22 ELECTROMAGNETIC INDUCTION

Transcription

1 CHAPTER ELECTROMAGNETIC INDUCTION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS. 3.5 m/s. (e) The work done by the hand equals the energy dissiated in the bulb. The energy dissiated in the bulb equals the ower used by the bulb times the time. Since the time is the same in each case, more work is done when the ower used is greater. The ower, however, is the voltage squared divided by the resistance of the bulb, according to Equation.6c, so that a smaller resistance corresonds to a greater ower. Thus, more work is done when the resistance of the bulb is smaller. 3. (c) The magnetic flux Ф that asses through a surface is BAcos (Equation.), where B is the magnitude of the magnetic field, A is the area of the surface, and is the angle between the field and the normal to the surface. Knowing Ф and A, we can calculate Bcos / A, which is the comonent of the field arallel to the normal or erendicular to the surface. 4. (b) The magnetic flux Ф that asses through a surface is BAcos (Equation.), where B is the magnitude of the magnetic field, A is the area of the surface, and is the angle between the field and the normal to the surface. It has the greatest value when the field strikes the surface erendicularly and a value of zero when the field is arallel to a surface 9 than to face 3. The field is more nearly erendicular to face 7 and is arallel to face. 5. (d) Faraday s law of electromagnetic induction states that the average emf induced in a coil of N loos is N / t (Equation.3), where Ф is the change in magnetic flux through one loo and t is the time interval during which the change occurs. Reducing the time interval t during which the field magnitude increases means that the rate of change of the flux will increase, which will increase (not reduce) the induced emf V 7. (c) According to Faraday s law, the magnitude of the induced emf is the magnitude of the change in magnetic flux divided by the time interval over which the change occurs (see Equation.3). In each case the field is erendicular to the coil, and the initial flux is zero since the coil is outside the field region. Therefore, the changes in flux are as follows: BL, BL (see Equation.). The corresonding time intervals are A B C

2 3 ELECTROMAGNETIC INDUCTION t t t, t t. Dividing gives the following results for the magnitudes of the A B C emfs: A BL, B L B, B L BL C t t t t. 8. (a) An induced current aears only when there is an induced emf to drive it around the loo. According to Faraday s law, an induced emf exists only when the magnetic flux through the loo changes as time asses. Here, however, there is no magnetic flux through the loo. The magnetic field lines roduced by the current are circular and centered on the wire, with the lanes of the circles erendicular to the wire. Therefore, the magnetic field is always arallel to the lane of the loo as the loo falls and never enetrates the loo. In other words, no magnetic flux asses through the loo. No magnetic flux, no induced emf, no induced current. 9. (d) When the switch is closed, current begins to flow counterclockwise in the larger coil, and the field that it creates aears inside the smaller coil. Using RHR- reveals that this field oints out of the screen toward you. According to Lenz s law, the induced current in the smaller coil flows in such a direction that it creates an induced field that ooses the growth of the field from the larger coil. Thus, the induced field must oint into the screen away from you. Using RHR- reveals that the induced current must, then, flow clockwise. The induced current exists only for the short eriod following the closing of the switch, when the field from the larger coil is growing from zero to its equilibrium value. Once the field from the larger coil reaches its equilibrium value and ceases to change, the induced current in the smaller coil becomes zero.. (b) The eak emf is roortional to the area A of the coil, according to Equation.4. Thus, we need to consider the areas of the coils. The length of the wire is L and is the same for each of the coil shaes. For the circle, the circumference is πr = L, so that the area is A circle L L r. For the square, the area is 4 A square L L. For the 4 6 L L L rectangle, the erimeter is (D + D) = L, so that the area is Arectangle. The circle has the largest area, while the rectangle has the smallest area, corresonding to answer b cm. (d) The back emf is roortional to the motor seed, so it decreases when the seed V decreases. The current I drawn by the motor is given by Equation.5 as I, where R V is the voltage at the socket, is the back emf, and R is the resistance of the motor coil. As decreases, I increases.

3 Chater Answers to Focus on Concets Questions 3 3. (c) According to Equation.7, the mutual inductance is St M. If the time interval I is cut in half and the change in the rimary current is doubled, while the induced emf remains the same, the mutual inductance must be reduced by a factor of four. 4. (b) The energy stored in an inductor is given by Equation. as Energy LI. Since the two inductors store the same amount of energy, we have L I L I. Thus, I L L.44. I L L / P 5. (e) According to Equation.8, we have LI N. Since Ф is the same for each coil, the number of turns is roortional to the roduct LI of the inductance and the current. For the coils secified in the table, this roduct is (LI) A = L I, (LI) B = L I /, (LI) C = 4L I. 6. (c) The current in the rimary is roortional to the current in the secondary according to Equation.3: IP ISNS / NP. The current in the secondary is the secondary voltage divided by the resistance, according to Ohm s law. Thus, when the resistance increases, the current in the secondary decreases and so does the current in the rimary. The wall socket delivers to the rimary the same ower that the secondary delivers to the resistance, assuming that no ower is lost within the transformer. The ower delivered to the resistance is given by Equation.5c as the square of the secondary voltage divided by the resistance. When the resistance increases, the ower decreases. Hence, the ower delivered to the rimary by the wall socket also decreases W 8. (a) The current in resistor (without the transformer) is the same as the current in resistor (with the transformer). In either event, the current I is I = V/R, where V is the voltage across the resistance R. Since the transformer is a ste-u transformer, the voltage alied across resistor is smaller than the voltage alied across resistor. The smaller voltage across resistor can lead to the same current as does the greater voltage across resistor only if R is less than R.

