26.11 Exercises 925 Section 26.2 The Magnetic Force Between Two Parallel Currents (14) Two long parallel wires are separated by a distance a = cm and carry antiparallel currents of the same magnitude, I1 = I2 = A (a) What is the magnitude of the magnetic field created by each wire at the location of the other? (b) What is the magnitude of the force per unit length that each wire exerts on the other? Is this force attractive or repulsive? (15) Two long parallel wires in the xy plane are separated by a distance 2a and carry equal currents I in opposite directions The origin of the x-axis is taken to be midway between the wires and x is the position of an arbitrary point P from that origin, see the cross-sectional view of Fig 26.29 (a) Derive an expression for the magnitude of the resultant magnetic field B (x) as a function of the position x (b) Plot B(x) for −60 mm < x < 60 mm for I = 20 A and a = 30 mm B1 B2 I a P x x I a Fig 26.29 See Exercise (15) (16) Figure 26.30 shows a very long wire which carries a current I1 = 10 A and a rectangular loop which carries a current I2 = 15 A Both the wire and the loop lie in one plane Take a = 0.1 m, b = 0.2 m, and c = 0.3 m (a) Find the magnitude and direction of the force exerted by the long wire on the wires and (b) Find the direction of the force exerted by the long wire on the wires and (c) Find the total force exerted by the long wire on the loop Section 26.3 Ampere’s Law (17) A long thin-walled conducting cylindrical shell of radius R carries a current I, see Fig 26.31 Use Ampere’s law to find the magnitude of the magnetic field inside and outside the shell 926 26 Sources of Magnetic Field Fig 26.30 See Exercise (16) I1 I2 c a Fig 26.31 See Exercise (17) b Cross-sectional view R I cylindrical conducting shell (18) Figure 26.32 shows two antiparallel currents of the same magnitude, I = 10 A → Evaluate the line integral B • d → s around the closed paths C1 , C2 , and C3 , where each line integral is taken with d → s in a counterclockwise direction Which path can be used to find the magnetic field at some point? Cross-sectional view Fig 26.32 See Exercise (18) C3 I I C1 C2 26.11 Exercises 927 (19) A very long coaxial cable consists of a central wire, surrounded by a rubber layer, which is surrounded by a concentric conducting shell of radius R = mm, which is surrounded by another rubber layer, see Fig 26.33 The current I1 in the inner wire is A out of the page and the current I2 in the outer conducting shell is A into the page Find the magnitude and direction of the magnetic field at = mm and rb = mm Cross-sectional view Coaxial cable Central wire R I2 I1 a b Insulator Conducting shell Outside insulator 2mm 2mm Fig 26.33 See Exercise (19) (20) A long wire of radius R = cm carries a steady current I = 50 A What are the magnitudes of the magnetic fields from the axis of the wire: (a) at a point cm from the center, (b) on the surface, and (c) at a point cm from the center? (21) A long conducting cylindrical shell of inner radius a and outer radius b carries a current I uniformly distributed across the cross-sectional area of the shell Find the magnitude of the magnetic field at points of radii: r < a, a < r < b, and r > b (22) A solenoid with n turns per unit length carries a current I, see Fig 26.34 Apply Ampere’s law to the rectangular path shown in the figure and derive an expression for the magnetic field B For a packed solenoid such as this, assume that B is uniform inside and B = outside (23) A solenoid of length = 0.25 m carries a current I = 10 A The solenoid consists of twenty closely packed layers, each of 500 turns What is the magnitude of the magnetic field inside the solenoid? (24) A solenoid has a length = 10 cm A superconducting fine wire (with almost zero resistance at low temperature) is wound in 10 layers such that n = × 104 turns per meter (a) What is the number of turns per layer? (b) What is the 928 26 Sources of Magnetic Field magnitude of the magnetic field B produced inside the solenoid when the current I in the wire is 60 A? Fig 26.34 See Exercise (22) I C d B I (25) Assume the wire of the solenoid of exercise 24 has a resistance of 104 at room temperature A 12 V battery is applied to the solenoid terminals Find B under these conditions (26) An insulating cylindrical shell has a radius R = 0.5 cm and length = 10 cm A fine wire of diameter d = 0.4 mm is wound in many layers to establish a magnetic field of magnitude π × 10−2 T inside the cylindrical solenoid when the current is A (a) Determine the number of layers of wire needed (b) Determine the length of the wire (27) Figure 26.35 shows a toroid with N turns that carries a current I The toroid has an inner radius a and an outer radius