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Home , Current density, Magnetic moment

Prof. Anchordoqui

Problems set # 8 Physics 169 April 14, 2015

1. The unit of magnetic flux is named for Wilhelm Weber. The practical-size unit of magnetic field is named for Johann Karl Friedrich . Both were scientists at Gottingen, Germany. Along with their individual accomplishments, together they built a telegraph in 1833. It consisted of a battery and switch, at one end of a transmission line 3 km long, operating an at the other end. (Andr´eAmp´er`esuggested electrical signaling in 1821; Samuel Morse built a telegraph line between Baltimore and Washington in 1844.) Suppose that Weber and Gausss transmission line was as diagrammed in Fig. 1. Two long, parallel wires, each having a mass per unit length of 40.0 g/m, are supported in a horizontal plane by strings 6.00 cm long. When both wires carry the same current I, the wires repel each other so that the angle θ between the supporting strings is 16.0◦. (i) Are the currents in the same direction or in opposite directions? (ii) Find the magnitude of the current.

Solution The separation between the wires is a a = 2 · 6.00cm · sin 8.00◦ = 1.67 cm. (i) Because the wires repel, the currents are in opposite directions. (ii) Because the magnetic force acts hori- 2 zontally, FB = µ0I ` = tan 8.00◦, yielding I2 = mg2πa tan 8.00◦ and so I = 67.8 A. Fg 2πamg µ0`

2. Figure 2 is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is 1.00 A out of the page and the current in the outer conductor is 3.00 A into the page. Determine the magnitude and direction of the magnetic field at points a and b.

Solution From ’s law, the magnetic field at point a is given by B = µ0Ia , where I is a 2πra a the net current through the of the circle of radius ra. In this case, Ia = 1.00 A out of the page 4π×10−7 T·m/A·1 A (the current in the inner conductor), so Ba = 2π·1.00×10−3 m = 200 µT toward top of the page. µ0Ib Similarly at point b: Bb = , where Ib is the net current through the area of the circle having 2πrb radius rb. Taking out of the page as positive, Ib = 1.00 A − 3.00 A = −2.00 A, or Ib = 2.00 A into 4π×10−7 T·m/A·2 A the page. Therefore, Bb = 2π·3.00×10−3 m = 133 µT toward the bottom of the page.

3. A long cylindrical conductor of radius R carries a current I as shown in Fig. 3. The current J, however, is not uniform over the cross section of the conductor but is a function of the radius according to J = br, where b is a constant. Find an expression for the magnetic field B

(i) at a distance r1 < R and (ii) at a distance r2 > R, measured from the axis.

H H Solution Use Amp´er`e’slaw, B~ · d~s = µ0I. For J~, this becomes B~ · d~s = 2 R ~ ~ R r1 µ0br1 µ0 J · dA. (i) For r1 < R, this gives B2πr1 = µ0 0 br2πrdr and B = 3 (for r1 < R or 3 R R 2πµ0bR inside the cylinder. (ii) When r2 > R, Amp´er`e’slaw yields 2πr2)B = µ0 0 br2πrdr = 3 or 3 B = µ0bR (for r > Ror outside the cylinder). 3r2 2 4. A toroid with a mean radius of 20.0 cm and 630 turns is filled with powdered steel whose χ is 100. The current in the windings is 3.00 A. Find B (assumed uniform) inside the toroid.

Solution: Assuming a uniform B inside the toroid is equivalent to assuming r  R (see Fig. 4); 630·3.00 then B0 = µ0H ≈ µ0NI as for a tightly wound . This leads to B0 = µ0 2π·0.200 = 0.00189 T. With the steel, B = (1 + χ)µ0H = 101 · 0.00189 T = 0.191 T.

5. In Bohrs 1913 model of the , the is in a circular orbit of radius 5.29 × 10−11 m and its speed is 2.19 × 106 m/s. (i) What is the magnitude of the magnetic due to the motion? (ii) If the electron moves in a horizontal circle, counterclockwise as seen from above, what is the direction of this vector?

ev Solution The current induced by the electron is I = 2πr , so it is straightforward to see that ev 2 −24 2 the Bohr model predicts the correct magnetic moment µ = IA = 2πr πr = 9.27 × 10 A · m . (ii) Because the electron has a negative charge, its [conventional] current is clockwise, as seen from above, and ~µ points downward.

