ELECTROMAGNETIC INDUCTION

Transcription

1 LCTOMAGNTIC INDUCTION IDNTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. ST UP: The flux through a coil of N turns is Φ = NA cos φ, and by Faraday s law the magnitude of the induced emf is = /. XCUT: (a) ΔΦ = NA = (5)(1. T)(.5 m)(.3 m) = 4.5 Wb (b) = / = (4.5 Wb)/(. s) =.3 V VALUAT: This induced potential is certainly large enough to be easily detectable. ΔΦ 9.. IDNTIFY: =. Φ = Acosφ. Φ is the flux through each turn of the coil. Δ t ST UP: φ i =. φ f = XCUT: (a) Φ,i = Acos = (6. 1 T)(1 1 m )(1) = 7. 1 Wb. The total flux through the coil is 8 5 NΦ,i = ()(7. 1 Wb) = Wb. Φ,f = Acos9 =. 5 NΦ i NΦ f Wb 4 (b) = = = V =.36 mv. Δt.4 s VALUAT: The average induced emf depends on how rapidly the flux changes IDNTIFY and ST UP: Use Faraday s law to calculate the average induced emf and apply Ohm s law to the coil to calculate the average induced current and charge that flows. ΔΦI (a) XCUT: The magnitude of the average emf induced in the coil is av = N. Initially, Δt f i Φ i = Acos φ = A. The final flux is zero, so Φ Φ NA av = N =. Δt Δt The average induced current is av NA I = =. The total charge that flows through the coil is NA NA Q= IΔ t = Δ t =. Δt Δt VALUAT: The charge that flows is proportional to the magnetic field but does not depend on the time Δ t. (b) The magnetic stripe consists of a pattern of magnetic fields. The pattern of charges that flow in the reader coil tell the card reader the magnetic field pattern and hence the digital information coded onto the card. (c) According to the result in part (a) the charge that flows depends only on the change in the magnetic flux and it does not depend on the rate at which this flux changes IDNTIFY and ST UP: Apply the result derived in xercise 9.3: Q= NA/. In the present exercise the flux changes from its maximum value of Φ = A to zero, so this equation applies. is the total resistance so here = 6. Ω+ 45. Ω= 15. Ω. 5 NA Q ( C)(15. Ω) XCUT: Q= says = = =.973 T. 4 NA 1(3. 1 m ) VALUAT: A field of this magnitude is easily produced IDNTIFY: Apply Faraday s law. ST UP: Let +z be the positive direction for A. Therefore, the initial flux is positive and the final flux is zero. ΔΦ (1.5 T) π (.1 m) XCUT: (a) and (b) = = =+ 34 V. Since is positive and A is toward us, 3 Δ t. 1 s the induced current is counterclockwise. VALUAT: The shorter the removal time, the larger the average induced emf. 9-1

2 9- Chapter IDNTIFY: Apply q.(9.4). I = /. ST UP: / = Ad/. N d d XCUT: (a) = = NA ( ) = NA ((.1 T/s) t+ (3. 1 T/s ) t ) = NA( (.1 T/s) + (1. 1 T/s ) t ) =.3 V + (3. 1 V/s ) t V 4 (b) At t = 5. s, =.3 V + (3. 1 V/s )(5. s) =.68 V I = = = A. 6 Ω VALUAT: The rate of change of the flux is increasing in time, so the induced current is not constant but rather increases in time IDNTIFY: Calculate the flux through the loop and apply Faraday s law. ST UP: To find the total flux integrate over the wih of the loop. The magnetic field of a long straight μi wire, at distance r from the wire, is =. The direction of is given by the right-hand rule. π r μi XCUT: (a) When =, into the page. π r μ i (b) = da= Ldr. π r b μil bdr μil (c) Φ = = ln( b/ a). a π = a r π μl di (d) = = ln( ba). π μ 7 (e) (.4 m) = ln(.36/.1)(9.6 A/s) = V. π VALUAT: The induced emf is proportional to the rate at which the current in the long straight wire is changing 9.8. IDNTIFY: Apply Faraday s law. ST UP: Let A be upward in Figure 9.8 in the textbook. XCUT: (a) d ind = = ( A ) d d (.57s ( 1 ) t 1 (.57s ) 1 ) t ind = Asin 6 = Asin 6 (1.4 T) e = ( π r )(sin 6 )(1.4 T)(.57 s ) e 1 (.57s 1 ) t (.57 s 1 ) t ind = π (.75 m) (sin 6 )(1.4 T)(.57 s ) e = (.1 V) e. 1 1 (b) = = 1 1 (.57 s ) t 1 (.1 V). (.1 V) = (.1 V) e. ln(1/1) = (.57 s )t and t = 4.4 s (c) is in the direction of A so Φ is positive. is getting weaker, so the magnitude of the flux is decreasing and / <. Faraday s law therefore says >. Since >, the induced current must flow counterclockwise as viewed from above. VALUAT: The flux changes because the magnitude of the magnetic field is changing IDNTIFY and ST UP: Use Faraday s law to calculate the emf (magnitude and direction). The direction of the induced current is the same as the direction of the emf. The flux changes because the area of the loop is changing; relate da/ to dc/, where c is the circumference of the loop. (a) XCUT: c= π r and A= πr so A= c /4π Φ = A = ( /4 π ) c dc = = c π At t = 9. s, c= 1.65 m (9. s)(.1 m/s) =.57 m = (.5 T)(1/ π )(.57 m)(.1 m/s) = 5.44 mv (b) ST UP: The loop and magnetic field are sketched in Figure 9.9. Figure 9.9 Take into the page to be the positive direction for A. Then the magnetic flux is positive.