b Apply Ampere’s law to the circular path C of radius r shown in the figure to derive an expression for the magnetic field B that is only confined to the space enclosed by the windings, and hence B = anywhere else Show that B is approximately uniform when (b − a)/2 R, where R = (a + b)/2 is the mean radius of the toroid (28) A plastic ring of mean radius cm is wound with N = 1,000 turns of wire If the current in this toroid is I = 0.6 A, find the magnitude of the magnetic field on the mean circumference (29) A tightly wound toroid of inner radius a = cm and outer radius b = cm has N = 3,300 turns of wire and carries a current I = A Find the magnitude of the magnetic field: (a) at any point on the circumference of a circle of radius r = 5.5 cm (b) on the mean circumference, which has a radius r = cm (30) An infinite conducting sheet lying in the xz plane carries a current in the positive z direction, see Fig 26.36 The current per unit length (or the linear current density) along the x-axis is λ (a) Use the Biot-Savart law and the symmetry of 26.11 Exercises 929 the problem to show that for every point P (such that y > 0) and every point P → (such that y < 0), the magnetic field B is parallel to the sheet and directed as shown in the figure (b) Apply Ampere’s law to the rectangular path shown in the figure to derive an expression for the magnitude of the magnetic field B Fig 26.35 See Exercise (27) I C B a r b I B λ is the current perunit length along the x direction and this current is directed out of the page along the z direction P d y x P′ B Fig 26.36 See Exercise (30) (31) Figure 26.37 shows two parallel infinite conducting sheets, each carries λ amperes of current per unit length out of the page Find the magnetic field at points a, b, and c Fig 26.37 See Exercise (31) λ a λ b y c x 930 26 Sources of Magnetic Field Section 26.4 Displacement Current and the Ampere–Maxwell Law (32) A capacitor has circular plates, each of radius R = cm At a particular instant, the capacitor is charging by a current of 0.2 A (a) What is the displacement current between the plates? (b) What is the rate of change of electric flux between the plates? (c) What is the magnitude of the magnetic field at r = cm from the capacitor’s axis in the region between the plates? (33) A capacitor has circular plates, each of radius R = 10 cm At a particular instant, the capacitor is charging by a current of 0.3 A (a) What is the rate of change of electric field between the plates? (b) Apply Ampere-Maxwell Law to find the magnitude of the magnetic field at r = cm from the capacitor’s axis in the region between the plates, see Fig 26.38 Fig 26.38 See Exercise (33) +q B −q axis R i r i id Section 26.5 The Origin of Magnetism (34) What is the value of the orbital angular momentum of an electron having orbital quantum number = 2? For this electronic state, what is the measured component of the orbital magnetic dipole moment μ ,z when its orbital magnetic quantum numbers are m = 0, m = 1, and m = −2? (35) An atomic electron has an orbital angular momentum with m = (a) What are the measured components Lz and μ ,z? (b) When an external magnetic field of magnitude B = 40 mT is applied along the z-axis, find the potential energy → U associated with the orientation of the orbital magnetic dipole moment μ (c) Repeat parts (a) and (b) for orbital angular momentum with m = −2 (36) What is the potential energy Us associated with the orientation of the spin → magnetic dipole moment μ s of an atomic electron when an external magnetic field of magnitude B = 0.5 T is applied along the z-axis Then what is the energy difference between parallel and antiparallel alignment of μs,z? 26.11 Exercises 931 → (37) An external magnetic field B of magnitude 35 T is produced along the z direction in a short period by a pulsed coil An electron whose μs,z was parallel to → B experiences a “spin flip” so that the final orientation of μs,z is antiparallel → to B Find the change in the orientation potential energy of the electron Section 26.7 Magnetic Materials Section 26.8 Diamagnetism and Paramagnetism (38) A small magnetic disk has a radius of 1.2 cm and thickness of 0.2 cm The disk has a uniform magnetization throughout its volume and along its axis The → magnetic moment of the disk is 10−2 A.m2 (a) What is the magnetization M of the disk? (b) If the magnetization is due to the alignment of N atoms along the disk axis, each with magnetic moment of µB (1 Bohr Magneton), what is the value of N? (39) Figure 26.39 shows a diamagnetic loop before and after applying an exter→ nal magnetic field B (a) What is the direction of the loop’s net magnetic → → dipole moment μ before and after the application of B? (b) Is the