4 3 ELECTROMAGNETIC INDUCTION CHAPTER ELECTROMAGNETIC INDUCTION PROBLEMS. REASONING During its fall, both the length of the bar and its velocity v are erendicular to the horizontal comonent B h of the earth s magnetic field (see the drawing for an overhead view). Therefore, the emf induced across the length L of the rod is given by vb L (Equation.), where v is the seed of the h rod. We will use Equation. to determine the magnitude B h of the horizontal comonent of the earth s magnetic field, and Right-Hand Rule No. from Section. to determine which end of the rod is ositive. SOLUTION a. Solving vbh L (Equation.) for B h, we find that West North B h South Overhead view East v F (velocity into age) B h V T vl m/s.8 m b. Consider a hyothetical ositive charge that is free to move inside the falling rod. The bar is falling downward, carrying the ositive charge with it, so that the velocity v of the charge is downward. In the drawing, which shows the situation as seen from above, downward is into the age. Alying Right-Hand Rule No. to the vectors v and B h, the magnetic force F on the charge oints to the east. Therefore, ositive charges in the rod would accelerate to the east, and negative charges would accelerate to the west. As a result, the east end of the rod acquires a ositive charge.. REASONING We can treat the car as if it were a straight rod of length L. m between the driver s side and the assenger s side. This rod and the vertically downward comonent of the earth s magnetic field are erendicular. The emf induced between the ends of this moving rod is vbl (Equation.), where v 5 m / s is the seed of 5 the car, and B 4.8 T is the vertically downward comonent of the earth s magnetic field. To determine which side of the car is ositive, we can use right-hand rule no. (See Section.), which will tell us the end of the rod at which ositive charge will accumulate.

5 Chater Problems 33 SOLUTION a. Using Equation., we find that the emf is 5 3 vbl 5 m / s 4.8 T. m.4 V b. Right-hand rule no. shows that ositive charge accumulates on the driver s side. 3. REASONING AND SOLUTION The motional emf generated by a conductor moving erendicular to a magnetic field is given by Equation. as = vbl, where v and L are the seed and length, resectively, of the conductor, and B is the magnitude of the magnetic field. The emf would have been vbl 7.6 m/s 5. T. m 78 V 4. REASONING The situation in the drawing given with the roblem statement is analogous to that in Figure.4b in the text. The blood flowing at a seed v corresonds to the moving rod, and the diameter of the blood vessel corresonds to the length L of the rod in the figure. The magnitude of the magnetic field is B, and the measured voltage is the emf induced by the motion. Thus, we can aly BvL (Equation.). SOLUTION Using Equation., we find that 3 3 BvL.6 T.3 m/s 5.6 m. V 5. SSM REASONING AND SOLUTION For the three rods in the drawing in the text, we have the following: Rod A: The motional emf is zero, because the velocity of the rod is arallel to the direction of the magnetic field, and the charges do not exerience a magnetic force. Rod B: The motional emf is, according to Equation., = vbl (.7 m/s)(.45 T)(.3 m) =.6 V The ositive end of Rod B is end. Rod C: The motional emf is zero, because the magnetic force F on each charge is directed erendicular to the length of the rod. For the ends of the rod to become charged, the magnetic force must be directed arallel to the length of the rod.

6 34 ELECTROMAGNETIC INDUCTION 6. REASONING a. The motional emf generated by the moving metal rod deends only on its seed, its length, and the magnitude of the magnetic field (see Equation.). The motional emf does not deend on the resistance in the circuit. Therefore, the emfs for the circuits are the same. b. According to Equation., the current I is equal to the emf divided by the resistance R of the circuit. Since the emfs in the two circuits are the same, the circuit with the smaller resistance has the larger current. Since circuit has one-half the resistance of circuit, the current in circuit is twice as large. / c. The ower P is P R (Equation.6c), where is the emf (or voltage) and R is the resistance. The emf roduced by the moving bar is directly roortional to its seed (see Equation.). Thus, the bar in circuit roduces twice the emf, since it s moving twice as fast. Moreover, the resistance in circuit is half that in circuit. As a result, the ower delivered to the bulb in circuit is SOLUTION a. The ratio of the emfs is, according to Equation. / 8 times greater than in circuit. vbl vbl b. Equation. states that the current is equal to the emf divided by the resistance. The ratio of the currents is I / R R I / R R 55 c. The ower, according to Equation.6c, is P = /R. The motional emf is given by Equation. as = vbl. The ratio of the owers is P R R v BL R P R v BL R R v R v R v R v R 8 7. REASONING Once the switch is closed, there is a current in the rod. The magnetic field alies a force to this current and accelerates the rod to the right. As the rod begins to move, however, a motional emf aears between the ends of the rod. This motional emf deends on the seed of the rod and increases as the seed increases. Equally imortant is the fact that the motional emf ooses the emf of the battery. The net emf causing the current in the