3 lectromagnetic Induction 9-3 XCUT: The positive flux is decreasing in magnitude; / is negative and is positive. y the righthand rule, for A into the page, positive is clockwise. VALUAT: ven though the circumference is changing at a constant rate, da/ is not constant and is not constant. Flux is decreasing so the flux of the induced current is and this means that I is clockwise, which checks IDNTIFY: A change in magnetic flux through a coil induces an emf in the coil. ST UP: The flux through a coil is Φ = NA cos φ and the induced emf is = /. XCUT: (a) and (c) The magnetic flux is constant, so the induced emf is zero. (b) The area inside the field is changing. If we let x be the length (along the 3.-cm side) in the field, then A = (.4 m)x. Φ = A = (.4 m)x = / = d[(.4 m)x]/ = (.4 m)dx/ = (.4 m)v = (1.5 T)(.4 m)(. m/s) =.1 V VALUAT: It is not a large flux that induces an emf, but rather a large rate of change of the flux. The induced emf in part (b) is small enough to be ignored in many instances IDNTIFY: A change in magnetic flux through a coil induces an emf in the coil. ST UP: The flux through a coil is Φ = NA cos φ and the induced emf is = /. XCUT: (a) = / = d[a( + bx)]/ = ba dx/ = bav (b) clockwise (c) Same answers except the current is counterclockwise. VALUAT: ven though the coil remains within the magnetic field, the flux through it increases because the strength of the field is increasing IDNTIFY: Use the results of xample 9.5. π rad/rev ST UP: max = NAω. av = max. ω = (44 rev/min) = 46.1 rad/s. π 6 s/min XCUT: (a) max = NAω = (15)(.6 T) π(.5 m) (46.1 rad/s) =.814 V (b) = av max (.815 V).519 V π = π = VALUAT: In max = NAω, ω must be in rad/s IDNTIFY: Apply the results of xample 9.5. ST UP: max = NAω max.4 1 V XCUT: ω = = = 1.4 rad/s NA (1)(.75 T)(.16 m) VALUAT: We may also express ω as 99.3 rev/min or 1.66 rev/s IDNTIFY: A change in magnetic flux through a coil induces an emf in the coil. ST UP: The flux through a coil is Φ = NA cos φ and the induced emf is = /. XCUT: The flux is constant in each case, so the induced emf is zero in all cases. VALUAT: ven though the coil is moving within the magnetic field and has flux through it, this flux is not changing, so no emf is induced in the coil IDNTIFY and ST UP: The field of the induced current is directed to oppose the change in flux. XCUT: (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise. (b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise. (c) The field is constant so the flux is constant and there is no induced emf and no induced current. VALUAT: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing IDNTIFY: y Lenz s law, the induced current flows to oppose the flux change that caused it. ST UP and XCUT: The magnetic field is outward through the round coil and is decreasing, so the magnetic field due to the induced current must also point outward to oppose this decrease. Therefore the induced current is counterclockwise. VALUAT: Careful Lenz s law does not say that the induced current flows to oppose the magnetic flux. Instead it says that the current flows to oppose the change in flux IDNTIFY and ST UP: Apply Lenz's law, in the form that states that the flux of the induced current tends to oppose the change in flux. XCUT: (a) With the switch closed the magnetic field of coil A is to the right at the location of coil. When the switch is opened the magnetic field of coil A goes away. Hence by Lenz's law the field of the current induced in coil is to the right, to oppose the decrease in the flux in this direction. To produce magnetic field that is to the right the current in the circuit with coil must flow through the resistor in the direction a to b.

4 9-4 Chapter 9 (b) With the switch closed the magnetic field of coil A is to the right at the location of coil. This field is stronger at points closer to coil A so when coil is brought closer the flux through coil increases. y Lenz's law the field of the induced current in coil is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil must flow through the resistor in the direction b to a. (c) With the switch closed the magnetic field of coil A is to the right at the location of coil. The current in the circuit that includes coil A increases when is decreased and the magnetic field of coil A increases when the current through the coil increases. y Lenz's law the field of the induced current in coil is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil must flow through the resistor in the direction b to a. VALUAT: In parts (b) and (c) the change in the circuit causes the flux through circuit to increase and in part (a) it causes the flux to decrease. Therefore, the direction of the induced current is the same in parts (b) and (c) and opposite in part (a) IDNTIFY: Apply Lenz s law. ST UP: The field of the induced current is directed to oppose the change in flux in the primary circuit. XCUT: (a) The magnetic field in A is to the left and is increasing. The flux is increasing so the field due to the induced current in is to the right. To produce magnetic field to the right, the induced current flows through from right to left. (b) The magnetic field in A is to the right and is decreasing. The flux is decreasing so the field due to the induced current in is to the right. To produce magnetic field to the right the induced current flows through from right to left. (c) The magnetic field in A is to the right and is increasing. The flux is increasing so the field due to the induced current in is to the left. To produce magnetic field to the left the induced current flows through from left to right. VALUAT: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing IDNTIFY and ST UP: Lenz's law requires that the flux of the induced current opposes the change in flux. XCUT: (a) Φ is " and increasing so the flux Φ ind of the induced current is and the induced current is clockwise. (b) The current reaches a constant value so Φ is constant. / = and there is no induced current. (c) Φ is " and decreasing, so Φ is ind " and current is counterclockwise. VALUAT: Only a change in flux produces an induced current. The induced current is in one direction when the current in the outer ring is increasing and is in the opposite direction when that current is decreasing. 9.. IDNTIFY: Use the results of xample 9.6. Use the three approaches specified in the problem for determining the direction of the induced current. I = /. ST UP: Let A be directed into the figure, so a clockwise emf is positive. XCUT: (a) = vl = (5. m/s)(.75 T)(1.5 m) = 5.6 V (b) (i) Let q be a positive charge in the moving bar, as shown in Figure 9.a. The magnetic force on this charge is F = q v, which points upward. This force pushes the current in a counterclockwise direction through the circuit. (ii) Φ is positive and is increasing in magnitude, so / >. Then by Faraday s law < and the emf and induced current are counterclockwise. (iii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase. Hence this current must flow in a counterclockwise sense, as shown in Figure 9.b. 5.6 V (c) = I. I =. A. = 5 Ω = VALUAT: All three methods agree on the direction of the induced current. Figure IDNTIFY: A conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving. ST UP: The induced emf is = vl sin φ, where φ is the angle between the velocity and the magnetic field.

