Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 35

25.5 Torque on a Current Loop 875 → → We assume that B makes an angle ≤ θ ≤ 90◦ with the vector area A , which is a vector perpendicular to the plane of the loop and has a magnitude equal to the area of the loop, see the side view of the loop shown in Fig 25.15 In Fig 25.15, the side is represented by a circle and the current passing through it is represented by a red dot, while the side has the current represented by a red → → cross From Eq 25.19, the magnitudes of F3 and F4 are the same and given by: F3 = F4 = IbB (25.24) The moment arm of F3 and F4 about O is (a/2) sin θ Thus, the magnitude of the net torque about the rotational axis OO is: τ = F3 (a/2) sin θ + F4 (a/2) sin θ = [F3 + F4 ](a/2) sin θ = [2IbB](a/2) sin θ (25.25) = IAB sin θ Fig 25.15 A side view of the F3 loop showing the two forces → A → F and F that produce a torque on the current loop I a /2 about point O B a sin O × I F4 → where A = ab is the area of the loop This equation shows that τmax = IAB when B → is perpendicular to the normal of the loop (θ = 90◦ ), and τmin = when B is parallel to the normal to the plane of the loop (θ = 0) The direction of the torque exerted on the loop can be expressed in terms of the vector area as follows: → → τ→ = I A × B (τ = IAB sin θ ) (25.26) → → The product I A is defined as the magnetic dipole moment μ (or simply the mag- netic moment) of the loop and has the SI unit ampere-meter (A.m2 ) Thus: 876 25 Magnetic Fields → → = IA μ (Single loop) (25.27) If we replace the single loop of current with a coil of N loops, or turns, then the → of the coil will be given by: magnetic dipole moment μ → → = NI A μ (Coil of N loops) (25.28) Using this definition, Eq 25.26 can be written as: → → ×B τ→ = μ (25.29) → → by using the right-hand rule, which is We can determine the direction of A and μ described in Fig 25.16 Fig 25.16 Using the right-hand rule for determining → →for a the direction of A and μ μ I A loop of wire carrying a current I I A μ 25.5.1 Electric Motors A motor is an apparatus that converts electrical energy into rotational energy A battery-powered motor uses the principle of torque exerted on a coil of wire wound onto a shaft that rotates 360◦ In order to allow the coil to continue rotating, the current through the coil must reverse the direction just as the coil reaches its vertical position As shown in Fig 25.17, several components are required to achieve this reversal First, an electric connection is made using two brushes These are contacts usually made of graphite Second, a ring that is split into two halves, called a split-ring commutator Brushes make contact with the commutator and allow current to flow into the coil As the coil rotates, so does the commutator, which is arranged so that each of its halves changes brushes just as the coil reaches the vertical position Changing brushes reverses the direction of the current in the coil As a result, the direction of the force on each side of the coil is reversed and the coil continues to rotate This process repeats at each half-turn, causing the coil to spin in the magnetic field 25.5 Torque on a Current Loop 877 Fig 25.17 The split-ring Coil commutators in an electric motor allow the current in the and thus enable the coil in the Shaft I wire coil to change direction Commutator motor to rotate continuously Brush I Brush + Insulators Example 25.6 A rectangular coil of sides a = cm and b = cm consists of N = 75 turns of wire and carries a current I = 10 mA A magnetic field of magnitude B = 0.2 T is applied parallel to the plane of the coil, see Fig 25.18 (a) Find the magnitude of the magnetic dipole moment of the coil (b) What is the magnitude of the torque acting on the coil? Fig 25.18 a I B b I Solution: (a) Using Eq 25.28, we have: μ = NIA = (75)(10 × 10−3 A)[(4 × 10−2 m)(8 × 10−2 m)] = 2.4 × 10−3 A.m2 → → (b) Since B is perpendicular to μ , then Eq 25.29 gives: τ = μB sin 90◦ = (2.4 × 10−3 A.m2 )(0.2 T) = 4.8 × 10−4 N.m 