7 Chater Problems 35 rod is the algebraic sum of the two emf contributions. Thus, as the seed of the rod increases and the motional emf increases, the net emf decreases. As the net emf decreases, the current in the rod decreases and so does the force that field alies to the current. Eventually, the seed reaches the oint when the motional emf has the same magnitude as the battery emf, and the net emf becomes zero. At this oint, there is no longer a net force acting on the rod and the seed remains constant from this oint onward, according to Newton s second law. This maximum seed can be determined by using Equation. for the motional emf, with a value of the motional emf that equals the battery emf. SOLUTION Using Equation. ( = vbl) and a value of 3. V for the emf, we find that the maximum seed of the rod is 3. V v BL.6 T. m 5 m/s 8. REASONING The average ower P delivered by the hand is given by P W / t, where W is the work done by the hand and t is the time interval during which the work is done. The work done by the hand is equal to the roduct of the magnitude F hand of the force exerted by the hand, the magnitude x of the rod s dislacement, and the cosine of the angle between the force and the dislacement. Since the rod moves to the right at a constant seed, it has no acceleration and is, therefore, in equilibrium. Thus, the force exerted by the hand must be equal to the magnitude F of the magnetic force that the current exerts on the rod. The magnitude of the magnetic force is given by F I LBsin (Equation.3), where I is the current, L is the length of the moving rod, B is the magnitude of the magnetic field, and is the angle between the direction of the current and that of the magnetic field. SOLUTION The average ower P delivered by the hand is W P (6.a) t The work W done by the hand in Figure.5 is given by W F hand xcos (Equation 6.). In this equation F hand is the magnitude of the force that the hand exerts on the rod, x is the magnitude of the rod s dislacement, and is the angle between the force and the dislacement. The force and dislacement oint in the same direction, so. Since the magnitude of the force exerted by the hand equals the magnitude F of the magnetic force, F hand = F. Substituting W F hand xcos into Equation 6.a and using the fact that F hand = F, we have that W Fhand x cos F xcos P () t t t The magnitude F of the magnetic force is given by F I LBsin (Equation.3). In this case, the current and magnetic field are erendicular to each other, so = 9 (see Figure.5). Substituting this exression for F into Equation gives

8 36 ELECTROMAGNETIC INDUCTION Fxcos I LBsin9 xcos P () t t The term x/t in Equation is the seed v of the rod. Thus, the average ower delivered by the hand is I LBsin 9 x cos P I LBsin 9v cos t.4 A.9 m. Tsin 9 m/scos W 9. SSM REASONING The minimum length d of the rails is the seed v of the rod times the time t, or d = vt. We can obtain the seed from the exression for the motional emf given in Equation.. Solving this equation for the seed gives v, where is the motional BL emf, B is the magnitude of the magnetic field, and L is the length of the rod. Thus, the length of the rails is d vt t. While we have no value for the motional emf, we do know BL that the bulb dissiates a ower of P = 6. W, and has a resistance of R = 4 Ω. Power is related to the emf and the resistance according to P (Equation.6c), which can be R solved to show that PR. Substituting this exression into the equation for d gives PR d t t BL BL SOLUTION Using the above exression for the minimum necessary length of the rails, we find that PR 6. W4 d t.5 s 5 m BL.4 T.6 m. REASONING AND SOLUTION a. Newton's second law gives the magnetic retarding force to be F = mg = IBL Now the current, I, is vbl I R R So vbl 4. m/s.5 T.3 m m Rg m/s.3 kg

9 Chater Problems 37 b. The change in height in a time t is h = vt. The change in gravitational otential energy is PE = mgh = mgvt = (.3 kg)(9.8 m/s )(4. m/s)(. s) =.8 J c. The energy dissiated in the resistor is the amount by which the gravitational otential energy decreases or.8 J.. SSM REASONING The definition of magnetic flux Ф is BAcos (Equation.), where B is the magnitude of the magnetic field, A is the area of the surface, and is the angle between the magnetic field vector and the normal to the surface. The values of B and A are the same for each of the surfaces, while the values for the angle are different. The z axis is the normal to the surface lying in the x, y lane, so that xy 35. The y axis is the normal to the surface lying in the x, z lane, so that xz 55. We can aly the definition of the flux to obtain the desired ratio directly. SOLUTION Using Equation., we find that xz x y BA cosxz cos 55 BA cos cos 35 x y.7. REASONING a. The magnetic flux though a surface of area A due to a uniform magnetic field of magnitude B is given by BAcos (Equation.) where is the angle between the direction of the magnetic field and the normal to the surface (see the drawing, which shows an edge-on view of the situation). The magnetic field B is horizontal, and the surface makes an angle of with the horizontal, so the normal to the surface is = 9. = 78. We will use Equation. to determine the surface area A. Surface B Normal to surface Edge-on view of first surface b. When exosed to a uniform magnetic field, the magnetic flux through a flat surface is greatest when the surface is erendicular to the direction of the magnetic field. When this occurs, the normal to the surface is arallel to the direction of the magnetic field, and the angle is zero. Therefore, to find the smallest surface area that has same amount of magnetic flux assing through it as the surface in art (a), we will take =. in Equation..

10 38 ELECTROMAGNETIC INDUCTION SOLUTION a. Solving BAcos (Equation.) for A yields A. Substituting = 78, we B cos find that Wb A.86 m B cos.47 T cos78 b. From A, with =., the minimum ossible area of the second surface is B cos Wb A.47 T cos..8 m 3. SSM REASONING The general exression for the magnetic flux through an area A is given by Equation.: BAcos, where B is the magnitude of the magnetic field and is the angle of inclination of the magnetic field B with resect to the normal to the area. The magnetic flux through the door is a maximum when the magnetic field lines are erendicular to the door and. so that max BA(cos. ) BA. SOLUTION When the door rotates through an angle, the magnetic flux that asses through the door decreases from its maximum value to one-third of its maximum value. Therefore,, and we have 3 max BA BA cos or cos or cos REASONING At any given moment, the flux that asses through the loo is given by BAcos (Equation.), where B is the magnitude of the magnetic field, A is the area of the loo, and = is the angle between the normal to the loo and the direction of the magnetic field (both directed into the age). As the handle turns, the area A of the loo changes, causing a change Δ in the flux assing through the loo. We can think of the loo as being divided into a rectangular ortion and a semicircular ortion. Initially, the area A of the loo is equal to the rectangular area A rec lus the area A r of the semicircle, where r is the radius of the semicircle: rec semi A A r. After half a revolution, the semicircle is once again within the lane of the loo, but now as a reduction of the area of the rectangular ortion. Therefore, the final area A of the loo is equal to the area of the rectangular ortion minus the area of the semicircle: A A r. rec SOLUTION The change Δ in the flux that asses through the loo is the difference between the final flux BAcos (Equation.) and the initial flux BA cos :