6. The magnetic moment of the is approximately 8.00 × 1022 A · m2. (i) If this were caused by the complete of a huge iron deposit, how many unpaired electrons would this correspond to? (ii) At two unpaired electrons per iron atom, how many kilograms of iron would this correspond to? [Hint: Iron has a density of 7, 900 kg/m3, and approximately 8.50 × 1028 iron /m3.]

8.00×1022 A·m2 45 Solution (i) Number of unpaired electrons = 9.27×10−24 A·m2 = 8.63 × 10 . Each iron atom 1 45 has two unpaired electrons, so the number of iron atoms required is 2 8.63 × 10 . (ii) Mass = 4.31×1045 atoms·7,900 kg/m3 20 8.50×1028 atoms/m3 = 4.01 × 10 kg.

7. Find the magnetic field on the axis at a distance a above a disk of radius R with σ rotating at an angular speed (clockwise when viewed from the point P ).

Solution Consider a thin ring between r and r + dr as shown in Fig. 5. The charge per length on the ring is q = σdr and each point on it is moving at a speed v = ω~r, so the current is I = vq = σωrdr. By symmetry, the magnetic field is in the z direction. Using the result from the 2 lectures, the contribution from the ring is dB = µ0(σωr dr) r . The magnetic field for the z 2 (r2+z2)3/2 3 entire disk is B = µ0σω R R r dr. Let u = r2, so du = 2rdr. This substitution (dont forget z 2 0 (r2+z2)3/2 2 2 2 R h 2 2 i µ0σω R R u µ0σω 2(u+2z ) µ0σω R +2z to change limits) gives Bz = 2 3/2 du = 2 1/2 = 2 2 1/2 − 2z . 4 0 (u+z ) 4 (u+z ) 0 2 (R +z )

8. A thin uniform ring of radius R and mass M carrying a charge +Q rotates about its axis with constant angular speed ω. Find the ratio of the magnitudes of its magnetic moment to its . (This is called the .)

Q Qω Solution: The current in the ring shown in Fig. 6 is i = T = 2π . The magnetic moment is ˆ 2 Qω ˆ QωR2 ˆ ~ ˆ 2 ˆ ~µ = Aik = πR 2π k = 2 k. The angular momentum is L = Iωk = MR ωk. So the gyromag- 2 netic ratio is |~µ| = QωR /2 = Q . |L~ | MR2ω 2M

9. A wire ring lying in the xy-plane with its center at the origin carries a counterclockwise I. There is a uniform magnetic field B~ = Bˆı in the +x-direction. The magnetic moment vector µ 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 on the loop?

Solution: The torque on a current loop in a uniform field is given by ~τ = ~µ × B~ , where µ = IA and the vector ˆı is perpendicular to the plane of the loop and right-handed with respect to the direction of current flow. The moment is given by ~µ = IA~ = πIR2kˆ. Therefore, ~τ = ~µ × B~ = πIR2kˆ × Bˆı = πIR2Bˆ. Instead of using the above formula, we can calculate the torque directly as follows. Choose a small section of the loop of length ds = Rdθ. Then the vector describing the current-carrying element is given by Id~s = IRdθ(− sin θˆı + cos θˆ). The force dF~ that acts on this current element is dF~ = Id~s × B~ = IRdθ(− sin θˆı + cos θˆ) × Bˆı = −IRB cos θdθkˆ. The force acting on the loop can be found by integrating the above expression, F~ = H dF~ = R 2π ˆ 2π ˆ 0 (−IRB cos θ)dθk = −IRB sin θ|0 k = 0. We expect this because the magnetic field is uniform and the force on a current loop in a uniform magnetic field is zero. Therefore we can choose any point to calculate the torque about. Let ~r be the vector from the center of the loop to the element Id~s. That is, ~r = R(cos θˆı + sin θˆ). The torque d~τ = ~r × dF~ acting on the current element is then d~τ = ~r × dF~ = R(cos θˆı + sin θˆ) × (−IRBdθ cos θkˆ) = −IR2Bdθ cos θ(sin θˆı − cos θˆ). Finally, H R 2π 2 integrate d~τ over the loop to find the total torque, ~τ = d~τ = − 0 IR Bdθ cos θ(sin θˆı−cos θˆ) = πIR2Bˆ. This agrees with our result above.