25.5.2 Galvanometers The basic component of analog ammeters, voltmeters, and ohmmeters is a galvanometer Figure 25.19 displays the main features of a type of galvanometer called the D’Arsonval galvanometer It consists of a coil of wire that has N loops, each of 878 25 Magnetic Fields cross-sectional area A That coil is attached to a pointer and a spring The coil is also suspended so that it can rotate freely in a radial magnetic field produced by a circular cross-sectional permanent magnet Fig 25.19 Sketch of the structure of a moving-coil galvanometer Scale φ Iron core Pointer I S N Coil Spring When a current I flows through the coil, the magnetic field exerts a torque on the coil given by Eq 25.29, and this torque has a magnitude given by: τ = μB = NIAB (25.30) This torque is opposed by the torque τs exerted by the spring, which is approximately proportional to the coil deflecting angle φ That is: τs = k φ (25.31) where k is the stiffness constant of the spring When the pointer is in equilibrium, we have τs = τ, and we get: φ= NAB I k or φ∝I (25.32) Thus, the angular deflection φ of the pointer is directly proportional to the current I in the coil 25.6 Non-Uniform Magnetic Fields One of the useful types of non-uniform magnetic fields is the “magnetic bottle” shown in Fig 25.20a Such magnetic bottles can be used to trap charged particles, because the magnetic field is strong at the ends and weak in the middle Charged particles spiral along the field lines back and forth almost indefinitely if they not collide 25.6 Non-Uniform Magnetic Fields 879 Therefore, this magnetic bottle can be used to confine a plasma (a gas consisting of electrons and ions) Such a confinement can help control nuclear fusion, a process that could supply us with energy indefinitely South magnetic pole Van Allen belt North geographic pole S N South geographic pole (a) North magnetic pole (b) Fig 25.20 (a) Trapping of charged particles in a non-uniform magnetic bottle (b) A sketch of the Van Allen belt, which consists of charged particles trapped by Earth’s non-uniform magnetic field The Earth behaves like a gigantic magnet Its north magnetic pole is actually near the geographic south pole, and its south magnetic pole is near the geographic north pole, see Fig 25.20b This non-uniform magnetic field traps charged particles (mostly electrons and protons) in a region of space known as Van Allen belt In this belt, charged particles spiral around the field lines from pole to pole in a period of few seconds The sun and stars are the sources of these particles (called cosmic rays) Most cosmic rays are deflected by the Earth’s magnetic field and never reach the atmosphere When some particles of the Van Allen belt are close to the poles, they collide with the atoms of the atmosphere causing them to emit light (Aurora Borealis or Aurora Australis) 25.7 Exercises Section 25.1 Magnetic Force on a Moving Charge (1) For each of the moving charges shown in Fig 25.21, find the direction of the → magnetic force, taking v→to be the velocity of the particle and B to be the magnetic field (2) Consider a uniform magnetic field directed vertically up along the page of this paper In which direction does an electron deflect if its velocity is directed: (a) into the paper, (b) up along the paper, (c) to the left, and (d) out of the paper 880 25 Magnetic Fields + - ×B× × × × ××+ ××× ××××× ××××× (a) (b) (c) B B ×B× × × × × × ×- × × ××××× ××××× (d) B B + B (e) - + - (f) (g) (h) B Fig 25.21 See Exercise (1) (3) When moving with a speed of 107 m/s in a magnetic field of magnitude 1.5 T, an electron experiences a magnetic force of magnitude 10−12 N What is the angle between the electron’s velocity and the field at this instant? → → (4) A proton that has a velocity v→ = (3 × 106 i + × 106 j ) (m/s) moves through → → → a magnetic field B = (0.3 i + 0.02 T j ) (T) Find the vector magnetic force exerted by the field on the proton, and then find the magnitude and direction of this force (5) Near the Earth’s surface