conventional current counterclockwise or clockwise? (c) What is the direction of the magnetic force on the loop? Fig 26.39 See Exercise (39) Diamagnetic loop B (40) The magnetic field inside a solenoid carrying a current is decreased by × 10−3 % when sample of liquid is inserted into its core What is the magnetic susceptibility of the liquid? (41) The number of turns for a toroid is N = 1,200 and the current it carries is I = 1.5 A The core has an average circumference of 100 cm and a crosssectional area of cm2 (a) If the core is air, find the magnetic field strength H, the magnetic field B◦ , and the total flux B in the toroid (b) If the core is filled with bismuth of magnetic susceptibility χ = −2 × 10−6 , find the mag- 932 26 Sources of Magnetic Field netization M of bismuth, the magnetic field B in bismuth, and the total flux B (42) A solenoid 0.4 m long is tightly wound with 800 turns of copper wire The current in the winding is I = A (a) If the core is air, find at the center of the core the magnetic field strength H and the magnetic field B◦ (b) If the solenoid has an aluminum core of magnetic susceptibility χ = 2.3 × 10−5 , find the magnetization M of aluminum and the magnetic field B (43) Repeat part (b) of Exercise 42 for a tungsten core of magnetic susceptibility χ = 6.8 × 10−5 (44) If all atoms in a material have their magnetic moments aligned, then the maximum magnetization is given by Mmax = nμatomic , where n is the number of atoms per unit volume Aluminum has a density of 2.7 × 103 kg/m3 , molecular mass of 27 kg/kmol, and atomic magnetic moment μatomic = 9.27 × 10−24 J/T = μB , where the quantity μB is called the Bohr magneton Find Mmax and μ◦ Mmax Section 26.9 Ferromagnetism (45) Assume that all iron atoms in an iron rod are completely aligned and each atom has an approximate magnetic dipole moment μIron = 1.9 × 10−23 J/T Iron has density of 7.8 × 103 kg/m3 and molecular mass of 55.85 kg/kmol (a) Find the maximum magnetization Mmax (b) Find the dipole moment of the rod if it is 10 cm long, cm wide, and 0.5 cm thick (c) When a magnetic field of 0.5 T is applied perpendicular to the rod, find the torque exerted by the field (46) A 4-A current flows through the wire of a toroid that has 250 turns per meter The toroid’s core is iron of magnetic permeability μm = 2,100 μ◦ , find the magnitude of the magnetic field in its core (47) A solenoid of 50 cm long, 1.5 cm in diameter, and 500 turns is filled with iron core When a current of 10 A flows through the wire of the solenoid, the magnetic field inside it reaches T What is the permeability of the iron? Faraday’s Law, Alternating Current, and Maxwell’s Equations 27 Experimentally, M Faraday and J Henry show that a changing magnetic field can establish a current in a circuit that has no battery 27.1 Faraday’s Law of Induction When we move a magnet toward a stationary loop that is connected to a galvanometer, see Fig 27.1a, the galvanometer’s needle deflects in one direction When the magnet stops, as shown in Fig 27.1b, no deflection is observed Now, when we move the magnet away from the loop, as shown in Fig 27.1c, the needle deflects in the opposite direction The current produced in this loop is called an induced current and the work done per unit charge in producing that current is called an induced emf This emf is due to the change in magnetic flux through the loop, and this process is known as Faraday’s law of induction and stated as: Faraday’s law of induction The magnitude of the induced emf |E| in a conducting loop is equal to the rate of change of the magnetic flux B through the loop Lenz’s Law Soon after Faraday proposed his law, Lenz devised a rule—now known as Lenz’s law, for determining the direction of an induced emf and the direction of an induced current in a loop This law states that: H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_27, © Springer-Verlag Berlin Heidelberg 2013 933 934 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations (a) I Motion Loop N S Galvanometer (b) At rest Loop N S Galvanometer (c) I Motion Loop N S Galvanometer Fig 27.1 A galvanometer registers an induced current in a loop when the magnet is moving with respect to the loop (parts a and c) In part b, the magnet is at rest, and no induced current is established Lenz’s law An induced current in a loop is created such that the internal magnetic field of the loop opposes the changes in the external magnetic flux To get a sense of Lenz’s law, let us consider the case of a bar magnet approaching the loop of Fig 27.2a During the motion toward the loop, the external magnetic field → B ext of the bar magnet increases the magnetic flux on the loop and thereby induces a current in the loop