11 Chater Problems 39 BAcos BA cos Bcos A A () Substituting A A r and A A r into Equation () yields rec rec B cos A A B cos Arec r A rec r r B cos Therefore the change in the flux assing through the loo during half a revolution of the semicircle is rbcos. m.75 T cos.94 Wb 5. REASONING According to Equation., the magnetic flux is the roduct of the magnitude B of the magnetic field, the area A of the surface, and the cosine of the angle between the direction of the magnetic field and the normal to the surface. The area of a circular surface is A r, where r is the radius. SOLUTION The magnetic flux through the surface is 3 BA cos B r cos.78 T. m cos 5. Wb 6. REASONING The magnetic flux Φ that asses through a flat single-turn loo of wire is BAcos (Equation.), where B is the magnitude of the magnetic field (the same for each loo since the field is uniform), A is the area of the loo, and ϕ is the angle between the magnetic field and the normal to the lane of the loo. Since both the square and the circle are erendicular to the field, we know that for both loos. We will aly Equation. to both the square and the circle. SOLUTION Using L to denote the length of each side of the square and R to denote the radius of the circle and recognizing that the areas of the square and circle are, resectively, L and πr, we have from Equation. that square BAsquare cos BL and circle BAcircle cos B R Dividing the right-hand equation by the left-hand equation, we obtain circle square BR BL R L We do have a value for either L or R, but we do know that the square and the circle both contain the same length of wire. Therefore, it follows that L 4L R or R Length of wire in square Circumference of circle ()

12 4 ELECTROMAGNETIC INDUCTION Substituting this result for the radius of the circle into Equation () gives circle square R L / 4 L L square 7. Wb 8.9 Wb circle 7. REASONING The general exression for the magnetic flux through an area A is given by Equation.: BAcos, where B is the magnitude of the magnetic field and is the angle of inclination of the magnetic field B with resect to the normal to the surface. SOLUTION Since the magnetic field B is arallel to the surface for the triangular ends and the bottom surface, the flux through each of these three surfaces is Wb. The flux through the. m by.3 m face is BAcos.5 T. m.3 m cos..9 Wb For the. m by.5 m side, the area makes an angle with the magnetic field B, where Therefore, 9 tan.3 m.4 m 53 BAcos.5 T. m.5 m cos53.9 Wb 8. REASONING An emf is induced during the first and third intervals, because the magnetic field is changing in time. The time interval is the same (3. s) for the two cases. However, the magnitude of the field changes more during the first interval. Therefore, the magnetic flux is changing at a greater rate in that interval, which means that the magnitude of the induced emf is greatest during the first interval. The induced emf is zero during the second interval, s. According to Faraday s law of electromagnetic induction, Equation.3, an induced emf arises only when the magnetic flux changes. During this interval, the magnetic field, the area of the loo, and the orientation of the field relative to the looare constant. Thus, the magnetic flux does not change, so there is no induced emf. During the first interval the magnetic field in increasing with time. During the third interval, the field is decreasing with time. As a result, the induced emfs will have oosite olarities during these intervals. If the direction of the induced current is clockwise during the first interval, it will be counterclockwise during the third interval. SOLUTION a. The induced emf is given by Equations.3 and.3:

13 Chater Problems 4 3. s: = s: BAcos B Acos t t t N N B B.4 T T NA t t 3. s s cos 5.5 m cos. V B B.4 T.4 T t t 6. s 3. s s: = NAcos 5.5 m cos V B B. T.4 T NA t t 9. s 6. s cos 5.5 m cos.5 V b. The induced current is given by Equation. as I = ξ/r. 3. s: s:. V I. A R.5.5 V I. A R.5 As exected, the currents are in oosite directions. 9. REASONING The magnitude of the emf induced in the loo can be found using Faraday s law of electromagnetic induction: N (.3) t t where N is the number of turns, and are, resectively, the final and initial fluxes, and t t is the elased time. The magnetic flux is given by BAcos (Equation.), where B is the magnitude of the magnetic field, A is the area of the surface, and is the angle between the direction of the magnetic field and the normal to the surface. SOLUTION Setting N = since there is only one turn, noting that the final area is A = m and the initial area is A =. m.35 m, and noting that the angle between the magnetic field and the normal to the surface is, we find that the magnitude of the emf induced in the coil is

14 4 ELECTROMAGNETIC INDUCTION BAcos BA cos N t t.65 T m cos.65 T. m.35 m cos.5 V.8 s. REASONING An emf is induced in the body because the magnetic flux is changing in time. According to Faraday s law, as given by Equation.3, the magnitude of the emf is N. This exression can be used to determine the time interval t during which t the magnetic field goes from its initial value to zero. The magnetic flux is obtained from Equation. as = BA cos, where = in this roblem. SOLUTION The magnitude of the induced emf is = BAcos B Acos N N t t Solving this relation for t gives.3 m cos.5 T NAcos B B t. V 4.8 s SSM REASONING According to Equation.3, the average emf induced in a coil of N loos is = N / t. SOLUTION For the circular coil in question, the flux through a single turn changes by BAcos 45 BAcos 9 BAcos 45 during the interval of t. s. magnitude of the emf as Therefore, for N turns, Faraday's law gives the = N BAcos 45 t Since the loos are circular, the area A of each loo is equal to r. Solving for B, we have t (.65 V)(. s) 5 B 8.6 T N r cos 45 (95) (.6 m) cos 45