5 lectromagnetic Induction 9-5 XCUT: (a) = vl sin φ = (5. m/s)(.45 T)(.3 m)(sin 9 ) =.675 V (b) The positive charges are moved to end b, so b is at the higher potential. (c) = V/L = (.675 V)/(.3 m) =.5 V/m. The direction of is from, b to a. (d) The positive charge are pushed to b, so b has an excess of positive charge. (e) (i) If the rod has no appreciable thickness, L =, so the emf is zero. (ii) The emf is zero because no magnetic force acts on the charges in the rod since it moves parallel to the magnetic field. VALUAT: The motional emf is large enough to have noticeable effects in some cases. 9.. IDNTIFY: The moving bar has a motional emf induced across its ends, so it causes a current to flow. ST UP: The induced potential is = vl and Ohm s law is = I. XCUT: (a) = vl = (5. m/s)(.75 T)(1.5 m) = 5.6 V (b) I = / = (5.6 V)/(5 Ω) =.3 A VALUAT: oth the induced potential and the current are large enough to have noticeable effects IDNTIFY: = vl ST UP: L = 5. 1 m. 1 mph =.447 m/s. XCUT: 1.5 V v = = = 46. m/s = 13 mph. L (.65 T)(5. 1 m) VALUAT: This is a large speed and not practical. It is also difficult to produce a 5. cm wide region of.65 T magnetic field IDNTIFY: = vl. 4 ST UP: 1 mph =.447 m/s. 1 G = 1 T..447 m/s 4 XCUT: (a) = (18 mph) (.5 1 T)(1.5 m) = 6. mv. This is much too small to be 1 mph noticeable..447 m/s 4 (b) = (565 mph) (.5 1 T)(64.4 m) =.813 mv. This is too small to be noticeable. 1 mph VALUAT: ven though the speeds and values of L are large, the earth s field is small and motional emfs due to the earth s field are not important in these situations IDNTIFY and ST UP: = vl. Use Lenz's law to determine the direction of the induced current. The force F ext required to maintain constant speed is equal and opposite to the force F I that the magnetic field exerts on the rod because of the current in the rod. XCUT: (a) = vl = (7.5 m/s)(.8 T)(.5 m) = 3. V (b) is into the page. The flux increases as the bar moves to the right, so the magnetic field of the induced current is out of the page inside the circuit. To produce magnetic field in this direction the induced current must be counterclockwise, so from b to a in the rod. 3. V (c) I =. A. = 1.5 Ω = FI = ILsin φ = (. A)(.5 m)(.8 T)sin 9 =.8 N. F I is to the left. To keep the bar moving to the right at constant speed an external force with magnitudef ext =.8 N and directed to the right must be applied to the bar. (d) The rate at which work is done by the force Fextis Fext v= (.8 N)(7.5 m/s) = 6. W. The rate at which thermal energy is developed in the circuit is I = (. A)(1.5 Ω ) = 6. W. These two rates are equal, as is required by conservation of energy. VALUAT: The force on the rod due to the induced current is directed to oppose the motion of the rod. This agrees with Lenz s law IDNTIFY: Use Faraday s law to calculate the induced emf. Ohm s law applied to the loop gives I. Use q.(7.19) to calculate the force exerted on each side of the loop. ST UP: The loop before it starts to enter the magnetic field region is sketched in Figure 9.6a. Figure 9.6a XCUT: For x < 3 L/ or x> 3 L/ the loop is completely outside the field region. Φ =, and =. Thus = and I =, so there is no force from the magnetic field and the external force F necessary to maintain constant velocity is zero.

6 9-6 Chapter 9 ST UP: The loop when it is completely inside the field region is sketched in Figure 9.6b. Figure 9.6b XCUT: For L/ < x< L/ the loop is completely inside the field region and Φ = L. d ut Φ = so = and I =. There is no force F = Il from the magnetic field and the external force F necessary to maintain constant velocity is zero. ST UP: The loop as it enters the magnetic field region is sketched in Figure 9.6c. Figure 9.6c XCUT: For 3 L/ < x< L/ the loop is entering the field region. Let x be the length of the loop that is within the field. Then Φ = Lx and = lv. The magnitude of the induced emf is = = Lv and the induced Lv current is I = =. Direction of I: Let A be directed into the plane of the figure. Then Φ is positive. The flux is positive and increasing in magnitude, so is positive. Then by Faraday s law is negative, and with our choice for direction of A a negative is counterclockwise. The current induced in the loop is counterclockwise. ST UP: The induced current and magnetic force on the loop are shown in Figure 9.6d, for the situation where the loop is entering the field. XCUT: FI = Il gives that the force F I exerted on the loop by the magnetic field is to the left and has Lv L v magnitude FI = IL= L=. Figure 9.6d The external force F needed to move the loop at constant speed is equal in magnitude and opposite in direction to F I so is to the right and has this same magnitude. ST UP: The loop as it leaves the magnetic field region is sketched in Figure 9.6e. Figure 9.6e XCUT: For L/ < x< 3 L/ the loop is leaving the field region. Let x be the length of the loop that is outside the field. Then Φ = L( L x ) and = Lv. The magnitude of the induced emf is = = Lv and the induced Lv current is I = =. Direction of I: Again let A be directed into the plane of the figure. Then Φ is positive and decreasing in magnitude, so is negative. Then by Faraday s law is positive, and with our choice for direction of A a positive is clockwise. The current induced in the loop is clockwise.

7 lectromagnetic Induction 9-7 ST UP: The induced current and magnetic force on the loop are shown in Figure 9.6f, for the situation where the loop is leaving the field. XCUT: FI = Il gives that the force F I exerted on the loop by the magnetic field is to the left and has Lv L v magnitude FI = IL = L =. Figure 9.6f The external force F needed to move the loop at constant speed is equal in magnitude and opposite in direction to F I so is to the right and has this same magnitude. (a) The graph of F versus x is given in Figure 9.6g. Figure 9.6g (b) The graph of the induced current I versus x is given in Figure 9.6h. Figure 9.6h VALUAT: When the loop is either totally outside or totally inside the magnetic field region the flux isn t changing, there is no induced current, and no external force is needed for the loop to maintain constant speed. When the loop is entering the field the external force required is directed so as to pull the loop in and when the loop is leaving the field the external force required is directed so as to pull the loop out of the field. These directions agree with Lenz s law: the force on the induced current (opposite in direction to the required external force) is directed so as to oppose the loop entering or leaving the field IDNTIFY: A bar moving in a magnetic field has an emf induced across its ends. ST UP: The induced potential is = vl sin φ. XCUT: Note that φ = 9 in all these cases because the bar moved perpendicular to the magnetic field. ut the effective length of the bar, L sin θ, is different in each case. (a) = vl sin θ = (.5 m/s)(1. T)(1.41 m) sin (37. ) =.55 V, with a at the higher potential because positive charges are pushed toward that end. (b) Same as (a) except θ = 53., giving 3.38 V, with a at the higher potential. (c) Zero, since the velocity is parallel to the magnetic field. (d) The bar must move perpendicular to its length, for which the emf is 4.3 V. For V b > V a, it must move upward and to the left (toward the second quadrant) perpendicular to its length. VALUAT: The orientation of the bar affects the potential induced across its ends IDNTIFY: Use q.(9.1) to calculate the induced electric field at a distance r from the center of the solenoid. Away from the ends of the solenoid, = μni inside and = outside. (a) ST UP: The end view of the solenoid is sketched in Figure 9.8. Let be the radius of the solenoid. Figure 9.8 Apply ú dl = to an integration path that is a circle of radius r, where r <. We need to calculate just the magnitude of so we can take absolute values.

8 9-8 Chapter 9 XCUT: dl = ( π r) ú Φ = πr = ú, π r d dl = implies ( πr) = πr = 1 d r d d di = μni, so = μn 1 di Thus = rμ n = (.5 m)(4π 1 T m/a)(9 m )(6. A/s) = V/m (b) r =.1 cm is still inside the solenoid so the expression in part (a) applies. 1 di = rμ n = (.1 m)(4π 1 T m/a)(9 m )(6. A/s) = V/m VALUAT: Inside the solenoid is proportional to r, so doubles when r doubles IDNTIFY: Apply qs.(9.9) and (9.1). ST UP: valuate the integral if q.(9.1) for a path which is a circle of radius r and concentric with the solenoid. The magnetic field of the solenoid is confined to the region inside the solenoid, so () r = for r > d d XCUT: (a) = A = π r1. 1 πr1 d r1 d (b) = = =. The direction of is shown in Figure 9.9a. πr1 πr1 1 π d d (c) All the flux is within r <, so outside the solenoid = = =. πr πr r (d) The graph is sketched in Figure 9.9b. d π d (e) At r =, = = π ( /) =. 4 d (f) At r =, = = π. d (g) At r =, = = π. VALUAT: The emf is independent of the distance from the center of the cylinder at all points outside it. ven though the magnetic field is zero for r >, the induced electric field is nonzero outside the solenoid and a nonzero emf is induced in a circular turn that has r >. Figure IDNTIFY: Use q.(9.1) to calculate the induced electric field and use this in q.(9.9) to calculate between two points. (a) ST UP: ecause of the axial symmetry and the absence of any electric charge, the field lines are concentric circles.

9 lectromagnetic Induction 9-9 (b) See Figure 9.3. is tangent to the ring. The direction of (clockwise or counterclockwise) is the direction in which current will be induced in the ring. Figure 9.3 XCUT: Use the sign convention for Faraday s law to deduce this direction. Let A be into the paper. Then Φ is positive. decreasing then means is negative, so by =, is positive and therefore clockwise. Thus is clockwise around the ring. To calculate apply ú dl = to a circular path that coincides with the ring. dl = ( π r) ú d Φ = πr ; = πr d d ( πr) = πr and = r = (.1 m)(.35 T/s) = V/m 3 3 (c) The induced emf has magnitude = ú dl = ( πr) = ( V/m)( π)(.1 m) = V. Then V A. I = = = 4. Ω (d) Points a and b are separated by a distance around the ring of = = = 3 4 ( πr) ( V/m)( π)(.1 m) V π r so 3 (e) The ends are separated by a distance around the ring of π r so = V as calculated in part (c). VALUAT: The induced emf, calculated from Faraday s law and used to calculate the induced current, is associated with the induced electric field integrated around the total circumference of the ring IDNTIFY: Apply q.(9.1) with Φ = μnia. ST UP: A= π r, where r =.11 m. In q.(9.11), r =.35 m. XCUT: d Φ d ( A) d ( nia) di di π r = = = μ = μna and = ( πr). Therefore, =. μna 6 di (8. 1 V/m) π (.35 m) = = 9.1 A/s. 1 μ(4 m ) π(.11 m) VALUAT: Outside the solenoid the induced electric field decreases with increasing distance from the axis of the solenoid IDNTIFY: A changing magnetic flux through a coil induces an emf in that coil, which means that an electric field is induced in the material of the coil. ST UP: According to Faraday s law, the induced electric field obeys the equation ú dl =. XCUT: (a) For the magnitude of the induced electric field, Faraday s law gives πr = d(πr )/ = πr d/.5 m (.5 T/s) = V/m rd = = (b) The field points toward the south pole of the magnet and is decreasing, so the induced current is counterclockwise. VALUAT: This is a very small electric field compared to most others found in laboratory equipment. ΔΦ IDNTIFY: Apply Faraday s law in the form av = N. Δt ST UP: The magnetic field of a large straight solenoid is = μni inside the solenoid and zero outside. Φ = A, where A is 8. cm, the cross-sectional area of the long straight solenoid.

10 9-1 Chapter 9 ΔΦ NA( ) NAμonI = N = =. Δt Δt Δt f i XCUT: av av 4 1 μ(1)(8. 1 m )(9 m )(.35 A) = =.4 s V. VALUAT: An emf is induced in the second winding even though the magnetic field of the solenoid is zero at the location of the second winding. The changing magnetic field induces an electric field outside the solenoid and that induced electric field produces the emf IDNTIFY: Apply q.(9.14). ST UP: XCUT: VALUAT: 11 P = F/m id = P = (3.5 1 F/m)(4. 1 V m/s ) t. id depends on the rate at which IDNTIFY: Apply q.(9.14), where P= KP. ST UP: XCUT: K = P P =.34. Φ is changing. 6 i D 1 1 A = gives t = 5. s / = 4( V m/s ) t. P = F/m. 1 id A 11 P = = =.7 1 F/m. The dielectric constant is / 4( V m/s )(6.1 1 s) ( ) VALUAT: The larger the dielectric constant, the larger is the displacement current for a given / IDNTIFY and ST UP: qs.(9.13) and (9.14) show that i C = i D and also relate i D to the rate of change of the electric field flux between the plates. Use this to calculate d / and apply the generalized form of Ampere s law (q.9.15) to calculate. id ic.8 A.8 A (a) XCUT: i = C id, so j = D 55.7 A/m A = A = πr = π(.4 m) = (b) d d j 55.7 A/m = = = = D D P so P C / N m j V/m s (c) ST UP: Apply Ampere s law ú dl = μ( ic+ id) encl (q.(8.)) to a circular path with radius r =. m. An end view of the solenoid is given in Figure y symmetry the magnetic field is tangent to the path and constant around it. Figure 9.36 XCUT: Thus dl = dl = dl = ( π r). ú ú i C = (no conduction current flows through the air space between the plates) The displacement current enclosed by the path is jd π r Thus ( πr) = μ( jdπr ) and μ j r D (4 π = = 1 T m/a)(55.7 A/m )(. m) = 7. 1 T 1 1 (d) = μ j r. Now r is the value in (c), so is 1 also: = (7. 1 T) = T D VALUAT: The definition of displacement current allows the current to be continuous at the capacitor. The magnetic field between the plates is zero on the axis (r = ) and increases as r increases. A IDNTIFY: q= CV. For a parallel-plate capacitor, C = P, where d P= KP. ic = dq/. jd = P. ST UP: = q/ P A so d/ = ic/ P A. 4 PA (4.7) P (3. 1 m )(1 V) 1 XCUT: (a) q = CV = V = = C. 3 d.5 1 m

11 lectromagnetic Induction 9-11 dq 3 (b) = ic = 6. 1 A. d ic ic 3 (c) jd = P = KP = = jc, so id = ic = 6. 1 A. KP A A VALUAT: i, D = i C so Kirchhoff s junction rule is satisfied where the wire connects to each capacitor plate IDNTIFY and ST UP: Use i C = q/ t to calculate the charge q that the current has carried to the plates in time t. The two equations preceeding q.(4.) relate q to the electric field and the potential difference between the plates. The displacement current density is defined by q.(9.16). 3 XCUT: (a) i C = A q = at t = The amount of charge brought to the plates by the charging current in time t is q= ic t = (1.8 1 A)(.5 1 s) = 9. 1 C 1 σ q 9. 1 C 5 = = = =.3 1 V/m 1 4 P P A ( C / N m )(5. 1 m ) V = d = = 5 3 (.3 1 V/m)(. 1 m) 46 V (b) = q/ P A 3 d dq / ic A 11 = = = = V/m s 1 4 PA PA ( C / N m )(5. 1 m ) Since i C is constant d/ does not vary in time. d (c) jd = P (q.(9.16)), with P replaced by P since there is vacuum between the plates.) 1 11 j D = ( C / N m )(4.7 1 V/m s) = 3.6 A/m i = j A= (3.6 A/m )(5. 1 m ) = A; i = i 4 3 D D D C VALUAT: i. C = i D The constant conduction current means the charge q on the plates and the electric field between them both increase linearly with time and i D is constant. d IDNTIFY: Ohm s law relates the current in the wire to the electric field in the wire. jd = P. Use q.(9.15) to calculate the magnetic fields. ST UP: Ohm s law says = ρj. Apply Ohm s law to a circular path of radius r. 8 ρi (. 1 Ω m)(16 A) XCUT: (a) = ρj = = =.15 V/m. 6 A.1 1 m 8 d d ρi ρ di. 1 Ω m (b) = = = 6 (4 A/s) = 38 V/m s. A A.1 1 m d 1 (c) jd = P = P (38 V/m s) = A/m (d) id = jd A= (3.4 1 A/m )(.1 1 m ) = A. q.(9.15) applied to a circular path of radius r 16 μid μ( A) 1 gives D = = =.38 1 T, and this is a negligible contribution. πr π(.6 m) μic μ (16 A) 5 C = = = T. πr π (.6 m) VALUAT: In this situation the displacement current is much less than the conduction current IDNTIFY: Apply Ampere's law to a circular path of radius r <, where is the radius of the wire. ST UP: The path is shown in Figure 9.4. ú dl = μ IC + P Figure 9.4

12 9-1 Chapter 9 XCUT: There is no displacement current, so ú dl = μic The magnetic field inside the superconducting material is zero, so ú dl =. ut then Ampere s law says that I C = ; there can be no conduction current through the path. This same argument applies to any circular path with r <, so all the current must be at the surface of the wire. VALUAT: If the current were uniformly spread over the wire s cross section, the magnetic field would be like that calculated in xample IDNTIFY: A superconducting region has zero resistance. ST UP: If the superconducting and normal regions each lie along the length of the cylinder, they provide parallel conducting paths. XCUT: Unless some of the regions with resistance completely fill a cross-sectional area of a long type-ii superconducting wire, there will still be no total resistance. The regions of no resistance provide the path for the current. VALUAT: The situation here is like two resistors in parallel, where one has zero resistance and the other is nonzero. The equivalent resistance is zero IDNTIFY: Apply q.(8.9): = +μ M. ST UP: For magnetic fields less than the critical field, there is no internal magnetic field. For fields greater than the critical field, is very nearly equal to. XCUT: (a) The external field is less than the critical field, so inside the superconductor = and ˆ (.13 T ) 5 (1.3 1 A/m) ˆ. μ i M = = = μ i Outside the superconductor, = = (.13 T)ˆ iand M =. (b) The field is greater than the critical field and = ˆ = (.6 T) i, both inside and outside the superconductor. VALUAT: elow the critical field the external field is expelled from the superconducting material IDNTIFY: Apply = +μ M. ST UP: When the magnetic flux is expelled from the material the magnetic field in the material is zero. When the material is completely normal, the magnetization is close to zero. XCUT: (a) When is just under c1 (threshold of superconducting phase), the magnetic field in the 3 c1 (55 1 T) iˆ 4 material must be zero, and M = = = ( A/m) iˆ. μ μ (b) When is just over c (threshold of normal phase), there is zero magnetization, and = ˆ c = (15. T) i. VALUAT: etween c1 and c there are filaments of normal phase material and there is magnetic field along these filaments IDNTIFY and ST UP: Use Faraday s law to calculate the magnitude of the induced emf and Lenz s law to determine its direction. Apply Ohm s law to calculate I. Use q.(5.1) to calculate the resistance of the coil. (a) XCUT: The angle φ between the normal to the coil and the direction of is 3.. = = ( Nπ r )( d/ ) and I = /. For t < and t > 1. s, d/ = = and I =. For t 1. s, d/ = (.1 T) π sinπ t = ( Nπr ) π(.1 T)sin πt = (.9475 V)sinπt ρl ρl for wire: ρ A π r L = Nc = N π r = (5)( π )(.4 m) = 15.7 m 8 3 w = = ; = Ω m, r =.15 1 m w = 358 Ω and the total resistance of the circuit is = 358 Ω+ 6 Ω= 3658 Ω I = / = (.59 ma)sin πt. The graph of I versus t is sketched in Figure 9.44a. Figure 9.44a

13 lectromagnetic Induction 9-13 (b) The coil and the magnetic field are shown in Figure 9.44b. Figure 9.44b increasing so Φ is " and increasing. Φ is so I is clockwise VALUAT: The long length of small diameter wire used to make the coil has a rather large resistance, larger than the resistance of the 6-Ω resistor connected to it in the circuit. The flux has a cosine time dependence so the rate of change of flux and the current have a sine time dependence. There is no induced current for t < or t > 1. s IDNTIFY: Apply Faraday s law and Lenz s law. V t/ C ST UP: For a discharging C circuit, it () = e, where V is the initial voltage across the capacitor. The resistance of the small loop is (5)(.6 m)(1. Ω /m) = 15. Ω. 6 XCUT: (a) The large circuit is an C circuit with a time constant of τ = C = (1 Ω )( 1 F) = μs. Thus, t / μs 1 i= (1 V)/(1 Ω ) e. At t = μs, we obtain i= (1 A)( e ) = 3.7 A. the current as a function of time is ( ) (b) Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the c+ aμib μib a small loop and referring to the solution of xercise 9.7 we obtain Φ = dr = ln 1. c + πr π c Therefore, μb a di the emf induced in the small loop at t = μs is = = ln 1 +. π c 7 (4π 1 Wb/A m )(. m) 3.7 A = ln(3.).81 mv. 6 =+ Thus, the induced current in the small π 1 s loop is i = =.81 mv = 54 μ A. 15. Ω (c) The magnetic field from the large loop is directed out of the page within the small loop.the induced current will act to oppose the decrease in flux from the large loop. Thus, the induced current flows counterclockwise. VALUAT: (d) Three of the wires in the large loop are too far away to make a significant contribution to the flux in the small loop as can be seen by comparing the distance c to the dimensions of the large loop IDNTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil. ST UP: Faraday s law says that the induced emf is = and the magnetic flux through a coil is defined as Φ = Acosφ. XCUT: In this case, Φ = A, where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure VALUAT: It is the rate at which the magnetic field is changing, not the field s magnitude, that determines the induced emf. When the field is constant, even though it may have a large value, the induced emf is zero. Figure IDNTIFY: Follow the steps specified in the problem. ST UP: Let the flux through the loop due to the current be positive. μi μ iπ a XCUT: (a) Φ = A = π a =. a

14 9-14 Chapter 9 d i a a di di (b) i μ π μ π = = = = i = i μπ a di (c) Solving i = μπ a for it ( ) t( / μπ a) yields it () = ie. t( / μπ a) (d) We want it () = i(.1) = ie, so ln(.1) = t( / μπ a) and μπ a μπ (.5 m) 5 t = ln(.1) = ln(.1) = s. (.1 Ω) VALUAT: (e) We can ignore the self-induced currents because it takes only a very short time for them to die out IDNTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil. ST UP: Faraday s law says that the induced emf is = and the magnetic flux through a coil is defined as Φ = Acos φ. XCUT: In this case, Φ = A, where A is constant. So the emf is proportional to the negative slope of the magnetic field. The result is shown in Figure VALUAT: It is the rate at which the magnetic field is changing, not the field s magnitude, that determines the induced emf. When the field is constant, even though it may have a large value, the induced emf is zero. Figure (a) IDNTIFY: (i) =. The flux is changing because the magnitude of the magnetic field of the wire decreases with distance from the wire. Find the flux through a narrow strip of area and integrate over the loop to find the total flux. ST UP: Consider a narrow strip of wih dx and a distance x from the long wire, as shown in Figure 9.49a. The magnetic field of the wire at the strip is = μ I/ π x. The flux through the strip is = b dx= ( μib/ π )( dx/ x) Figure 9.49a μib XCUT: The total flux through the loop is Φ = = π μib r + a Φ = ln π r dr μib a = = v π r ( r + a) μiabv = π r r+ a ( ) (ii) IDNTIFY: = vl for a bar of length l moving at speed v perpendicular to a magnetic field. Calculate the induced emf in each side of the loop, and combine the emfs according to their polarity. r+ a r dx x

15 lectromagnetic Induction 9-15 ST UP: The four segments of the loop are shown in Figure 9.49b. XCUT: The emf in each side μi of the loop is 1 = vb, π r μi = vb, = 4 = π rr ( + a) Figure 9.49b oth emfs 1 and are directed toward the top of the loop so oppose each other. The net emf is μivb 1 1 μiabv = 1 = = π r r+ a πr( r+ a) This expression agrees with what was obtained in (i) using Faraday s law. (b) (i) IDNTIFY and ST UP: The flux of the induced current opposes the change in flux. XCUT: is. Φ is and decreasing, so the flux Φ ind of the induced current is and the current is clockwise. (ii) IDNTIFY and ST UP: Use the right-hand rule to find the force on the positive charges in each side of the loop. The forces on positive charges in segments 1 and of the loop are shown in Figure 9.49c. Figure 9.49c XCUT: is larger at segment 1 since it is closer to the long wire, so F is larger in segment 1 and the induced current in the loop is clockwise. This agrees with the direction deduced in (i) using Lenz s law. (c) VALUAT: When v = the induced emf should be zero; the expression in part (a) gives this. When a the flux goes to zero and the emf should approach zero; the expression in part (a) gives this. When r the magnetic field through the loop goes to zero and the emf should go to zero; the expression in part (a) gives this IDNTIFY: Apply Faraday s law. ST UP: For rotation about the y-axis the situation is the same as in xamples 9.4 and 9.5 and we can apply the results from those examples. XCUT: (a) otating about the y-axis: the flux is given by = = A = = max ω (35. rad/s)(.45 T)(6. 1 m).945 V. d (b) otating about the x-axis: Φ = and =. (c) otating about the z-axis: the flux is given by Φ = Acosφ and max = = ωa = (35. rad/s)(.45 T)(6. 1 m) =.945 V. Φ = Acosφ and VALUAT: The maximum emf is the same if the loop is rotated about an edge parallel to the z-axis as it is when it is rotated about the z-axis IDNTIFY: Apply the results of xample 9.4, so max = NωA for N loops. ST UP: For the minimum ω, let the rotating loop have an area equal to the area of the uniform magnetic field, so A = (.1 m). XCUT: N = 4, = 1.5 T, A = (.1 m) and max = 1 V gives ω = / NA = ( rad/s)(1 rev/ π rad)(6 s/1 min) = 19 rpm. max VALUAT: In max = ωa, ω is in rad/s.

16 9-16 Chapter IDNTIFY: Apply the results of xample 9.4, generalized to N loops: max = NωA. v= rω. ST UP: In the expression for, max ω must be in rad/s. 3 rpm = 3.14 rad/s 9. V XCUT: (a) Solving for A we obtain A = = = 18 m. 5 ωn (3.14 rad/s)( turns)(8. 1 T) (b) Assuming a point on the coil at maximum distance from the axis of rotation we have A 18 m v= rω = ω = (3.14 rad/s) = 7.5 m s. π π VALUAT: The device is not very feasible. The coil would need a rigid frame and the effects of air resistance would be appreciable. ΔΦ IDNTIFY: Apply Faraday s law in the form av = N to calculate the average emf. Apply Lenz s law to Δt calculate the direction of the induced current. ST UP: Φ = A. The flux changes because the area of the loop changes. ΔΦ ΔA πr π(.65/ m) XCUT: (a) av = = = = (.95 T) =.16 V. Δt Δt Δt.5 s (b) Since the magnetic field is directed into the page and the magnitude of the flux through the loop is decreasing, the induced current must produce a field that goes into the page. Therefore the current flows from point a through the resistor to point b. VALUAT: Faraday s law can be used to find the direction of the induced current. Let A be into the page. Then Φ is positive and decreasing in magnitude, so / <. Therefore > and the induced current is clockwise around the loop IDNTIFY: y Lenz s law, the induced current flows to oppose the flux change that caused it. ST UP: When the switch is suddenly closed with an uncharged capacitor, the current in the outer circuit immediately increases from zero to its maximum value. As the capacitor gets charged, the current in the outer circuit gradually decreases to zero. XCUT: (a) (i) The current in the outer circuit is suddenly increasing and is in a counterclockwise direction. The magnetic field through the inner circuit is out of the paper and increasing. The magnetic flux through the inner circuit is increasing, so the induced current in the inner circuit is clockwise (a to b) to oppose the flux increase. (ii) The current in the outer circuit is still counterclockwise but is now decreasing, so the magnetic field through the inner circuit is out of the page but decreasing. The flux through the inner circuit is now decreasing, so the induced current is counterclockwise (b to a) to oppose the flux decrease. (b) The graph is sketched in Figure VALUAT: ven though the current in the outer circuit does not change direction, the current in the inner circuit does as the flux through it changes from increasing to decreasing. Figure IDNTIFY: Use Faraday s law to calculate the induced emf and Ohm s law to find the induced current. Use q.(7.19) to calculate the magnetic force F I on the induced current. Use the net force F FI in Newton s nd law to calculate the acceleration of the rod and use that to describe its motion.

17 (a) ST UP: The forces in the rod are shown in Figure 9.55a. XCUT: Lv I = Figure 9.55a = = lectromagnetic Induction 9-17 Lv Use = to find the direction of I: Let A be into the page. Then Φ >. The area of the circuit is increasing, so >. Then < and with our direction for A this means that and I are counterclockwise, as shown in the sketch. The force F I on the rod due to the induced current is given by FI = Il. This gives F I to the left with magnitude FI = IL = ( Lv / ) L = L v/. Note that F I is directed to oppose the motion of the rod, as required by Lenz s law. VALUAT: The net force on the rod is F F I, so its acceleration is a= ( F FI )/ m= ( F Lv/ )/ m. The rod starts with v = and a = F/m. As the speed v increases the acceleration a decreases. When a = the rod has reached its terminal speed v. t The graph of v versus t is sketched in Figure 9.55b. (ecall that a is the slope of the tangent to the v versus t curve.) Figure 9.55b F Lvt / F (b) XCUT: v= vt when a = so = and vt =. m L VALUAT: A large F produces a large v. t If is larger, or is smaller, the induced current is larger at a given v so F I is larger and the terminal speed is less IDNTIFY: Apply Newton s nd law to the bar. The bar will experience a magnetic force due to the induced current in the loop. Use a= dv/ to solve for v. At the terminal speed, a =. ST UP: The induced emf in the loop has a magnitude Lv. The induced emf is counterclockwise, so it opposes the voltage of the battery,. XCUT: (a) The net current in the loop is I = Lv. The acceleration of the bar is F IL sin(9 ) ( Lv) L a = = =. To find vt ( ), set dv ( Lv) L = a = and solve for v using the method m m m m of separation of variables: v dv t L v / /3.1 s (1 e L t m t ) (1 m/s)(1 e = = = ) ( Lv) m L The graph of v versus t is sketched in Figure Note that the graph of this function is similar in appearance to that of a charging capacitor. (b) Just after the switch is closed, v = and I = / =.4 A, F = IL=.88 N and a= F/ m= 3. m/s. [1 V (1.5 T)(.8 m)(. m/s)](.8 m)(1.5 T) (c) When v=. m/s, a= =.6 m/s. (.9 kg)(5. Ω) (d) Note that as the speed increases, the acceleration decreases. The speed will asymptotically approach the terminal speed 1 V 1 m/s, L = (1.5 T)(.8 m) = which makes the acceleration zero.

18 9-18 Chapter 9 VALUAT: The current in the circuit is counterclockwise and the magnetic force on the bar is to the right. The energy that appears as kinetic energy of the moving bar is supplied by the battery. Figure IDNTIFY: Apply = vl. Use F = ma applied to the satellite motion to find the speed v of the satellite. mm ST UP: The gravitational force on the satellite is Fg = G, where m is the mass of the satellite and r is the r radius of its orbit. 5 mm v 3 Gm 3 XCUT: = 8. 1 T, L=. m. G = m and r= 4 1 m + gives v = = m/s. r r r 5 3 Using this v in = vl gives = (8. 1 T)( m/s)(. m) = 1. V. VALUAT: The induced emf is large enough to be measured easily IDNTIFY: The induced emf is = vl, where L is measured in a direction that is perpendicular to both the magnetic field and the velocity of the bar. ST UP: The magnetic force pushed positive charge toward the high potential end of the bullet. 5 XCUT: (a) = Lv = (8 1 T)(.4 m)(3 m/s) = 96 μv. Since a positive charge moving to the east would be deflected upward, the top of the bullet will be at a higher potential. (b) For a bullet that travels south, v and are along the same line, there is no magnetic force and the induced emf is zero. (c) If v is horizontal, the magnetic force on positive charges in the bullet is either upward or downward, perpendicular to the line between the front and back of the bullet. There is no emf induced between the front and back of the bullet. VALUAT: Since the velocity of a bullet is always in the direction from the back to the front of the bullet, and since the magnetic force is perpendicular to the velocity, there is never an induced emf between the front and back of the bullet, no matter what the direction of the magnetic field is IDNTIFY: Find the magnetic field at a distance r from the center of the wire. Divide the rectangle into narrow strips of wih dr, find the flux through each strip and integrate to find the total flux. ST UP: xample 8.8 uses Ampere s law to show that the magnetic field inside the wire, a distance r from the axis, is r () = μ Ir π. XCUT: Consider a small strip of length W and wih dr that is a distance r from the axis of the wire, as shown μiw in Figure The flux through the strip is = () r W dr = r dr. The total flux through the rectangle is π μiw μiw Φ = = rdr. = π 4π VALUAT: Note that the result is independent of the radius of the wire. Figure 9.59

19 lectromagnetic Induction IDNTIFY: Apply Faraday s law to calculate the magnitude and direction of the induced emf. ST UP: Let A be directed out of the page in Figure 9.5 in the textbook. This means that counterclockwise emf is positive. 3 XCUT: (a) Φ = A = πr (1 3( t t) + ( t t) ). d 3 πr 6 πr t t (b) = = πr (1 3( t/ t) + ( t/ t) ) = ( 6( t/ t) + 6( t/ t) ). = =. At t t t t π (.4 m) 5. 1 s 5. 1 s t = 5. 1 s, = =.665 V. is positive so it is.1 s.1 s.1 s counterclockwise..665 V (c) I = total = r+ = r = 3 1 Ω= 1. Ω. total I 3. 1 A (d) valuating the emf at t = s and using the equations of part (b), =.676 V, and the current flows clockwise, from b to a through the resistor. t t t (e) = when =. 1 = and t = t =.1 s. t t t VALUAT: At t = t, =. At 3 t = 5. 1 s, is in the +k ˆ direction and is decreasing in magnitude. Lenz s law therefore says is counterclockwise. At t =.11 s, is in the +k ˆ direction and is increasing in magnitude. Lenz s law therefore says is clockwise. These results for the direction of agree with the results we obtained from Faraday s law (a) and (b) IDNTIFY and Set Up: The magnetic field of the wire is given by μi = and varies along the length of the π r bar. At every point along the bar has direction into the page. Divide the bar up into thin slices, as shown in Figure 9.61a. Figure 9.61a XCUT: The emf d induced in each slice is given by d = v dl. v is directed toward the wire, so μ I d = v dr = v dr. The total emf induced in the bar is π r b d+ L μiv μiv d+ Ldr μiv d+ L Vba = d = dr [ ln( r) ] a d = = d πr π d r π Iv Iv V = μ μ ba (ln( d L) ln( d)) ln(1 L/ d) π + = π + VALUAT: The minus sign means that V ba is negative, point a is at higher potential than point b. (The force F = qv on positive charge carriers in the bar is towards a, so a is at higher potential.) The potential difference increases when I or v increase, or d decreases. (c) IDNTIFY: Use Faraday s law to calculate the induced emf. ST UP: The wire and loop are sketched in Figure 9.61b. XCUT: As the loop moves to the right the magnetic flux through it doesn t change. Thus = = and I =. Figure 9.61b VALUAT: This result can also be understood as follows. The induced emf in section ab puts point a at higher potential; the induced emf in section dc puts point d at higher potential. If you travel around the loop then these two induced emf s sum to zero. There is no emf in the loop and hence no current.