10. A square coil with sides equal to 25.0 cm carries a current of 2.00 A. It lies in the z = 0 plane in a magnetic field B~ = 0.40ˆı + 0.30kˆ T with the current counter-clockwise when viewed from a point on the positive z-axis. If the coil has 6 turns what is (i) the torque acting on the coil, and (ii) the potential of the coil/field system?

Solution If the current is counter-clockwise when viewed from a point on the positive z-axis, then the magnetic moment of the coil is in the positive z direction (i.e., along kˆ). It follows that ~µ = niAkˆ = 6 × (2.00 A) × (0.25 m)2kˆ = (0.75 A · m2)kˆ. (i) The torque is ~τ = ~µ × B~ = (0.75 A · m2)kˆ × (0.40ˆı + 0.30kˆ) T = (0.30 N · m)ˆ. (ii) The is U = −~µ · B~ = (0.75 A · m2)kˆ · (0.40ˆı + 0.30kˆ) T = 0.225 J. Problems 959

Wire 3 is located such that when it carries a certain × I1 current, each wire experiences no net force. Find (a) the 5.00 cm position of wire 3, and (b) the magnitude and direction of the current in wire 3. P 13.0 cm 20. The unit of magnetic flux is named for Wilhelm Weber. The practical-size unit of magnetic field is named for 12.0 cm Johann Karl Friedrich Gauss. Both were scientists at Göttingen, Germany. Along with their individual accomplishments, together they built a telegraph in 1833. × I2 It consisted of a battery and switch, at one end of a Figure P30.15 transmission line 3 km long, operating an electromagnet at the other end. (André Ampère suggested electrical signaling in 1821; Samuel Morse built a telegraph line between Baltimore and Washington in 1844.) Suppose that Section 30.2 The Magnetic Force Between Two Weber and Gauss’s transmission line was as diagrammed in Parallel Conductors Figure P30.20. Two long, parallel wires, each having a mass per unit length of 40.0 g/m, are supported in a horizontal 16. Two long, parallel conductors, separated by 10.0 cm, carry plane by strings 6.00 cm long. When both wires carry the currents in the same direction. The first wire carries same current I, the wires repel each other so that the angle ϭ ϭ current I1 5.00 A and the carries I 2 8.00 A. ␪ between the supporting strings is 16.0°. (a) Are the (a) What is the magnitude of the magnetic field created by currents in the same direction or in opposite directions? I1 at the location of I 2? (b) What is the force per unit (b) Find the magnitude of the current. length exerted by I1 on I 2? (c) What is the magnitude of the magnetic field created by I 2 at the location of I1? y (d) What is the force per length exerted by I 2 on I1? 17. In Figure P30.17, the current in the long, straight wire is 960 CHAPTER 30 • Sources of the Magnetic6.00 cm I1 ϭ 5.00 A and the wire lies in the plane of the rectangu- 23. Figure P30.23 is a cross-sectional θview of a coaxial cable. of the can and I the upward current, uniformly distributed lar loop, which carries the current I2 ϭ 10.0 A. The dimen- The center conductor is surrounded by16.0 a rubber° layer, over its curved wall. Determine the magnetic field (a) just sions are c ϭ 0.100 m, a ϭ 0.150 m, and ᐉ ϭ 0.450 m. Find which is surrounded by an outer conductor, which is inside the wall and (b) just outside. (c) Determine the surroundedz by another rubber layer. In a particular pressure on the wall. the magnitude and direction of the net force exerted on application, the current in the inner conductor is 1.00 A 28. Niobium metal becomes a superconductor when cooled the loop by the magnetic field created by the wire. out of the page and the current in the outer conductorx is below 9 K. Its is destroyed when the 3.00 A into the page. Determine the magnitude and surface magnetic field exceeds 0.100 T. Determine the direction of the magnetic field at points a and b. FigureFigure 1: ProblemP30.20 1. maximum current a 2.00-mm-diameter niobium wire can carry and remain superconducting, in the absence of any external magnetic field. × 29. A long cylindrical conductor of radius R carries a current I × × as shown in Figure P30.29. The current density J, however, Section 30.3 Ampère’s Law3.00 A is not uniform over the cross section of the conductor but ϭ a b is a function of the radius according to J br, where b is a 21. Four long, parallel× 1.00 A conductors. × carr. y equal currentsconstant. Findof an expression for the magnetic field B (a) at a distance r 1 Ͻ R and (b) at a distance r 2 Ͼ R, I ϭ 5.00 A. Figure P30.21 is an end view of the conductors.measured from the axis. I1 × × The current direction is into× the page at points A and B I2 I (indicated by the crosses) 1and mm 1 mmout1 mm of the page at C and D