at the equator, the magnetic and electric fields are about 50 µT due North and 100 N/C downwards, respectively Find the net force on an electron traveling with velocity 107 m/s due East Section 25.2 Motion of a Charged Particle in a Uniform Magnetic Field (6) In a uniform magnetic field of magnitude of 10−4 T, an ion that has a charge q = +2e completes two revolutions in 1.51 ms Find the mass and the type of the ion (7) A proton travels with a speed of × 107 m/s perpendicular to a uniform magnetic field of magnitude T (a) What is the radius of the proton’s circular path? (b) What is the period of the motion? (c) Find the magnitude of the magnetic force on the proton (8) An alpha particle has a charge q = e and mass m mp , where mp is the mass of a proton The alpha particle has a kinetic energy of MeV and enters a uniform magnetic field of 1.5 T directed perpendicular to its velocity (a) Find the speed 25.7 Exercises 881 of the alpha particle (b) Find the magnetic force acting on the particle due to the field (c) Find the radius of the particle’s path (d) Find the acceleration of the particle due the magnetic force (9) An electron of speed × 106 m/s enters a uniform magnetic field of magnitude 0.01 T at an angle of 36.87◦ (a) Determine the radius of the electron’s helical path (b) Determine the period of one helical path (c) Determine the pitch of the electron’s helical path → (10) Figure 25.22 shows a region of uniform magnetic field B of magnitude 0.5 T which extends for a width W = 0.4 m Consider a proton moving with a velocity → → v of magnitude × 107 m/s, where v→ is perpendicular to B If the incident angle θ◦ at the lower boundary is 60◦ , the proton emerges from the lower boundary as shown in the left part of the figure However, if the incident angle θ◦ at the lower boundary is 0◦ , the proton emerges from the upper boundary as shown in the right part of the figure (a) At what angle θ and distance d does the proton exit from the lower boundary? (b) At what angle θ and distance d does the proton exit from the upper boundary? (c) At what critical incident angle θ◦ does the proton barely touch the upper boundary? d B θ B W W θ° d (a) θ θ° = (b) Fig 25.22 See Exercise (10) Section 25.3 Charged Particles in Electric and Magnetic Fields (11) A uniform magnetic field of magnitude 0.02 T is perpendicular to a uniform electric field of magnitude 750 V/m What is the speed of an electron that goes undeflected when moving perpendicular to both fields? → → (12) Assume that a keV electron travels in a uniform electric field E = 385 j → → (kV/m) and a uniform magnetic field B = Bz k Find the value of Bz such that 882 25 Magnetic Fields → the electron would have a velocity v→ = vx i and would move undeflected in the presence of the two fields (13) Figure 25.23 shows the path of an electron in a region of uniform magnetic field Each of the plates is uniformly charged (a) Which plate is at the higher electric potential for each pair? (b) What is the direction of the magnetic field in this region? (c) For both pairs of plates, if the magnitude of the electric field between the plates is × 104 V/m and the magnitude of the magnetic field is mT, find the radius of the two semicircles Fig 25.23 See Exercise (13) Semicircle Straight Electron's path Straight Semicircle (14) In the mass spectrometer shown schematically in Fig 25.6, the magnitude of the electric and magnetic fields in the velocity-selector region are kV/m and 40 mT, respectively The magnitude of the magnetic field in the deflecting chamber is 75 mT (a) What is the speed of ions in the velocity selector? (b) What is the radius of the path in the deflecting chamber for a singly-charged ion having a mass of 6.49 × 10−26 kg? (15) Two single ions of the boron isotopes (of masses 10 u and 11 u) are studied in the mass spectrometer shown schematically in Fig 25.6 Assume that the values B = B = 250 mT and E = 60 kV/m are used in this experiment (a) What is the speed of the ions in the velocity selector? (b) What is the spacing between the marks produced on the photographic plate by the ions of boron? (16) A strip of copper of thickness t = 0.4 mm and width d = mm is placed in a → uniform magnetic field B of magnitude 1.5 T perpendicular to the strip, see Fig 25.24 When a current I = 20 A passes though the strip, a Hall potential difference VH is generated across the width of the strip The number of charge carriers per unit volume for copper is 8.47 × 1028 electrons/m3 (a) Find the Hall coefficient RH for the copper strip (b) How large is the Hall voltage VH across the strip? (c) Find the magnitude of the Hall electric field EH 25.7 Exercises 883 B d I t Voltmeter Δ VH Fig 25.24 See Exercise (16) (17) A silver slab of thickness t = 1.5 mm and width d = 2.5 mm carries a current → I = A in a region in which there is a uniform magnetic field B of magnitude 1.25 T perpendicular to the slab The Hall voltage VH across the slab is found to be 0.356 µV (a) Calculate the density of the charge carriers in the slab (b) Compare your answer in part (a) to the density of atoms in the silver slab, which has a density ρ = 10.5 × 103 kg/m3 and a molar mass M = 107.9 kg/kmol What is the conclusion that you can find from this comparison? (c) Find the magnitude of the Hall electric field EH (18) A metal strip of thickness t = mm and width d = cm carries a current → I = 12.5 A in a region in which there is a uniform magnetic field B of magnitude 1.6 T perpendicular to the strip, as shown in Fig 25.25 The Hall voltage VH across the strip is measured to be 2.135 µV (a) Calculate the drift speed of the electrons in the strip (b) Find the density of the charge carriers in the strip (c) Which point is at the higher potential, a or b? Fig 25.25 See Exercise (18) B b d t a I 884 25 Magnetic Fields Section 25.4 Magnetic Force on a Current-Carrying Conductor (19) A 1.5 m long straight stiff wire carries a current of A and makes an angle 30◦ with a uniform magnetic field of 0.35 T Find the magnitude of the force on the wire (20) The L-shaped wire shown in Fig 25.26 lies in the xy plane In the presence of → → a uniform magnetic field B = 1.5 k (T), the wire carries a current of 2.5 A from point a to point c (a) Find the net force exerted on the wire (b) Show that this net force is the same as if the wire were a straight segment from point a to point c Fig 25.26 See Exercise (20) y c B a cm I b cm x Z (21) For the circuit shown in Fig 25.27, find the magnitude and direction of the force on each side, and find the resultant force Fig 25.27 See Exercise (21) c L B a 60° I b (22) A straight horizontal wire has a length L = 20 cm and mass m = 0.02 kg The wire is by connecting it by massless flexible leads to an emf source A uniform magnetic field of magnitude B = 1.6 T is perpendicular to the wire, as shown in Fig 25.28 Find the necessary current needed to suspend the wire and hence remove the tension in the flexible wire (23) If B = 0.2 T and I = A in Fig 25.29, find the force exerted on each segment of the wire 890 26 Sources of Magnetic Field (a) dB (b) P B r rˆ I r rˆ θ P θ q ds → Fig 26.1 (a) The differential magnetic field vector d B at point P, which is located by a position vector → r drawn from a differential current element I d → s to P (b) In case of a point charge q moving with a → velocity → v , the magnetic field B is related to the product q→ v → To find the magnetic field B created at some point by a current of an extended circuit, we integrate Eq 26.1 over all current elements as follows: → B = μ◦ I 4π d→ s ×→ rˆ r2 (26.3) It is useful to compare the Biot-Savart law with Coulomb’s Law as follows: → → Biot-Savart law (d B ) Coulomb’s law (d E ) → → d B is due to differential current element I d→ s , a vector d E is due to differential charge dq, a scalar 1/r distance dependence 1/r distance dependence Proportional to electric current I Proportional to electric charge dq Lateral, perpendicular to the → r direction Radial, in the → r direction Some Applications of the Biot-Savart Law In some situations, the integrand of Eq 26.3 needs lengthy mathematical steps For those interested, several mathematical and integration techniques are given at the end of this book In this section we avoid the complexity arising from integrating Eq 26.3 and only present the results for some cases Magnetic Field on the Extension of a Straight Wire I P (26.4) 26.1 The Biot-Savart Law 891 Magnetic Field Surrounding a Thin Straight Wire P B a θ1 (26.5) θ2 I Magnetic Field Surrounding a Very Long Straight Wire B P B a I (26.6) a B × P' Magnetic Field Due to a Curved Wire Segment R R B× θ I θ B I (26.7) Magnetic Field at the Center of a Circular Wire Loop R B× R I B I (26.8) 892 26 Sources of Magnetic Field Magnetic Field on the Axis of a Circular Wire Loop y I x R B O (26.9) x P z → Sketch of B Along the Axis of a Loop and a bar Magnet The magnetic pattern of a circular current loop N N Looks like I S The magnetic pattern of a bar magnet S Example 26.1 A point charge q = µC is moving in a straight line with a velocity v→ = × → 104 i (m/s) When the charge is at the location P(3 m, m, 0), find the magnetic field produced by this point charge at the origin o, see Fig 26.2 Fig 26.2 y P (3 m, m, 0) q + 4m θ v = ×104 i (m /s) r o x 3m z 26.1 The Biot-Savart Law 893 Solution: For a point charge q moving with a velocity v→, Eq 26.1 leads to: → B = μ◦ q v→ × → rˆ 4π r2 From the figure, we can find r and → rˆ (from the point charge) as follows: → → → r = [−3 i − j ](m) r= Thus: (−3 m)2 + (−4 m)2 = m → → → → → [−3 i − j ](m) r = = −0.6 i − 0.8 j r = r 5m → ˆ → Substituting the above results into the equation for B we obtain: → → → → → μ◦ q (v i ) × [−0.6 i − 0.8 j ] μ◦ q v (0.8 k ) μ◦ q v→ × → rˆ = = − 4π r2 4π r2 4π r2 −6 (6 × 10 C)(5 × 10 m/s)(0.8) → = −(10−7 T.m/A) k (5 m)2 B = → = −9.6 × 10−10 k (T) Indeed this is a very small value for the magnetic field produced by this charge, which is equivalent to the charge of about × 1013 protons Example 26.2 Two very long parallel straight wires carry currents that are perpendicular to the page Wire carries a current I1 = A out of the page and passes through the origin o of the x-axis, while wire carries a current I2 = A into the page and passes through the x-axis at a distance d = 0.6 m from the origin (a) On the x-axis, show the directions of the magnetic fields, to right of wire , between the two wires, and to the left of wire (b) To the right of wire , find a distance a at which the resultant magnetic field is zero Solution: (a) Using the right hand rule presented in the figure of Eq 26.6, we can → → draw the direction of B1 of wire and B2 of wire on the three regions of the x-axis as shown in Fig 26.3: 894 26 Sources of Magnetic Field → → (b) From Eq 26.6 the magnitudes of the magnetic-field vectors B and B at point P are: μ◦ I1 μ◦ I2 , and B2 = 2π(d + a) 2π a B1 = B1 For I1 > I2 B2 I1 B2 × x P O B1 B1 I2 a d B2 Fig 26.3 → → When the magnitudes of the opposite two vectors B1 and B are equal, the resultant magnetic field becomes zero Therefore, we have: μ◦ I1 μ◦ I2 = 2π(d + a) 2π a ⇒ ⇒ Thus: a= d I1 −1 I2 I1 I2 = ⇒ d+a a I1 a −1 =d I2 = 0.6 m 3A −1 2A aI1 = I2 (d + a) = 1.2 m Since I1 > I2 , P is the only point at which Bnet = on the x-axis Example 26.3 Two straight wires and , each of length L = cm, are connected by a quarter circular arc wire of radius R = cm, as shown in Fig 26.4 Determine the magnitude and direction of the magnetic field at the center P of the arc, when the current I is A Solution: There is no contribution to the field at point P from the lower wire since P is on the extension of the wire, i.e B1 = From Eq 26.7, the quarter circular arc wire has a magnetic field: B2 = μ◦ I (Directed out of the page) 8R , 26.1 The Biot-Savart Law θ1 895 θ2 R I P L L Fig 26.4 According to Eq 26.5, point P is at a distance R = cm from the straight wire and subtends two angles with the wire, θ1 and θ2 From the figure, we get: cos θ1 = L/ L + R2 = 4/5 and cos θ2 = cos 90◦ = Thus: B3 = μ◦ I μ◦ I (cos θ1 + cos θ2 ) = (Directed out of the page) 4π R 5π R The total magnetic field is the superposition of the fields from the three wires Thus, the resultant magnetic field is: μ◦ I μ◦ I + 8R 5π R (4π × 10−7 T.m/A)(2A) = × 10−2 m B = B1 + B2 + B3 = + = μ◦ I R 1 + 5π 1 + 5π = 1.58 × 10−5 T = 15.8 µT (Directed out of the page) 26.2 The Magnetic Force Between Two Parallel Currents Figure 26.5 shows a portion of length of two long straight parallel wires separated by a distance a and carrying currents I1 and I2 in the same direction Since each wire lies in the magnetic field established by the other, each will experience a magnetic force → Wire sets up a magnetic field B2 perpendicular to wire According to → → → Eq 25.19, the magnetic force on a length of wire is F1 = I1 × B2 , where the → → → → direction of F1 is toward wire Since ⊥ B2 , the magnitude of F1 is F1 = I1 B2 When we substitute with the magnitude of B2 given by Eq 26.6, we get: 896 26 Sources of Magnetic Field μ◦ I2 2π a F1 = I1 B2 = I1 = μ◦ I1 I2 2π a Fig 26.5 Two parallel wires (26.10) r carrying currents in the same r F1 r direction attract each other a Wire sets up a magnetic → F2 field B2 at wire and wire → I1 B2 r I2 B1 sets up a magnetic field B at wire → → We can show that the magnetic force F2 on wire has the same magnitude as F1 but is opposite in direction, i.e the two wires attract each other We denote the magnitude of the force between the two wires by the symbol FB and write this magnitude per unit length as: FB = μ◦ I I2 2π a (26.11) If the two currents were antiparallel (i.e the wires were parallel but the currents were opposite in direction), then the wires would repel Spotlight Parallel currents attract and antiparallel currents repel Example 26.4 A battery of 12 V is connected to a resistor of resistance R = by two parallel wires each of length L = 50 cm and separated by a distance a = cm, see Fig 26.6 All the connecting wires have negligible resistance Find the magnitude of the magnetic force between the two wires Will the wires repel or attract each other? Solution: According to the figure, the battery sets a clockwise current I in the circuit, and the current in the parallel two wires have the same value but opposite direction The value of this current is: I= 12 V V = =4A R From Eq 26.11, the magnetic force between the two wires is: 26.2 The Magnetic Force Between Two Parallel Currents 897 L I ΔV = 12 V + − R a I Fig 26.6 FB = 4π × 10−7 T.m/A (4 A)2 μ◦ I L= × 50 × 10−2 m 2π a 2π × 10−2 m = × 10−5 N Since the currents in the two wires are antiparallel, the wires will repel each other, but with a very small force due the smallness of μ◦ 26.3 Ampere’s Law When Oersted traced the magnetic field near a long vertical wire carrying a current I by a compass, he found that its needle deflects in a direction tangent to any circu→ lar path concentric with the wire, i.e the needle points in the direction of B , see Fig 26.7 Fig 26.7 The compass needle deflects in a direction tangent to a circle of radius r, → which is the direction of B r I B r created by I The same results can be obtained when we use the Biot-Savart Law to calculate the magnetic field around a long straight wire carrying a current The magnitude of → B was given by Eq 26.6 The work of Oersted and Biot-Savart was continued by Ampere Ampere’s work lead to what is now known as Ampere’s law, a law used in the cases of steady currents, which can be stated as follows: 898 26 Sources of Magnetic Field Ampere’s law The line integral of the tangential magnetic field around a closed path is proportional to the net conduction steady current I enclosed by the path That is: → B • d→ s = μ◦ I (Ampere’s law) (26.12) As a check for the long wire of Fig 26.7, let us consider an element d → s on the → → circular path and integrate the product B • d → s over this closed path Since B is → → → parallel to d s , then B • d s = B ds Thus: → B • d→ s = B ds = B ds = B(2π r) (26.13) By Ampere’s law, this result should be equal to μ◦ I Therefore: B= μ◦ I 2π r (26.14) This result is in complete agreement with Eq 26.6 obtained by using the Biot-Savart law; however, Ampere’s law saves considerable effort when we deal with problems that have some symmetry Some Applications of Ampere’s Law In these applications, we avoid solving the integrand of Eq 26.12 and only present the results of some well-known cases Magnetic Field Inside and Outside a Long Straight Wire I (out of page) B r R r I Amperian loops B (26.15) 26.3 Ampere’s Law 899 Magnetic Field of a Solenoid of n Turns per Unit Length Packed solenoid I Solenoid (26.16) I N S I I Magnetic Field of a Toroid of N Total Turns (or n turns/m) Amperian loop B (26.17) r I Magnetic Field Produced by an Infinite Current Sheet Current per unit length λ along the x direction (out of page) (26.18) y B x B 900 26 Sources of Magnetic Field Example 26.5 A long wire of radius R = 10 mm carries a current I = A What are the magnitudes of the magnetic field at a point mm and a point 50 mm from the axis of the wire? Solution: For a point inside the wire we use Eq 26.15 for r ≤ R : B= (4π × 10−7 T.m/A)(3 A) μ◦ I r = × (5 × 10−3 m) = × 10−5 T 2π R2 (2π )(10 × 10−3 m)2 For a point outside the wire we use Eq 26.15 for r ≥ R : B= μ◦ I (4π × 10−7 T.m/A)(3 A) = = 1.2 × 10−5 T 2π r (2π )(50 × 10−3 m) Example 26.6 A solenoid of length L = 0.5 m carries a current I = A The solenoid consists of six closely-packed layers, each of 800 turns What is the magnitude of the magnetic field inside the solenoid? Solution: The diameter of winding does not enter into the solenoid Eq 26.16 The number of turns per unit length is: n= × 800 turns (No of layers)(No of turns per layer) = = 9,600 turns/m L 0.5 m Since n is large, then from Eq 26.16 we have: B = μ◦ n I = (4π × 10−7 T.m/A)(9,600 turns/m)(2 A) = 2.41 × 10−2 T Example 26.7 In a fusion reactor, a toroid has inner and outer radii a = 0.5 m and b = 1.5 m, respectively The toroid has 900 turns and carries a current of 12 kA What is the magnitude of the magnetic field at a point located on a circle having the average radius of the toroid? 26.3 Ampere’s Law 901 Solution: With R = (a + b)/2 = (0.5 + 1.5)/2 = m, Eq 26.17 gives: B= 26.4 (4π × 10−7 T.m/A)(900 turns)(12 × 103 A) μ◦ N I = = 2.16 T 2π R (2π )(1 m) Displacement Current and the Ampere-Maxwell Law Ampere’s law is incomplete when the conduction current is not steady We can show this by considering the region near a parallel-plate capacitor while the capacitor is charging, see Fig 26.8a A variable conduction current i = dq/dt reaches one plate and the same conduction current i leaves the other plate There is no current flow across the space between the plates Experiments show the establishment of a magnetic field between the two plates as well as on both sides of the plates In → addition, experiments show that the value of B • d → s is the same for the three circular loops labeled , , and in Fig 26.8a But according to Ampere’s law, → B • d→ s must be zero for loop , because the conduction current is zero B i +q B −q Gaussian surface B R i +q −q E i E i dA (a) id (b) Fig 26.8 (a) The displacement current id between the plates of a capacitor (b) The Gaussian surface that encloses the varying charge q Maxwell solved this problem by postulating an additional term to the right side of Ampere’s law that is related to the changing electric field between the plates of the capacitor This term is referred to as the displacement current id between the plates This current is defined as: id = ◦ d E dt (26.19) The displacement current id between the plates is equivalent to the conduction current i in the wires, i.e id = i, and hence produces the same magnetic effects observed experimentally, see Fig 26.8a Maxwell added the displacement current id to the varying conduction current i and expressed Ampere’s law as follows: 902 26 Sources of Magnetic Field → B • d→ s = μ◦ (i + id ) = μ◦ i+ ◦ d E dt (Ampere–Maxwell law) (26.20) When there is a conduction current but no change in electric flux (only like loops and ), the second term is zero When there is a change in electric flux but no conduction current (only like loop ), the first term is zero Spotlight Magnetic fields are produced both by conduction currents i and by displacement currents id , created by a time varying electric flux To establish the relation Eq 26.20, we apply Gauss’s law for the Gaussian surface shown in Fig 26.8b According to Gauss’s law, see Eq 21.7, this surface encloses a net charge q, and we have: E = → → E • dA = q ◦ (26.21) → As q changes, E changes too, and the rate at which q changes gives the displacement current postulated by Maxwell Thus: id = dq = dt ◦ d E dt (26.22) Example 26.8 The circular capacitor of Fig 26.8 a has a radius R = 10 cm and a charge q = (4 × 10−4 C) sin(2 × 104 t) that varies with time t In the region between the plates, find the displacement current and the maximum value of the magnetic field at radius r = 15 cm Solution: From Eq 26.22, we find the displacement current as: id = d dq = [(4 × 10−4 C) sin(2 × 104 t)] = (8 A) cos(2 × 104 t) dt dt For a maximum displacement current (id )max of A at a point between the plates, we use Eq 26.15 for r ≥ R to find Bmax : Bmax = (4π × 10−7 T.m/A)(8 A) μ◦ (id )max = = 1.07 × 10−5 T 2π r (2π )(15 × 10−2 m) 26.5 Gauss’s Law for Magnetism 26.5 903 Gauss’s Law for Magnetism As in the case of an electric flux, we calculate the magnetic flux throughout a particular surface S, see Fig 26.9, as follows: B = → → B • dA (26.23) The SI unit for the magnetic flux is tesla-square meter, which is called weber (abbreviated Wb) Thus, weber = Wb = Tm2 θ dA B dA → Fig 26.9 The differential surface vector area d A is perpendicular to the differential area d A and pointing → → → → outwards When the magnetic field B makes an angle θ with d A , the differential flux d B is B • d A Since magnetic fields form closed loops, i.e the magnetic field lines not begin or end at any point, and for a closed surface the number of lines entering that surface equals the number of lines leaving it Thus, the net magnetic flux over a closed surface is zero This is known as Gauss’s law for magnetism and can be stated as: Gauss’s Law for Magnetism The net magnetic flux throughout any closed surface is always zero: → → B • dA = (Gauss’s law for magnetism) (26.24) Example 26.9 Find the net magnetic flux through the closed surfaces S1 and S2 of Fig 26.10, which are represented by dashed lines intersecting the page Fig 26.10 S2 B S S1 N 904 26 Sources of Magnetic Field Solution: According to Gauss’s law for magnetism, we must have: → → → B • dA = 0, and S1 → B • dA = S2 Notice that surface S2 encloses only the north pole of the magnet, and that the south pole is associated with the left boundary of S2 26.6 The Origin of Magnetism We have seen how to generate a magnetic field by allowing an electric current to pass through a wire Moreover, we found that the magnetic pattern of a circular current → loop has a North Pole and a South Pole with a magnetic dipole moment μ producing a magnetic pattern that looks like the magnetic pattern produced by a bar magnet (Searches for magnetic monopoles in cosmic rays or elsewhere have been negative.) In addition, there are two subatomic ways that produce a magnetic field in space, each one involving a magnetic dipole moment These require an understanding of quantum physics, which is beyond the scope of this study Therefore, we shall only begin our study by presenting the results of the classical model of atoms and electrons Orbital Magnetic Dipole Moments of Atoms In the classical Bohr model of hydrogen atoms, we assume that an electron of mass me and charge −e moves around a fixed nucleus with a constant speed v in a circular orbit of radius r, see Fig 26.11 L A r μ I −e r Fig 26.11 The classical model of a hydrogen atom, where an electron moves with a constant speed in a circular orbit about a nucleus The direction of the associated current is opposite to the direction of the electron’s motion