The induced current produces its own internal magnetic field → → B int that counteracts the increase in the external magnetic flux In Fig 27.2b, B int opposes the decrease in the external flux Based on Faraday’s law and Lenz’s law, the induced emf in a coil of N loops of the same area is given by: E = −N d B dt (Faraday’s law) (27.1) 940 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations b B I L a ΦB x BLa x x x a BLa b I x BLx x 0 a BL BL BL x (a + b) x B L(a+b-x) 0 x BLa 0 a b a a b a b x b x a a b (a+ b) x BL x a −B L b (a + b) x −B L Fig 27.7 27.3 Electric Generators Electric generators are devices that convert rotational energy to electric energy A generator consists of a coil of wire wound on an armature that can rotate in a magnetic field between the poles of the magnet The magnetic flux through the coil changes with time Thus, according to Faraday’s law, an induced emf and current will be created in the coil → If θ is the angle between the magnetic field B and the normal to the plane of the coil, then the magnetic flux through each loop of the coil will be given by: → → B = B • A = BA cos θ (27.9) If the shaft of the generator rotates with constant angular frequency ω (in rad/s), then the relation between the angular position θ (in rad) and the frequency ω is θ = ωt Therefore, B = BA cos ωt Hence, according to Faraday’s law, the induced emf of a coil of N loops will be: E = −N d B dt = −N d (BA cos ωt) = N B A ω sin ωt dt which can be written in compact (E ≡ v and E◦ ≡ V ) form as follows: v = V sin ωt where V = N BAω (27.10) The output emf is sinusoidal with amplitude (or peak) V From this relation we see → that v = when ωt = or ωt = π, and this occurs when B is perpendicular to the plane of the coil Furthermore, v = V when ωt = π/2 or ωt = 3π/2, and this occurs → when B is in the plane of the coil 27.3 Electric Generators 941 The Direct Current (dc) Generator Figure 27.8 illustrates the simplest form of the direct current (dc) generator The ends of the coil are connected to a split-ring commutator (as shown in Fig 27.8a) that rotate with the coil Those splits are in contact with two brushes that act as the output terminals of the generator The output is always of the same polarity and varies with time as shown in Fig 27.8b Coil N Shaft I Commutator I S Brush A B × I I Side view Brush External rotator Insulators o t (a) (b) Fig 27.8 (a) A sketch of a dc generator An emf is induced in a coil when it rotates with constant angular → frequency ω in a magnetic field B (b) The direct induced emf is plotted as a function of time The Alternating Current (ac) Generator Figure 27.9 illustrates the simplest form of an alternating current (ac) generator The ends of the coil are connected to slip-rings (as shown in Fig 27.9a) that rotate with the coil Those slips are in contact with two brushes that act as the output terminals of the generator The output varies sinusoidally with time See Fig 27.9b Coil N Shaft I Slip rings I S I Brush A θ =ωt B × I Side view Brush t External rotator (a) (b) Fig 27.9 (a) A sketch of an ac generator An emf is induced in a coil when it rotates with constant angular frequency ω in a magnetic field (b) The alternating, induced emf is plotted as a function of time 942 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations 27.4 Alternating Current When electric generators at electric power plants produce alternating emf we get alternating current, or ac (uppercase letters can be used) with the usual symbol Alternating current reverses direction many times per second, which means that electrons in a wire will repeatedly move in one direction and then reverse their direction Since the output emf of an ac generator is sinusoidal, as shown in Fig 27.9b, then the current it produces is also sinusoidal Ohm’s law, Eq 24.8, is also valid for alternating voltage and current Based on Eq 27.10, when a sinusoidal voltage v exists across a resistance R, see Fig 27.10a, then the alternating current i (we use the small letter i for ac) through the resistor is: i= V v = sin ωt = I sin ω t R R where I= V R (27.11) where I is the peak current, see Fig 27.10b From this figure we see that the average current is zero This does not mean that no heat is produced in the resistor On the contrary, electrons produce heat when they move back and forth in the resistor The power delivered at time t to a resistor of resistance R, see Fig 27.10c, is: P(t) = i2 R = I R sin2 ωt (27.12) which indicates that the power is always positive because the current is squared The quantity sin2 ωt varies between and and we can prove that its mean (or average) value is 21 , i.e sin2 ωt = 1/2 Therefore, using P = i2 R or P = v /R2 as the average power delivered to the resistor, we get: P = I R/2 or P = V /2R Power P(t ) = i R Current R i = I sin ω t (27.13) P = 12 I R = I rms R I2R +I t t I (a) (b) (c) Fig 27.10 (a) A resistor connected to an ac source (b) Alternating current in a resistor as a function of time (c) Power and average power delivered to a resistor as a function of time 27.4 Alternating Current 943 As introduced in Sect 13.1, the notation rms stands for root-mean-square, which here means the square root of the mean value of the square of the current Irms = i2 or the voltage Vrms = E2 Thus (remember E ≡ v): I V Irms = √ = 0.707 I and Vrms = √ = 0.707 V 2 (27.14) This means that an alternating current whose maximum value is A delivers to a resistor the same power as a direct current of 0.707 A Thus, Ohm’s law and the average power delivered to a resistor give: Vrms = Irms R R = V /R P = Irms Vrms = Irms rms and (27.15) Alternating-current instruments are usually calibrated to read the rms values of the √ √ current defined by Irms = I/ and voltage defined by Vrms = V / More than 81% of countries around the globe use Vrms in the range from 220 to 240 V Example 27.5 Find the resistance and the peak current in a 1,000-W heater connected to a 220-V ac line Solution: Using Eq 27.15, we can find the rms current: Irms = P 1,000 W = 4.55 A = Vrms 220 V Then, the peak current and the resistance of the heater will be: 27.5 I= √ Irms = 6.43 A R= Vrms 220 V = 48.35 = Irms 4.55 A Transformers A transformer is a device used to increase or decrease an ac voltage Transformers are widely used in: reducing the high voltage from the electric power plant to a usable household electric ac outlet (120 or 220 V), in chargers for mobiles, laptops, and other electronic devices, in cars to increase the voltage to a high voltage needed to spark the plugs, in CRT monitors, and on many electrical applications 944 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations An ideal transformer consists of two resistanceless coils known as the primary and secondary coils, wound around an iron core, see Fig 27.11 We assume that the primary coil has NP turns and the secondary coil has NS turns If all magnetic lines are confined to the iron core, then at any instant the magnetic flux per turn produced by the current in the primary coil will pass through the secondary coil Fig 27.11 An ac input of voltage vP (with peak VP ) is Primary coil connected to the primary coil NS B Secondary coil turns of a transformer to get an ac output in the secondary coil VS VP The figure shows a step-up NP transformer where NP = and NS = Input turns Output When an ac voltage vP (with peak VP ) is applied to the primary coil, the magnetic flux change is the same in each turn of the primary and secondary coils Thus, according to Faraday’s law, the resulting induced emf in the primary and secondly coils will be: EP = −NP d B dt and ES = −NS d B dt (27.16) where EP and ES have the same frequency as the ac input source vP Since the flux per turn B is the same, the ratio of the secondary emf to the primary emf at any instant is ES /EP = NS /NP If the windings have zero resistance, the induced emf EP and ES are exactly balanced by the terminal voltage across the primary voltage vP (with peak VP ) and the secondary voltage vS (with peak VS ), respectively Thus: NS VS = VP NP (27.17) This transformer equation relates the output (the secondary) to the input (the primary) and can apply for the amplitude or the rms values If NS > NP , then VS > VP and this kind of transformer is called a step-up transformer Similarly, if NS < NP , then VS < VP and this kind of the transformer is called a step-down transformer For step-up or step-down ideal transformers, the power output is equal to the power input Using Eq 24.24, we have: 27.5 Transformers 945 IP VP = IS VS or NP IS = IP NS (27.18) Example 27.6 The transformer used to charge a laptop reduces 220-V ac to 19 V ac Assume that the primary coil contains 400 turns and the charger supplies A to the laptop Find: (a) the number of turns in the secondary coil and the current in the primary coil, and (b) the power transformed Solution: (a) Using Eq 27.17, we have: VS NS = VP NP ⇒ NS = NP VS (400)(19 V) = 35 turns = VP 220 V Using Eq 27.18, we have: IP VP = IS VS ⇒ I P = IS VS (5 A)(19 V) = 431.8 mA = VP 220 V (b) The power transformed is: P = IS VS = (5 A)(19 V) = 95 W 27.6 Induced Electric Fields By Faraday’s law a changing magnetic flux induces both an emf and a current in a conducting loop But in Chap 24, see Eq 24.11, we related current to an electric field that applies electric forces on charged particles Similarly, we relate an induced current to an electric field such as: Spotlight A changing magnetic field in