(indicated by the dots).Figure P3Calculate0.23 the magnitude andr 2 ᐉ Figure 2: Problem 2. r 1 direction of the magnetic field at point P, located at the R center of24. theThe squaremagnetic fi eldof 40.0 edgecm away length from a long0.200 straightm. wire carrying current 2.00 A is 1.00 ␮T. (a) At what distance is Figure P30.29 it 0.100 ␮T? (b) What If? At one instant, the two conductors in a long household extension cord carry 30. In Figure P30.30, both currents in the infinitely long wires equal 2.00-A currents in opposite directions. The two wires are in the negative x direction. (a) Sketch the magnetic areA 3.00× mm apart. Find the magnetic field 40.0C cm away field pattern in the yz plane. (b) At what distance d along from the middle of the straight cord, in the plane of the the z axis is the magnetic field a maximum? two wires. (c) At what distance is it one tenth as large? (d) The center wire in a coaxial cable carries current c a 2.00 A in one direction and the sheath around it carries z current 2.00 A in the opposite direction. What magnetic field does the cable create at points outside?0.200 m a 25. A packed bundle of 100 long, straight, insulated wires Figure P30.17 P a forms a cylinder of radius R ϭ 0.500 cm. (a) If each wire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on a wire located I 18. Two long, parallel wires are attracted to each other by a 0.200 cm from the center of the bundle? (b) What If ? Would aB wire× on the outer edge of the bundle experience a force x force per unit length of 320 ␮N/m when they are sepa- greater or smaller than the value calculated in partD (a)? y 0.200 m I 26. The magnetic coils of a tokamak fusion reactor are in the rated by a vertical distance of 0.500 m. The current in the Figure P30.30 shape of a toroidFigure having an innerP30.21 radius of 0.700 m and an upper wire is 20.0 A to the right. Determine the location outer radius of 1.30 m. The toroid has 900 turns of large- diameter wire, each of which carries a current of 14.0 kA. of the line in the plane of the two wires along which the 22. A long straightFind the wire magnitude lies of onthe magnetic a horizontal field inside the table toroid and carries a Section 30.4 The of a Solenoid total magnetic field is zero. current of along1.20 (a)␮ theA. inner In radius a vacuum, and (b) the outer a radius. moves parallel 27. Consider a column of passing through 31. What current is required in the windings of a long 19. Three long wires (wire 1, wire 2, and wire 3) hang verti- to the wireplasma (opposite (ionized gas).the Filamentscurrent) of current with withina constant the speedsolenoid thatof has 1 000 turns uniformly distributed over a column are magnetically attracted to one another. They length of 0.400 m, to produce at the center of the solenoid 4 Ϫ4 cally. The distance between wire 1 and wire 2 is 20.0 cm. 2.30 ϫ 10 canm/s crowd at together a distance to yield a verd yabove great current the density wire. Determinea magnetic field of magnitude 1.00 ϫ 10 T? and a very strong magnetic field in a small region. 32. Consider a solenoid of length ᐉ and radius R, containing On the left, wire 1 carries an upward current of 1.50 A. To the value ofSometimes d. You the currentmay canignore be cut off the momentarily magnetic by this fieldN dueclosely tospaced turns and carrying a steady current the right, wire 2 carries a downward current of 4.00 A. the Earth. pinch effect. (In a metallic wire a pinch effect is not I. (a) In terms of these parameters, find the magnetic important, because the current-carrying electrons repel field at a point along the axis as a function of distance one another with electric forces.) The pinch effect can be a from the end of the solenoid. (b) Show that as ᐉ demonstrated by making an empty aluminum can carry a becomes very long, B approaches ␮0NI/2ᐉ at each end of large current parallel to its axis. Let R represent the radius the solenoid. 960 CHAPTER 30 • Sources of the Magnetic Field

