity z!" z!" 2(z2 R2 )3/ 2 + so ! ! ! " B ! ds = 0 (1.7) semi-circle atinfinity Therefore combining Eqs. (1.7), (1.6), and (1.5) we have that ! ! B ! ds = µ I (1.8) "" 0 From the figure above we see that I = I (1.9) enc Therefore Ampere’s Law is satisfied. Solving 7-7 Problem 3: Magnetic fields of Current Loops (a) Find the magnetic field at point P due to current loop shown in the figure below. Solution: (b) There is no magnetic field due to the straight segments because point P is along the ! lines. Using the general expression for dB obtained in (a), for the outer segment, we have ! ! µ Id" $ µ I ' B = 0 kˆ = 0 kˆ out # 4! b %& 4b () 0 Similarly, the contribution to the magnetic field from the inner segment is 0 ! µ Id" % µ I ( B = 0 kˆ = $ 0 kˆ in # 4! a &' 4a )* ! Therefore the net magnetic field at Point P is ! ! ! µ I " 1 1% B = B + B = ! 0 ! kˆ net out in 4 #$ a b&' . Solving 7-8 Problem 4: Torque on Circular Current Loop A wire ring lying in the xy-plane with its center at the origin carries a counterclockwise current I. There is an external uniform magnetic field ! B = B ˆi + B ˆj such that B > 0 and x y y ! B < 0 .The magnetic moment vector µ x is perpendicular to the plane of the loop and has magnitude µ = IA and the direction is given by right-hand-rule with respect to the direction of the current. What is the direction and magnitude of the torque on the loop? Solution: The torque on a current loop in a uniform field is given by ! ! ! , ! = µ " B ! ! where µ is the magnetic dipole moment with magnitude µ = IA and points perpendicular to the plane of the loop and right-handed with respect to the direction of current flow.