a conducting loop, or even in any hypothetical closed path, induces an electric field We can examine this statement by considering a circular copper loop of radius r → placed in a uniform magnetic field B perpendicular to the loop, see Fig 27.12 If the magnetic field increases with time, then according to Faraday’s law, an induced emf and an induced current are created in the loop But this induced current implies the → → existence of an induced electric field E The work done by E to move a test charge 946 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations q◦ around the loop is W = q◦ E On the other hand, according to Eq 22.3, this work is given for a closed loop by: → E • d→ s W = q◦ Fig 27.12 A copper loop in → (27.19) Copper loop a uniform magnetic field If B changes with time, an induced × × E × electric field is produced tangent to the loop × E × r q°× R × E × I B increases × Thus, equating the two expressions of work, we get: q◦ → E • d→ s = q◦ E (27.20) Using E = −d B /dt, then Faraday’s law of induction can be expressed in terms of the induced electric field as follows: → E • d→ s =− d B (Faraday’s law) dt (27.21) The striking feature of Eq 27.21 is that the electric field is induced even if there is no conducting loop of r < R In addition, an induced electric field is established even if r > R, see Fig 27.13 Copper loop × × E × × E r × B increases R × × × I × (a) E × E × E Hypothetical circular path × × E × × × × × E B increases × B increases R × × × × r × × E (b) R × r × E (c) Fig 27.13 (a) Same as Fig 27.10 when the conducting loop is in place (b) Induced electric field is established even for a hypothetical path of r < R (c) Same as (b), but when r > R 27.6 Induced Electric Fields 947 Example 27.7 In Fig 27.13, find an expression for the magnitude of the induced electric field E for r < R and r > R When R = cm and the magnitude of the magnetic field increases at a rate given by dB/dt = 0.2 T/s, evaluate E for r = cm and r = 10 cm Solution: We evaluate the integral of Eq 27.21 for any radius: → E • d→ s = The flux B and its rate d B The flux B B /dt ds = E(2π r) = 2π rE through the circular path of r < R are: = BA = B(π r ) and d and its rate d B E ds = E B /dt B /dt = π r dB/dt through the circular path of r > R are: = BA = B(π R2 ) and d B /dt = π R2 dB/dt Thus, dropping the minus sign of Eq 27.21 leads to: E= E= 27.7 r dB 0.05 m (for r < R) and E|r=5 cm = 0.2 T/s = × 10−3 V/m dt (0.08 m)2 R2 dB (for r > R) and E|r=10 cm = 0.2 T/s = 6.4×10−3 V/m 2r dt × (0.1 m) Maxwell’s Equations of Electromagnetism From our previous studies, we can collect all the relationships between electric and magnetic fields and their sources This collection consists of four equations, called Maxwell’s equations Maxwell used these equations to predict the existence of electromagnetic waves → → The first two equations involve a surface integral of E and B over a closed sur→ → face The third and fourth two equations involve a line integral of B and E over a closed loop 948 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations The first is simply Gauss’s law for electric fields, Eq 21.7, which states that “the net electric flux through any closed surface is equal to the net charge inside the surface divided by the permittivity of free space ◦ ” That is: → → E •.A = → qin (Gauss’s law for E ) ◦ (27.22) The second is the analogous relationship for magnetic field, Eq 26.24, which states “The net magnetic flux throughout any closed surface is always zero” That is: → → → B •d A = (Gauss’s law for B ) (27.23) The third equation is Ampere–Maxwell law, Eq 26.20, which states “Magnetic fields are produced both by varying conduction currents i and by displacement currents id , created by a time varying electric flux” That is: → B • d→ s = μ◦ (i + id ) = μ◦ i + ◦ d E (Ampere−Maxwell law) dt (27.24) The fourth equation is Faraday’s, Eq 27.21, which states that “A changing magnetic field in a conducting loop, or even in any hypothetical closed path, induces an electric field” That is: → E • d→ s =− d B (Faraday’s law) dt (27.25) → In all Maxwell’s equations, E is the total electric field, which comes from → an electrostatic field E s caused by static charges and magnetically-induced, non→ electrostatic field E ind It is worth noting that in empty space, where there is no charge and conduction current, the first two equations are identical in form In addition to that, replacing → → → → E by E • dA and B by B • dA , we can write the third and fourth equations in different but equivalent forms Thus, in empty space Maxell’s equations reduce to: → → → E • dA = 0, → B • d→ s = μ◦ → B • dA = d ◦ dt → → E • dA , → E • d→ s =− d dt → → B • dA (27.26) 27.7 Maxwell’s