23. Figure P30.23 is a cross-sectional view of a coaxial cable. of the can and I the upward current, uniformly distributed The center conductor is surrounded by a rubber layer, over its curved wall. Determine the magnetic field (a) just which is surrounded by an outer conductor, which is inside the wall and (b) just outside. (c) Determine the surrounded by another rubber layer. In a particular pressure on the wall. application, the current in the inner conductor is 1.00 A 28. Niobium metal becomes a superconductor when cooled out of the page and the current in the outer conductor is below 9 K. Its superconductivity is destroyed when the 3.00 A into the page. Determine the magnitude and surface magnetic field exceeds 0.100 T. Determine the direction of the magnetic field at points a and b. maximum current a 2.00-mm-diameter niobium wire can carry and remain superconducting, in the absence of any external magnetic field. × 29. A long cylindrical conductor of radius R carries a current I × × as shown in Figure P30.29. The current density J, however, 3.00 A is not uniform over the cross section of the conductor but ϭ a b is a function of the radius according to J br, where b is a × 1.00 A . × . constant. Find an expression for the magnetic field B (a) at a distance r 1 Ͻ R and (b) at a distance r 2 Ͼ R, measured from the axis. × × × Example: Flat Coil of Wire Disk I 1 mm 1 mm 1 mm Another direction in which we could expand upon our ring solution is radially – say we want to know the on-axis field due to a flat coil of current. Figure P30.23 (dIr)r2 2 dB r 0 zˆ (from the ring) 2 2 3/ 2 2 r z r 1 R Chapter 30 201 Where 24. The magnetic field 40.0 cm away from a long straight wire F N I bg2 rB 20100130a .. mfa Tf dI dI NI NI P30.40carryingBnI current 2.00 A isI 1.00 so I␮T. (a) At what distance is K 277 mAdI dr G J 7 dl dr R FigureR P30.29 it 0.100 ␮T? (b) HWhat2 r K If? At oneN instant,50004 ejthe 10two WbA maf 470 Figure 3: Problem 3. NI R r 2 dr conductors in a long household extension cord carry B r 0 zˆ 30. In Figure2 R P30.30,2 2 both3/ 2 currents in the infinitely long wires P30.41equal 2.00-AAssuming currents a uniform in opposite B inside directions. the toroid The is two equivalent wires to assuming 0 r z are in the negative x direction. (a) Sketch the magnetic are 3.00 mm apart. Find the magneticNI field 40.0 cm away rR ; then B as for a tightly wound solenoid. field pattern in the yz plane. (b) At what distance d along from the middle of the00 straight cord, in the plane of the Problem 5.47 (a) – Charged Rotating Disk 2 R the z axis is the magnetic field a maximum? two wires. (c) At630 what 3. 00distance is it one tenth as large? Find the magnetic field on the axis at a distance a above a disk of radius R with a fa f charge density rotating at an angular speed (clockwise when viewed from the (d) TheB 00center wire in a coaxial000189. cable T carries current 2.00 A in one direction20200af. and the sheath around it carries point P). z current With2.00 Athe in steel, the oppositeBB direction.1101000189 B What magnetic. T ConsiderB 0191. a thin T ring between r andFigure r + dr 4: Problemas shown 4. below. The charge per length on m 00bgafbgthe ring is dr and each pointFIG. on it isP30.41 moving at a speed v r, so the current is field does the cable create at points outside? I v r dr. a 25. A packed bundle of 100 long, straight,27 insulated wires32 24 TM afaf400... K 100%ej 800 10 atoms maf 500 ..ej 927 10 J T P KJa P30.42forms a Ccylinder of radius R ϭ 0.500 cm. (a) If each wire 297. 