20 9- Chapter IDNTIFY: = vl, where v is the component of velocity perpendicular to the field direction and perpendicular to the bar. ST UP: Wires A and C have a length of.5 m and wire D has a length of (.5 m) =.77 m. XCUT: Wire A: v is parallel to, so the induced emf is zero. Wire C: v is perpendicular to. The component of v perpendicular to the bar is v cos45. = (.35 m/s)(cos 45 )(.1 T)(.5 m) =.148 V. Wire D: v is perpendicular to. The component of v perpendicular to the bar is v cos45. = (.35 m/s)(cos 45 )(.1 T)(.77 m) =.1 V. VALUAT: The induced emf depends on the angle between v and and also on the angle between v and the bar (a) IDNTIFY: Use the expression for motional emf to calculate the emf induced in the rod. ST UP: The rotating rod is shown in Figure 9.63a. The emf induced in a thin slice is d = v dl. Figure 9.63a XCUT: Assume that is directed out of the page. Then v is directed radially outward and dl = dr, so v dl = v dr v= rω so d = ωr dr. The d for all the thin slices that make up the rod are in series so they add: = d = ωrdr = ωl = (8.8 rad/s)(.65 T)(.4 m) =.165 V L 1 1 VALUAT: increases with ω, or L. (b) No current flows so there is no I drop in potential. Thus the potential difference between the ends equals the emf of.165 V calculated in part (a). (c) ST UP: The rotating rod is shown in Figure 9.63b. Figure 9.63b 1 1 XCUT: The emf between the center of the rod and each end is = ω( L/ ) = (.165 V) =.41 V, 4 with the direction of the emf from the center of the rod toward each end. The emfs in each half of the rod thus oppose each other and there is no net emf between the ends of the rod. VALUAT: ω and are the same as in part (a) but L of each half is 1 L for the whole rod. is proportional to L, so is smaller by a factor of IDNTIFY: The power applied by the person in moving the bar equals the rate at which the electrical energy is dissipated in the resistance. ( Lv) ST UP: From xample 9.7, the power required to keep the bar moving at a constant velocity is P =. XCUT: (a) ( Lv) [(.5 T)(3. m)(. m s)] = = =.9 Ω. P 5 W (b) For a 5 W power dissipation we would require that the resistance be decreased to half the previous value. (c) Using the resistance from part (a) and a bar length of. m, ( Lv) [(.5 T)(. m)(. m s)] P = = =.11 W..9 Ω VALUAT: When the bar is moving to the right the magnetic force on the bar is to the left and an applied force directed to the right is required to maintain constant speed. When the bar is moving to the left the magnetic force on the bar is to the right and an applied force directed to the left is required to maintain constant speed.

21 lectromagnetic Induction (a) IDNTIFY: Use Faraday s law to calculate the induced emf, Ohm s law to calculate I, and q.(7.19) to calculate the force on the rod due to the induced current. ST UP: The force on the wire is shown in Figure XCUT: When the wire has speed v the induced emf is = va and the va induced current is I = / = Figure 9.65 The induced current flows upward in the wire as shown, so the force F = Il exerted by the magnetic field on the induced current is to the left. F opposes the motion of the wire, as it must by Lenz s law. The magnitude of the force is F = Ia= a v/. (b) Apply F = ma to the wire. Take +x to be toward the right and let the origin be at the location of the wire at t =, so x =. Fx = max says F = max F a v ax = = m m Use this expression to solve for v(t): dv a v dv a ax = = and = m v m v dv a t = v v m at ln( v) ln( v) = m v a t a t/ m ln = and v = ve v m Note: At t =, v= v and v when t Now solve for x(t): dx a t/ m a t/ m v= = ve so dx= ve x t a t / m = dx v e a t m ( 1 ) t m a t / m mv / x= v e e = a a Comes to rest implies v =. This happens when t. mv t gives x=. Thus this is the distance the wire travels before coming to rest. a VALUAT: The motion of the slide wire causes an induced emf and current. The magnetic force on the induced current opposes the motion of the wire and eventually brings it to rest. The force and acceleration depend on v and are constant. If the acceleration were constant, not changing from its initial value of a /, x = a v m then the stopping distance would be x = v / ax = mv/ a. The actual stopping distance is twice this IDNTIFY: Since the bar is straight and the magnetic field is uniform, integrating dε = v dl along the length of the bar gives = ( v ) L ST UP: v= (4. m/s) iˆ. L = (.5 m)(cos36.9 iˆ+ sin36.9 ˆj ). = ( v ) L= (4. m/s) iˆ ((.1 T) iˆ. T ˆj (.9 T) kˆ) L. XCUT: (a) ( ) = (( ) ˆ j ( ) k ˆ ) ( i ˆ + ˆ j ).378 V/m.94 V/m (.5 m)(cos 36.9 sin 36.9 ). = (.378 V/m)(.5 m)sin 36.9 =.567 V. (b) The higher potential end is the end to which positive charges in the rod are pushed by the magnetic force. v has a positive y-component, so the end of the rod marked + in Figure 9.66 is at higher potential.

22 9- Chapter 9 VALUAT: Since v has nonzero ĵ and ˆk components, and L has nonzero î and ĵ components, only the ˆk component of contributes to. In fact, = vl = (4. m/s)(.9 T)(.5 m)sin36.9 =.567 V. x z y Figure IDNTIFY: Use q.(9.1) to calculate the induced electric field at each point and then use F = q. ST UP: Apply ú dl = to a concentric circle of radius r, as shown in Figure 9.67a. Take A to be into the page, in the direction of. Figure 9.67a d XCUT: increasing then gives Φ >, so d l ú is negative. This means that is tangent to the circle in the counterclockwise direction, as shown in Figure 9.67b. dl = ( π r) Figure 9.67b ú d Φ = d π r d 1 d ( πr) = πr so = r point a The induced electric field and the force on q are shown in Figure 9.67c. 1 d F = q = qr F is to the left (F is in the same direction as since q is positive.) Figure 9.67c point b The induced electric field and the force on q are shown in Figure 9.67d. d 1 F = q = qr F is toward the top of the page. Figure 9.67d point c r = here, so = and F =. VALUAT: If there were a concentric conducting ring of radius r in the magnetic field region, Lenz s law tells us that the increasing magnetic field would induce a counterclockwise current in the ring. This agrees with the direction of the force we calculated for the individual positive point charges IDNTIFY: A bar moving in a magnetic field has an emf induced across its ends. The propeller acts as such a bar. ST UP: Different parts of the propeller are moving at different speeds, so we must integrate to get the total induced emf. The potential induced across an element of length dx is d = vdx, where is uniform.