Equations of Electromagnetism 949 The most remarkable feature about these Maxwell’s equations is that time→ → → → varying of E induces a field B and time-varying of B induces a field E in neighboring regions of space Maxwell recognized that Eq 27.26 predict the existence of electromagnetic disturbance of electric and magnetic fields that propagate from one point to another, even if no matter is present This disturbance is called an electromagnetic wave (EMW), and this provides the physical basis for light and all the rest of the electromagnetic spectrum The properties of electromagnetic waves can be deduced from Maxwell’s equations, but the mathematical treatment is beyond the scope of this book Instead, one can focus attention on an electromagnetic wave traveling in the x-direction By doing so, we can show that the line integral of the last two forms of Eq 27.26 lead to the following two differential equations: ∂ 2E − μ◦ ∂x ◦ ∂ 2B ∂ 2E = and − μ◦ ∂t ∂x ◦ ∂ 2B =0 ∂t (27.27) These two differential equations have the identical form as the general wave equation introduced in Chap 14, see Eq 14.58, for one-dimensional wave motion, but with speed c, given by: c= √ μ◦ Taking μ◦ = 4π × 10−7 T.m/A and (27.28) ◦ ◦ = 8.8542 × 10 −12 C2 /N.m2 , we find that c = 2.99792 × 108 m/s, which is precisely the speed of light in empty space In → → addition E and B are perpendicular to one another, and both are perpendicular to the wave velocity → c , see Fig 27.14a The simplest solution to Eq 27.27 is a sinusoidal wave, where E and B vary with x and t according to the two expressions: E = E◦ cos(kx − ωt) and B = B◦ cos(kx − ωt) (27.29) where E◦ and B◦ are the peak values of the electric and magnetic waves, respectively, while k and ω are defined in Chap 14 Figure 27.14b displays the various types of electromagnetic waves 950 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations Frequency (Hz) Wavelength 1022 pm 1021 Gamma 20 10 rays 1019 X-rays 1018 nm 1017 1016 Ultraviolet 1015 m Visible 1014 Infrared light 13 10 y E B x c z 1012 1011 1010 109 108 107 106 105 104 103 ~400 nm Violet Blue Green Yellow Orange Red ~700 nm mm cm Microwaves 1m TV, FM Radio waves AM km Long waves (a) (b) Fig 27.14 (a) Propagation of EMW (b) The EMW spectrum 27.8 Exercises Section 27.1 Faraday’s Law of Induction → → → → (1) A loop of wire with a vector area A = (0.2 i + 0.3 j + 0.6 k ) m2 is placed → → in a uniform magnetic field B = 0.2 j (T) (a) Find the magnetic flux through → → the loop (b) What is the angle between B and A ? → → (2) If the magnetic field of Exercise changes to B = 0.2 k (T) in a period of 0.5 s, find the magnitude of the average induced emf and current in the loop, assuming the loop has a resistance of 1.5 (3) In Fig 27.15, the south pole of the magnet approaches the loop What is the direction of the induced current in the resistor of the circuit? (4) In Fig 27.16, the north pole of the magnet recedes from the loop What is the direction of the induced current in the resistor of the circuit? 27.8 Exercises 951 Fig 27.15 See Exercise (3) Motion Loop S N R Fig 27.16 See Exercise (4) Motion S Loop N R (5) The magnetic flux through a loop of wire changes from −20 to +20 Wb in 0.2 s What is the average induced emf in the loop? (6) Figure 27.17 shows a circuit containing a battery and a resistor whose resistance can vary Two loops are located inside and outside the circuit If the resistance is slowly decreased, what is the direction of the induced current in the two loops? Fig 27.17 See Exercise (6) I (7) A circular loop of radius r = cm is perpendicular to a uniform magnetic field that is pointing out of the page and has an initial magnitude Bi = 0.6 T During a time interval of 0.2 s the field is changed to a final magnitude Bf = 0.2 T and now points into the page What is the average induced emf in the loop? (8) A square loop of wire has a side a = cm and is perpendicular to a uniform magnetic field of magnitude B = 0.8 T The orientation of the loop changes in a period of 0.4 s until the surface of its plane is parallel to the field What is the average induced emf in the loop? 952 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations (9) The plane of a circular loop of radius r = 10 cm is perpendicular to a uniform magnetic field of initial magnitude B = 0.8 T The field’s magnitude then decreases at a constant rate of dB/dt = −10−3 T/s (a) What is the magnitude of the field at any time? (b) What is the induced emf produced in the loop? (10) For each situation in Fig 27.18, show the direction of the induced current B increases B decreases Rotating loop ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× ××××× (g) (h) Moving in loop B (a) Moving out loop B (b) (e) (f) Shrinking loop B (c) Expanding loop B (d) Rotating loop Fig 27.18 See Exercise (10) (a) Moving in loop (b) Moving out loop (c) Shrinking loop (d) Expanding → → loop (e) B increases (f) B decreases (g) Rotating loop (h) Rotating loop (11) The diameter of the circular loop of Fig 27.18c decreases from 20 to cm in 0.6 s The magnitude of the magnetic field in that figure has a value B = 0.5 T (a) What is the direction of the induced current? (b) What is the average induced emf in the loop? (c) What is the average magnitude of the induced current if the loop’s resistance is ? (12) The radius of the circular loop of Fig 27.18d increases from cm to 15 cm in 0.2 s The magnitude of the magnetic field in that figure has a value B = 0.25 T (a) What is the direction of the induced current? (b) What is the average induced emf in the loop? (c) What is the average magnitude of the induced current if the loop’s resistance is 1.5 ? (13) The plane of the circular loop of Fig 27.18g has an area A = cm2 In 0.2 s it rotates to make an angle θ = 60◦ with the field lines The magnitude of the 27.8 Exercises 953 magnetic field in that figure has a value B = 0.75 T (a) What is the direction of the induced current? (b) What is the average induced emf in the loop? (c) What is the average magnitude of the induced current if the loop’s resistance is 2.5 ? (14) A solenoid of length L = 0.25 m and radius r = cm has 100 turns A coil of N = 20 turns and resistance R = is wound tightly around the solenoid, see Fig 27.19 The current in the direction shown in the figure increases uniformly from to A in 0.5 s (a) What is the direction of the induced current in the coil during this period of time? (b) What is the magnitude of the induced emf in the coil? (c) What is the magnitude of the induced current in the coil? (d) Redo parts a, b, and c assuming this time that the solenoid’s core is made entirely out of iron whose magnetic permeability μm is 1,000μ◦ and also assume that the direction of the solenoid’s current is reversed Fig 27.19 See Exercise (14) Solenoid I Coil B r S N I L (15) A thin circular gold ring has a diameter of cm, a resistance of 60 µ , a mass of 20 g, and a specific heat of 129 J/kg.C◦ In a period of 40 ms, the ring moves from a zero-field location to a location where a magnetic field has magnitude B = 0.75 T and points perpendicular to the ring (a) What is the average magnitude of the induced emf in the ring? (b) Find the thermal energy dissipated in the ring (c) Find the temperature rise in the ring if all thermal energy converts to heat (16) A coil of radius rc = 10 cm consists of N = 30 turns of copper wire The wire of the coil has a radius of rw = 1.5 mm and a resistivity ρw = 1.68 × 10−8 m A uniform magnetic field perpendicular to the plane of the coil changes at a rate dB/dt of 8.5 × 10−3 T/s (a) What is the induced emf produced in the coil? (b) What is the resistance of the wire of the coil? (c) What is the induced current 954 27 Faraday’s Law, Alternating Current, and Maxwell’s Equations in the coil? (d) What is the rate at which the thermal energy is dissipated in the wire of the coil? (17) A circular loop of wire has a radius r = 15 cm and resistance R = 80 The loop is initially in a uniform magnetic field perpendicular to the plane of the loop and has a magnitude B = 0.5 T The loop is removed from the field in 150 ms (a) What is the average induced emf produced in the loop? (b) Find the electric energy delivered by the process if it is equal to the energy dissipated in the loop (18) A vertical rectangular loop of width a and height b is at a distance x from a vertical long wire carrying a current I, see Fig 27.20 (a) Find the magnetic flux through the loop (b) If the rectangular loop is pulled away from the wire with a speed v, find the instantaneously-induced emf produced in the loop and the instantaneous force required Fig 27.20 See Exercise (18) Resistance R b I x a Section 27.2 Motional emf (19) The rod in Fig 27.4 has a length L = 12 cm and moves with a speed v = m/s in a uniform magnetic field of magnitude B = 1.5 T Find the induced motional emf developed in the rod (20) An induced emf of V is established by moving a rod 0.8 m long at a speed of m/s perpendicular to a uniform magnetic field Find the magnitude of that field (21) A jet plane is flying horizontally at 250 m/s in a region where the vertical component of the Earth’s magnetic field is 80 µT The distance between the tips of the two wings of the plane is 30 m What is the electric potential difference induced between the two wing tips? (22) The rod in Fig 27.5 has a length L = 25 cm and moves with a constant speed v = 10 m/s in a uniform magnetic field of magnitude B = 1.5 T The resistor