104 B 500. T Tm23 carries 2.00 A, what are the magnitude and direction of the z magnetic force per unit length acting on a wire located I B 6 P30.430.200 cmBHM from0 afthe center so of H the bundle?M (b)262 .What 10 If ? AmWould a wire on the outer edge of the 0bundle experience a force xdr r greater or smaller than the value calculated in part (a)? y P30.44 In BHM we have 2. 00 T M . But also Mxn . Then Bxn where nI is the 26. The magnetic coils0 af of a tokamak fusion reactor0 are in the B 0 B number of atoms per volume and x is the number of electronsBy symmetry, per theatom magnetic contributing. fieldFigure isFigure in the 5: ProblemP3z direction.0.3 7.0 Using the result from Ex. 5.6 shape of a toroid having an inner radius of 0.700 m and an (Eq. 5.38), the contribution from the ring is outer radius of 1.30 m. The Btoroid has 900 turns of large- 200. T Then x r dr 202. r2 . 28 3247 dB 0 . diameter wire, each of which0 carriesBn 8.50 a current 10 mof 14.0927.kA. 10 N m T 4 10 Tz m A 3 2 ejejej2 r2 z2 Find the magnitude of the magnetic field inside the toroid along (a) the inner radius and (b) the outer radius. SectionThe magnetic 30.4 field forT thehe entire Magnetic disk is Field of a Solenoid P30.45 (a) Comparing Equations 30.29 and 30.30, we see that the applied field is describedR by3 31. What current is required0 r drin the windings of a long 27. Consider a column of electric current passing throughB C Bz . 0 2 2 3 2 BH00 . Then Eq. 30.35 becomes MCsolenoid0 H , and thatthe definitionhas 1 0002 ofturns 0susceptibilityr uniformlyz distributed over a plasma (ionized gas). Filaments of current within the T T length2 of 0.400 m, to produce at the center of the solenoid column are magnetically attractedM to oneC another. They Let u r , so du 2r dr. This substitution (don’t forget to change limits) gives (Eq. 30.32) is . a magnetic field of magnitude 1.00 ϫ 10Ϫ4 T? can crowd together to yield a verHy greatT current0 density and a very strong magnetic field in a small region. 32. Consider a solenoid of length ᐉ and radius R, containing Sometimes the current can be cut off4 momentarily by this N closely spaced turns and carrying a steady current2 T ej270. 10af 300 K KA pinch effect. (In a metallic wire a pinch effect is not4 I. (a) In terms of these parameters, find the magnetic (b) C 7 645. 10 important, because the0 current-carr410 Tying m Aelectrons repel Tm field at a point along the axis as a function of distance one another with electric forces.) The pinch effect can be a from the end of the solenoid. (b) Show that as ᐉ demonstrated by making an empty aluminum can carry a becomes very long, B approaches ␮0NI/2ᐉ at each end of large current parallel to its axis. Let R represent the radius the solenoid. Section 30.9 The Magnetic Field of the Earth

7 NI ej4105000600 afaf.. P30.46 (a) BB0 12. 6 T h coil 2R 0300. B 12. 6 T (b) BBsin B h 56. 0 T h sin sin13 . 0

22 2 FIG. P30.46 800. 10 A m P30.47 (a) Number of unpaired electrons 863. 1045 . 927. 1024 A m2 Each iron atom has two unpaired electrons, so the number of iron atoms required is 1 863. 1045 . 2 ej ejej431. 1045 atoms 7900 kg m3 (b) Mass 401. 1020 kg ej850. 1028 atoms m3

Problem 7: Gyromagnetic Ratio

A thin uniform ring of radius R and mass M carrying a charge +Q rotates about its axis with constant angular speed ! as shown in the figure below. Find the ratio of the magnitudes of its magnetic dipole moment to its angular momentum. (This is called the gyromagnetic ratio.)

Solution: The current in the ring is Figure 6: Problem 8. Q Q! I = = . T 2"

The magnetic moment is ! Q" Q"R2 µ = AI kˆ = ! R2 kˆ = kˆ . 2! 2

The angular momentum is

! L I kˆ MR2 kˆ . = momnet! = !

So the gyromagnetic ratio is

! µ Q!R2 / 2 Q ! = = L MR2! 2M