15 Chater Problems 43. REASONING According to Ohm s law (see Section.), the resistance of the wire is equal to the emf divided by the current. The emf can be obtained from Faraday s law of electromagnetic induction. SOLUTION The resistance R of the wire is R (.) I According to Faraday s law of electromagnetic induction, the induced emf is N N t t t (.3) where N is the number of loos in the coil, and are, resectively, the final and initial fluxes, and t t is the elased time. Substituting Equation.3 into Equation. yields N 4. Wb 9. Wb t t R.5 s 5. I I 3 A 3. REASONING We will use Faraday s law of electromagnetic induction, Equation.3, to find the emf induced in the loo. Once this value has been determined, we can emloy Equation.3 again to find the rate at which the area changes. SOLUTION a. The magnitude of the induced emf is given by Equation.3 as BAcos B Acos B B = N N NAcos t t t t t t.t T.35 m.55 mcos V.45 s s b. When the magnetic field is constant and the area is changing in time, Faraday s law can be written as = BAcos BA cos N N t t t t A A A N B cos N B cos t t t

16 44 ELECTROMAGNETIC INDUCTION Solving this equation for the emf, we find that A t and substituting in the value of.38 V for the magnitude of A t.38 V = =.43 m / s N Bcos.T cos REASONING According to Faraday s law the emf induced in either single-turn coil is given by Equation.3 as N t t since the number of turns is N =. The flux is given by Equation. as BA cos BA cos BA where the angle between the field and the normal to the lane of the coil is, because the field is erendicular to the lane of the coil and, hence, arallel to the normal. With this exression for the flux, Faraday s law becomes BA B A t t t In this exression we have recognized that the area A does not change in time and have searated it from the magnitude B of the magnetic field. We will aly this form of Faraday s law to each coil. The current induced in either coil deends on the resistance R of the coil, as well as the emf. According to Ohm s law, the current I induced in either coil is given by I () R SOLUTION Alying Faraday s law in the form of Equation to both coils, we have B B B B square Asquare L and circle Acircle r t t t t The area of a square of side L is L, and the area of a circle of radius r is πr. The rate of change B/t of the field magnitude is the same for both coils. Dividing these two exressions gives B L square t L (3) circle B r r t The same wire is used for both coils, so we know that the erimeter of the square equals the circumference of the circle: ()

17 Chater Problems 45 L 4L r or r 4 Substituting this result into Equation (3) gives square L square circle circle r or.8 V.63 V (4) Using Ohm s law as given in Equation (), we find for the induced currents that square Isquare square R I circle circle circle R Here we have used the fact that the same wire has been used for each coil, so that the resistance R is the same for each coil. Using the result in Equation (4) gives I I square circle square or Isquare Icircle 3. A.5 A circle 5. REASONING The magnitude of the average emf induced in the triangle is given by N (see Equation.3), which is Faraday s law. This exression can be used t directly to calculate the magnitude of the average emf. Since the triangle is a single-turn coil, the number of turns is N =. According to Equation., the magnetic flux Ф is BA cos BA cos BA () where B is the magnitude of the field, A is the area of the triangle, and is the angle between the field and the normal to the lane of the triangle (the magnetic field is erendicular to the lane of the triangle). It is the change in the flux that aears in Faraday s law, so that we use Equation () as follows: BA BA BA where A = m is the initial area of the triangle just as the bar asses oint A, and A is the area after the time interval t has elased. The area of a triangle is one-half the base (d AC ) times the height (d CB ) of the triangle. Thus, the change in flux is AC CB BA B d d The base and the height of the triangle are related, according to d CB = d AC tan θ, where θ = 9º. Furthermore, the base of the triangle becomes longer as the rod moves. Since the rod moves with a seed v during the time interval t, the base is d AC = vt. With these substitutions the change in flux becomes

18 46 ELECTROMAGNETIC INDUCTION tan B d ACdCB B d AC dac B vt tan SOLUTION Substituting Equation () for the change in flux into Faraday s law, we find that the magnitude of the induced emf is B vt tan N N N Bv t tan t t.38 T.6 m/s 6. s tan9.4 V () 6. REASONING An emf is induced in the coil because the magnetic flux through the coil is changing in time. The flux is changing because the angle between the normal to the coil and the magnetic field is changing. The amount of induced current is equal to the induced emf divided by the resistance of the coil (see Equation.). According to Equation., the amount of charge q that flows is equal to the induced current I multilied by the time interval t = t t during which the coil rotates, or q = I t t ). SOLUTION According to Equation., the amount of charge that flows is q = It. The current is related to the emf in the coil and the resistance R by Equation. as I = /R. The amount of charge that flows can, therefore, be written as q I t t R The emf is given by Faraday s law of electromagnetic induction as = BAcos BAcos t t N N where we have also used Equation., which gives the definition of magnetic flux as BAcos. With this emf, the exression for the amount of charge becomes BAcos BAcos N t q I t t R Solving for the magnitude of the magnetic field yields C NBA cos cos Rq B NA cos cos 5.5 m cos 9 cos 3 R.6 T

19 Chater Problems SSM REASONING According to Equation.3, the average emf induced in a single loo (N = ) is / t. Since the magnitude of the magnetic field is changing, the area of the loo remains constant, and the direction of the field is arallel to the normal to the loo, the change in flux through the loo is given by (B)A. Thus the magnitude of the induced emf in the loo is given by B A/ t. Similarly, when the area of the loo is changed and the field B has a given value, we find the magnitude of the induced emf to be BA/ t. SOLUTION a. The magnitude of the induced emf when the field changes in magnitude is B 3 A (.8 m )(. T/s) 3.6 V t b. At a articular value of B (when B is changing), the rate at which the area must change can be obtained from 3 BA A 3.6 V 3 or. m / s t t B.8 T In order for the induced emf to be zero, the magnitude of the magnetic field and the area of the loo must change in such a way that the flux remains constant. Since the magnitude of the magnetic field is increasing, the area of the loo must decrease, if the flux is to remain constant. Therefore, the area of the loo must be shrunk. 8. REASONING The magnitude B I of the magnetic field at the center of a circular coil with N turns and a radius r, carrying a current I is given by I (.6) BI N r 7 where 4 T m/a is the ermeability of free sace. We use the symbol r to denote the radius of the coil in order to distinguish it from the coil s resistance R. The coil s induced current I is caused by the induced emf. We will determine the induced current with Ohm s law (Equation.), the magnitude of the induced emf, and the resistance R of the coil: I (.) R The magnitude of the induced emf is found from Faraday s law: N (Equation.3), where Δ/Δt is the rate of change of the flux caused by the t

20 48 ELECTROMAGNETIC INDUCTION external magnetic field B. At any instant, the flux through each turn of the coil due to the external magnetic field is given by BA cos BA cos BA (.) where A is the cross-sectional area of the coil. The angle between the direction of the external magnetic field and the normal to the lane of the coil is zero, because the external magnetic field is erendicular to the lane of the coil. SOLUTION In this situation, the cross-sectional area A of the coil is constant, but the magnitude B of the external magnetic field is changing. Therefore, substituting Equation. into N (Equation.3) yields t BA B N N NA t t t Substituting Equation () into Equation., we obtain B NA t NA B I R R R t Substituting Equation () into Equation (.6) yields Lastly, we substitute I NA B A B BI N N N r r R t rr t A r into Equation (3), because the coil is circular. This yields () () (3) BI N r r R B r B N t R t 7 4 T m/a4. m T/s.4 T.48 (4) 9. REASONING A greater magnetic flux asses through the coil in art b of the drawing. The flux is given as BAcos (Equation.), where B is the magnitude of the magnetic field, A is the area of the surface through which the field asses, and is the angle between the normal to the lane of the surface and the field. In art b the coil area has two arts, and so does the flux. There is the larger semicircular area r larger on the horizontal surface and the smaller semicircular r smaller area erendicular to the horizontal surface. The field is arallel to the normal for the larger area and erendicular to the normal for the smaller area.

21 Chater Problems 49 ω ω (a) (b) Hence, the flux is BA cos BA cos b larger larger smaller smaller larger cos smaller cos9 larger B r B r B r In art a the entire area of the coil lies on the horizontal surface and is the area between the two semicircles. The lane of the coil is erendicular to the magnetic field. The flux, therefore, is BA cos B r r cos B r r () a larger smaller larger smaller which is smaller than the flux in art b of the drawing. According to Lenz s law the induced magnetic field associated with the induced current ooses the change in flux. To oose the increase in flux from art a to art b of the drawing, the induced field must oint downward. In this way, it will reduce the effect of the increasing coil area available to the uward-ointing external field. An alication of RHR- shows that the induced current must flow clockwise in the larger semicircle of wire (when viewed from above) in order to create a downward-ointing induced field. The average induced current can be determined according to Ohm s law as the average induced emf divided by the resistance of the coil. The eriod T of the rotational motion is related to the angular frequency ω according to (Equation.6). The shortest time interval t that elases between arts a and b of T the drawing is the time needed for one quarter of a turn or 4 T. SOLUTION According to Ohm s law, the average induced current is I R where is the average induced emf and R is the resistance of the coil. According to Faraday s law, Equation.3, the average induced emf is, where we have set t N =. We can determine the change Ф in the flux as b a, where the fluxes in ()

22 5 ELECTROMAGNETIC INDUCTION arts a and b of the drawing are available from Equations () and (). The time interval t is T, as discussed in the REASONING. Using Equation.6, we have for the eriod that 4 T. Thus, we find for the current that b t 4 T I R R R.35 T. m.5 /.5 rad/s a larger larger smaller smaller smaller R T R T R 4 / 4 4 B r B r r B r B r.84 A 4 3. REASONING We can obtain the magnitude B of the magnetic field with the aid of Faraday s law, which is N (Equation.3), where is the emf induced in the coil, t N 85 is the number of turns in the coil, ΔΦ is the change in magnetic flux, and Δt is the time interval during which the flux changes. The flux is BAcos (Equation.3), 4 where A 4.7 m is the area of each turn of the coil and is the angle between the normal to the coil and the field. The emf can be related to the induced charge that flows in the coil by using Ohm s law IR (Equation.), where R 45. is the q total resistance of the circuit and I is the current. The current is given by I t 3 (Equation.), where q 8.87 C is the amount of charge that flows in the time interval Δt. We have no value for Δt. However, it will be eliminated algebraically from our calculation, as we will see. SOLUTION Using Equation.3, we can write the change in flux as BAcos BAcos BA Final flux when B is resent Initial flux when B is zero With this change in flux, Faraday s law (without the minus sign, since we seek only the magnitude of the field) becomes BA t N N or B t t NA

23 Chater Problems 5 Combining Ohm s law and Equation. for the current, we obtain Substituting this result for into the exression for B, we obtain q R t 3 t t q R 8.87 C45. B NA NA NA m q IR R. t.459 T 3. SSM REASONING AND SOLUTION Consider one revolution of either rod. The magnitude of the emf induced across the rod is A B L = B = t t The angular seed of the rods is = /t, so = BL. The rod tis have oosite olarity since they are rotating in oosite directions. Hence, the difference in otentials of the tis is V = BL so V 4.5 V BL T.68 m rad/s 3. REASONING Our solution is based on Lenz s law. According to Lenz s law, the induced emf has a olarity that leads to an induced current whose direction is such that the induced magnetic field ooses the original flux change. We will also have need of right-hand rule no. (RHR-) as we determine which end of the resistor is the ositive end. SOLUTION As the loo rotates through one-half a revolution, the area through which the magnetic field asses decreases. The magnetic flux, being roortional to the area, also decreases. The induced magnetic field must oose this decrease in flux, so the induced magnetic field must strengthen the original magnetic field. Thus, the induced magnetic field oints in the same direction as the original magnetic field, or into the lane of the age. According to RHR-, the induced current flows clockwise around the loo (and right-to-left through the resistor) so that its magnetic field will be directed into the age. Since conventional current flows through a resistor from higher otential toward lower otential, the right end of the resistor must be the ositive end.

24 5 ELECTROMAGNETIC INDUCTION 33. SSM REASONING The external magnetic field is erendicular to the lane of the horizontal loo, so it must oint either uward or downward. We will use Lenz s law to decide whether the external magnetic B field oints u or down. This law redicts that the direction of the induced magnetic field B ind ooses the change in the magnetic flux through the loo due to the external field. SOLUTION The external magnetic field B is increasing in magnitude, so that the magnetic flux through the loo also increases with time. In order to oose the increase in magnetic flux, the induced magnetic field B ind must be directed oosite to the external magnetic field B. The drawing shows the loo as viewed from above, with an induced current I ind flowing clockwise. According to Right-Hand Rule No. (see Section.7), this induced current creates an induced magnetic field B ind that is directed into the age at the center of the loo (and all other oints of the loo s interior). Therefore, the external magnetic field B must be directed out of the age. Because we are viewing the loo from above, out of the age corresonds to uward toward the viewer. I ind B ind (into age) B (out of age) 34. REASONING The magnetic field roduced by I extends throughout the sace surrounding the loo. Using RHR-, it can be shown that the magnetic field is arallel to the normal to the loo. Thus, the magnetic field enetrates the loo and generates a magnetic flux. According to Faraday s law of electromagnetic induction, an emf is induced when the magnetic flux through the loo is changing in time. If the current I is constant, the magnetic flux is constant, and no emf is induced in the loo. However, if the current is decreasing in time, the magnetic flux is decreasing and an induced current exists in the loo. Lenz s law states that the induced magnetic field ooses the change in the magnetic field roduced by the current I. The induced magnetic field does not necessarily oose the magnetic field itself. Thus, the induced magnetic field does not always have a direction that is oosite to the direction of the field roduced by I. SOLUTION At the location of the loo, the magnetic field roduced by the current I is directed into the age (this can be verified by using RHR-). The current is decreasing, so the magnetic field is decreasing. Therefore, the magnetic flux that enetrates the loo is decreasing. According to Lenz s law, the induced emf has a olarity that leads to an induced current whose direction is such that the induced magnetic field ooses this flux change. The induced magnetic field will oose this decrease in flux by ointing into the age, in the same direction as the field roduced by I. According to RHR-, the induced current must flow clockwise around the loo in order to roduce such an induced field. The current then flows from left-to-right through the resistor.

25 Chater Problems SSM REASONING The current I roduces a magnetic field, and hence a magnetic flux, that asses through the loos A and B. Since the current decreases to zero when the switch is oened, the magnetic flux also decreases to zero. According to Lenz s law, the current induced in each coil will have a direction such that the induced magnetic field will oose the original flux change. SOLUTION a. The drawing in the text shows that the magnetic field at coil A is erendicular to the lane of the coil and oints down (when viewed from above the table to). When the switch is oened, the magnetic flux through coil A decreases to zero. According to Lenz's law, the induced magnetic field roduced by coil A must oose this change in flux. Since the magnetic field oints down and is decreasing, the induced magnetic field must also oint down. According to Right-Hand Rule No. (RHR-), the induced current must be clockwise around loo A. b. The drawing in the text shows that the magnetic field at coil B is erendicular to the lane of the coil and oints u (when viewed from above the table to). When the switch is oened, the magnetic flux through coil B decreases to zero. According to Lenz's law, the induced magnetic field roduced by coil B must oose this change in flux. Since the magnetic field oints u and is decreasing, the induced magnetic field must also oint u. According to RHR-, the induced current must be counterclockwise around loo B. 36. REASONING According to Lenz s law, the induced current in the triangular loo flows in such a direction so as to create an induced magnetic field that ooses the original flux change. SOLUTION a. As the triangle is crossing the +y axis, the magnetic flux down into the lane of the aer is increasing, since the field now begins to enetrate the loo. To offset this increase, an induced magnetic field directed u and out of the lane of the aer is needed. By alying RHR- it can be seen that such an induced magnetic field will be created within the loo by a counterclockwise induced current. b. As the triangle is crossing the x axis, there is no flux change, since all arts of the triangle remain in the magnetic field, which remains constant. Therefore, there is no induced magnetic field, and no induced current aears. c. As the triangle is crossing the y axis, the magnetic flux down into the lane of the aer is decreasing, since the loo now begins to leave the field region. To offset this decrease, an induced magnetic field directed down and into the lane of the aer is needed. By alying RHR- it can be seen that such an induced magnetic field will be created within the loo by a clockwise induced current. d. As the triangle is crossing the +x axis, there is no flux change, since all arts of the triangle remain in the field-free region. Therefore, there is no induced magnetic field, and no induced current aears.

26 54 ELECTROMAGNETIC INDUCTION 37. REASONING The current I in the straight wire roduces a circular attern of magnetic field lines around the wire. The magnetic field at any oint is tangent to one of these circular field lines. Thus, the field oints erendicular to the lane of the table. Furthermore, according to Right- Hand Rule No., the field is directed u out of the table surface in region above the wire and is directed down into the table surface in region below the wire (see the drawing at the right). To deduce the direction of any induced current in the circular loo, we consider Faraday s law and the change that occurs in the magnetic flux through the loo due to the field of the straight wire. SOLUTION As the current I decreases, the magnitude of the field that it roduces also decreases. However, the directions of the fields in regions and do not change and remain as discussed in the reasoning. Since the fields in these two regions always have oosite directions and equal magnitudes at any given radial distance from the straight wire, the flux through the regions add u to give zero for any value of the current. With the flux remaining constant as time asses, Faraday s law indicates that there is no induced emf in the coil. Since there is no induced emf in the coil, there is no induced current. 38. REASONING Our solution is based on Lenz s law. According to Lenz s law, the induced emf has a olarity that leads to an induced current whose direction is such that the induced magnetic field ooses the original flux change. As we refer to the drawing at the right, we will also have need of right-hand rule no. (RHR-). Note that the horizontal arrows in the drawing indicate induced current that exists in the ring. I S N B induced B induced I S N SOLUTION a. When the magnet is above the ring its magnetic field oints down through the ring and is increasing in strength as the magnet falls. According to Lenz s law, an induced magnetic field aears that attemts to reduce the increasing field. Therefore, the induced field must oint u. Using RHR-, we can see that the induced current in the ring is as shown in the drawing. Because of the induced current, the ring looks like a magnet with its north ole at the to (use RHR-). The north ole of the loo reels the falling magnet and retards its motion. When the magnet is below the ring, its magnetic field still oints down through the ring but is decreasing as the magnet falls. According to Lenz s law, the induced magnetic field attemts to bolster the decreasing field and, therefore, must oint down. The induced current in the ring is as shown in the drawing, and the ring then looks like a magnet with its

27 Chater Problems 55 north ole at the bottom, attracting the south ole of the falling magnet and retarding its motion. b. The motion of the magnet is unaffected, since no induced current can flow in the cut ring. No induced current means that no induced magnetic field can be roduced to reel or attract the falling magnet. 39. REASONING AND SOLUTION a. Location I As the loo swings downward, the normal to the loo makes a smaller angle with the alied field. Hence, the flux through the loo is increasing. The induced magnetic field must oint generally to the left to counteract this increase. The induced current flows x y z Location II The angle between the normal to the loo and the alied field is now increasing, so the flux through the loo is decreasing. The induced field must now be generally to the right, and the current flows z y x b. Location I The argument is the same as for location II in art a. z y x Location II The argument is the same as for location I in art a. x y z 4. REASONING When the motor is running at normal seed, the current is the net emf divided by the resistance R of the armature wire. The net emf is the alied voltage V minus the back emf develoed by the rotating coil. We can use this relation to find the back emf. When the motor is just turned on, there is no back emf, so the current is just the alied voltage divided by the resistance of the wire. To limit the starting current to 5. A, a resistor R is laced in series with the resistance of the wire, so the equivalent resistance is R + R. The current to the motor is equal to the alied voltage divided by the equivalent resistance. SOLUTION a. According to Equation.5, the back emf generated by the motor is V I R. V 7. A.7 5 V b. When the motor has been just turned on, the back emf is zero, so the current is

28 56 ELECTROMAGNETIC INDUCTION V. V V I 67 A R.7 c. When a resistance R is laced in series with the resistance R of the wire, the equivalent resistance is R + R. The current to the motor is Solving this exression for R gives I V R R V. V V R R I 5. A 4. SSM REASONING We can use the information given in the roblem statement to determine the area of the coil A. Since it is square, the length of one side is A. SOLUTION According to Equation.4, the maximum emf induced in the coil is NAB. Therefore, the length of one side of the coil is 75. V A NB (48)(.7 T)(79. rad/s).5 m 4. REASONING The emf of the generator is NAB sint (Equation.4), where N 5 is the number of turns in the coil, A.85 m is the area er turn, B is the magnitude of the magnetic field, is the angular seed in rad/s, and t is the time. The angular seed is f (Equation.6), where f 6. Hz is the frequency in cycles er second or Hertz. Equation.4 indicates that the maximum emf is max NAB, since sin t has a maximum value of. SOLUTION Substituting f into max NAB and solving for the result for B reveals that B max 55 V. T NA f 5.85 m 6. Hz 43. REASONING The number N of turns in the coil of a generator is given by N AB (Equation.4), where is the eak emf, A is the area er turn, B is the magnitude of the magnetic field, and ω is the angular seed in rad/s. We have values for A, and B. Although we are not given the eak emf, we know that it is related to the rms emf, which is known: