ELECTROMAGNETIC INDUCTION 29.1 IDENTIFY: The changing magnetic field causes a changing magnetic flux through the loop This induces an emf in the loop which causes a current to flow in it dΦB , Φ B = BA cos φ , φ = 0° A is constant and B is changing SET UP: ε = dt EXECUTE: (a) ε = A ε dB = (0.0900 m )(0.190 T/s) = 0.0171 V dt 0.0171V = 0.0285 A 0.600 Ω EVALUATE: These are small emfs and currents by everyday standards dΦB IDENTIFY: ε = Φ B = BA cos φ Φ B is the flux through each turn of the coil dt (b) I = 29.2 29 R = SET UP: φ i = 0° φf = 90° EXECUTE: (a) Φ B ,i = BA cos0° = (6.0 × 10−5 T)(12 × 10−4 m )(1) = 7.2 × 10−8 Wb The total flux through the coil is N Φ B,i = (200)(7.2 × 10−8 Wb) = 1.44 × 10−5 Wb Φ B ,f = BA cos90° = N Φ i − N Φ f 1.44 × 10−5 Wb = = 3.6 × 10−4 V = 0.36 mV Δt 0.040 s EVALUATE: The average induced emf depends on how rapidly the flux changes IDENTIFY and SET 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 ΔΦ B (a) EXECUTE: The magnitude of the average emf induced in the coil is ε av = N Initially, Δt (b) ε av = 29.3 Φ Bf − Φ Bi NBA The average induced current = Δt Δt ε NBA NBA ⎛ NBA ⎞ The total charge that flows through the coil is Q = I Δt = ⎜ is I = av = ⎟ Δt = R R Δt R ⎝ RΔt ⎠ EVALUATE: The charge that flows is proportional to the magnetic field but does not depend on the time Δt (b) The magnetic stripe consists of a pattern of magnetic fields The pattern of charges that flow in the reader coil tells 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 IDENTIFY and SET UP: Apply the result derived in Exercise 29.3: Q = NBA/R In the present exercise the Φ Bi = BA cos φ = BA The final flux is zero, so ε av = N 29.4 flux changes from its maximum value of Φ B = BA to zero, so this equation applies R is the total resistance so here R = 60.0 Ω + 45.0 Ω = 105.0 Ω © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-1 29-2 Chapter 29 EXECUTE: Q = 29.5 29.6 NBA QR (3.56 × 10−5 C)(105.0 Ω) says B = = = 0.0973 T R NA 120(3.20 × 10−4 m ) EVALUATE: A field of this magnitude is easily produced IDENTIFY: Apply Faraday’s law G SET UP: Let + z be the positive direction for A Therefore, the initial flux is positive and the final flux is zero G ΔΦ B − (1.5 T)π (0.120 m) A EXECUTE: (a) and (b) ε = − ε is positive and is =− = + 34 V Since Δt 2.0 × 10−3 s toward us, the induced current is counterclockwise EVALUATE: The shorter the removal time, the larger the average induced emf IDENTIFY: Apply Eq (29.4) I = ε /R SET UP: d Φ B /dt = AdB/dt EXECUTE: (a) ε = Nd Φ B d d = NA ( B ) = NA ((0.012 T/s)t + (3.00 × 10−5 T/s )t ) dt dt dt ε = NA((0.012 T/s) + (1.2 × 10−4 T/s4 )t ) = 0.0302 V + (3.02 × 10−4 V/s3 )t (b) At t = 5.00 s, ε = 0.0302 V + (3.02 × 10−4 V/s3 )(5.00 s)3 = 0.0680 V 29.7 29.8 ε 0.0680 V = 1.13 × 10−4 A R 600 Ω EVALUATE: The rate of change of the flux is increasing in time, so the induced current is not constant but rather increases in time IDENTIFY: Calculate the flux through the loop and apply Faraday’s law SET UP: To find the total flux integrate dΦ B over the width of the loop The magnetic field of a long G μ I straight wire, at distance r from the wire, is B = The direction of B is given by the right-hand rule 2π r μ0i EXECUTE: (a) B = , into the page 2π r μi (b) d Φ B = BdA = Ldr 2π r b μ iL b dr μ0iL (c) Φ B = ∫ d Φ B = ∫ = ln(b/a ) a 2π a r 2π d Φ B μ0 L di (d) ε = ln(b /a ) = 2π dt dt μ (0.240 m) (e) ε = ln(0.360/0.120)(9.60 A/s) = 5.06 × 10−7 V 2π EVALUATE: The induced emf is proportional to the rate at which the current in the long straight wire is changing IDENTIFY: Apply Faraday’s law G SET UP: Let A be upward in Figure E29.8 in the textbook dΦB EXECUTE: (a) ε ind = = d ( B⊥ A) dt dt I= = ε ind = A sin 60° −1 ⎞ −1 dB d⎛ = A sin 60° ⎜ (1.4 T ) e− (0.057s )t ⎟ = (π r )(sin 60°)(1.4 T)(0.057 s −1 )e− (0.057s )t dt dt ⎝ ⎠ ε ind = π (0.75 m) (sin 60°)(1.4 T)(0.057 s −1 )e−(0.057s ε = (0.12 V) (b) ε = 10 10 (0.12 V) 10 −1 )t = (0.12 V) e − (0.057 s = (0.12 V) e −(0.057 s −1 )t −1 )t ln(1/10) = −(0.057 s −1 )t and t = 40.4 s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29-3 G G (c) B is in the direction of A so Φ B is positive B is getting weaker, so the magnitude of the flux is decreasing and d Φ B /dt < Faraday’s law therefore says ε > Since ε > 0, the induced current must flow 29.9 counterclockwise as viewed from above EVALUATE: The flux changes because the magnitude of the magnetic field is changing IDENTIFY and SET 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/dt to dc/dt , where c is the circumference of the loop (a) EXECUTE: c = 2π r and A = π r so A = c 2/4π Φ B = BA = ( B/4π )c d Φ B ⎛ B ⎞ dc =⎜ ⎟c dt ⎝ 2π ⎠ dt At t = 9.0 s, c = 1.650 m − (9.0 s)(0.120 m/s) = 0.570 m ε = ε = (0.500 T)(1/2π )(0.570 m)(0.120 m/s) = 5.44 mV (b) SET UP: The loop and magnetic field are sketched in Figure 29.9 Take into the page to be the G positive direction for A Then the magnetic flux is positive Figure 29.9 EXECUTE: The positive flux is decreasing in magnitude; d Φ B /dt is negative and ε is positive By the G right-hand rule, for A into the page, positive ε is clockwise EVALUATE: Even though the circumference is changing at a constant rate, dA/dt is not constant and ε 29.10 is not constant Flux ⊗ is decreasing so the flux of the induced current is ⊗ and this means that I is clockwise, which checks IDENTIFY: Rotating the coil changes the angle between it and the magnetic field, which changes the magnetic flux through it This change induces an emf in the coil G ΔΦ B SET UP: ε av = , Φ B = BA cos φ φ is the angle between the normal to the loop and B, so Δt φ i = 90.0° − 37.0° = 53.0° and φf = 0° 29.11 NBA cos φf − cos φ i (80)(1.10 T)(0.250 m)(0.400 m) = cos0° − cos53.0° = 58.4 V Δt 0.0600 s EVALUATE: The flux changes because the orientation of the coil relative to the magnetic field changes, even though the field remains constant IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil SET UP: The flux through a coil is Φ B = NBA cosφ and the induced emf is ε = − d Φ B /dt EXECUTE: ε av = EXECUTE: (a) ε = d Φ B /dt = d [ A( B0 + bx)]/dt = bA dx /dt = bAv 29.12 (b) clockwise (c) Same answers except the current is counterclockwise EVALUATE: Even though the coil remains within the magnetic field, the flux through it changes because the strength of the field is changing IDENTIFY: Use the results of Examples 29.3 and 29.4 ⎛ 2π rad/rev ⎞ SET UP: ε max = NBAω ε av = ε max ω = (440 rev/min) ⎜ ⎟ = 46.1 rad/s π ⎝ 60 s/min ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-4 Chapter 29 EXECUTE: (a) ε max = NBAω = (150)(0.060 T)π (0.025 m) (46.1 rad/s) = 0.814 V (b) ε av = π ε max = π (0.815 V) = 0.519 V EVALUATE: In ε max = NBAω , ω must be in rad/s 29.13 IDENTIFY: Apply the results of Example 29.3 SET UP: ε max = NBAω EXECUTE: ω = 29.14 29.15 29.16 29.17 ε max 2.40 × 10−2 V = 10.4 rad/s (120)(0.0750 T)(0.016 m) EVALUATE: We may also express ω as 99.3 rev/min or 1.66 rev/s IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil SET UP: The flux through a coil is Φ B = NBA cosφ and the induced emf is ε = − d Φ B /dt EXECUTE: The flux is constant in each case, so the induced emf is zero in all cases EVALUATE: Even 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 IDENTIFY and SET UP: The field of the induced current is directed to oppose the change in flux EXECUTE: (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 EVALUATE: 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 IDENTIFY: By Lenz’s law, the induced current flows to oppose the flux change that caused it SET UP and EXECUTE: 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 EVALUATE: 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 IDENTIFY and SET UP: Apply Lenz’s law, in the form that states that the flux of the induced current tends to oppose the change in flux EXECUTE: (a) With the switch closed the magnetic field of coil A is to the right at the location of coil B 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 B 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 B must flow through the resistor in the direction a to b (b) With the switch closed the magnetic field of coil A is to the right at the location of coil B This field is stronger at points closer to coil A so when coil B is brought closer the flux through coil B increases By Lenz’s law the field of the induced current in coil B 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 B 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 B The current in the circuit that includes coil A increases when R is decreased and the magnetic field of coil A increases when the current through the coil increases By Lenz’s law the field of the induced current in coil B 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 B must flow through the resistor in the direction b to a EVALUATE: In parts (b) and (c) the change in the circuit causes the flux through circuit B 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) NBA = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29.18 29.19 29-5 IDENTIFY: Apply Lenz’s law SET UP: The field of the induced current is directed to oppose the change in flux in the primary circuit EXECUTE: (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 B is to the right To produce magnetic field to the right, the induced current flows through R 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 B is to the right To produce magnetic field to the right the induced current flows through R 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 B is to the left To produce magnetic field to the left the induced current flows through R from left to right EVALUATE: 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 IDENTIFY and SET UP: Lenz’s law requires that the flux of the induced current opposes the change in flux EXECUTE: (a) Φ B 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 B is constant d Φ B /dt = and there is no induced current (c) Φ B is : and decreasing, so Φ ind is : and current is counterclockwise 29.20 EVALUATE: 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 IDENTIFY: The changing flux through the loop due to the changing magnetic field induces a current in the wire Energy is dissipated by the resistance of the wire due to the induced current in it dΦB dB SET UP: The magnitude of the induced emf is ε = , P = I R, I = ε /R = π r2 dt dt G EXECUTE: (a) B is out of page and Φ B is decreasing, so the field of the induced current is directed out of the page inside the loop and the induced current is counterclockwise dΦB dB (b) ε = The current due to the emf is = π r2 dt dt I= ε R = π r dB R dt = π (0.0480 m) 0.160 Ω (0.680 T/s) = 0.03076 A The rate of energy dissipation is P = I R = (0.03076 A)2 (0.160 Ω) = 1.51 × 10−4 W 29.21 EVALUATE: Both the current and resistance are small, so the power is also small IDENTIFY: The changing flux through the loop due to the changing magnetic field induces a current in the wire dΦB dB SET UP: The magnitude of the induced emf is ε = , I = ε /R = π r2 dt dt G EXECUTE: B is into the page and Φ B is increasing, so the field of the induced current is directed out of the page inside the loop and the induced current is counterclockwise dΦB dB ε = = π r2 = π (0.0250 m) (0.380 T/s3 )(3t ) = (2.238 × 10−3 V/s )t dt dt I= ε R = (5.739 × 10−3 A/s )t When B = 1.33 T, we have 1.33 T = (0.380 T/s3 )t , which gives t = 1.518 s At this t, I = (5.739 × 10−3 A/s )(1.518 s) = 0.0132 A 29.22 EVALUATE: As the field changes, the current will also change IDENTIFY: The magnetic flux through the loop is decreasing, so an emf will be induced in the loop, which will induce a current in the loop The magnetic field will exert a force on the loop due to this current SET UP: The motional ε is ε = vBL, I = ε /R, and FB = ILB © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-6 Chapter 29 EXECUTE: I = 29.23 ε = BLv B L2 3.00 m/s FB = ILB = v = (3.50 T)2 (0.0150 m) = 0.0138 N R R 0.600 Ω R G B is into the page and Φ B is decreasing, so the field of the induced current is into the page inside G G G the loop and the induced current is clockwise Using F = Il × B , we see that the force on the left-hand end of the loop to be to the left EVALUATE: The force is very small by everyday standards IDENTIFY: A conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving SET UP: The induced emf is ε = vBL sin φ , where φ is the angle between the velocity and the magnetic field EXECUTE: (a) ε = vBL sin φ = (5.00 m/s)(0.450 T)(0.300 m)(sin 90°) = 0.675 V (b) The positive charges are moved to end b, so b is at the higher potential G (c) E = V/L = (0.675 V)/(0.300 m) = 2.25 V/m The direction of E is from b to a 29.24 (d) The positive charges are pushed to b, so b has an excess of positive charge (e) (i) If the rod has no appreciable thickness, L = 0, 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 EVALUATE: The motional emf is large enough to have noticeable effects in some cases IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil SET UP: The flux through a coil is Φ B = NBA cosφ and the induced emf is ε = − d Φ B /dt EXECUTE: (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 30.0-cm side) in the field, then A = (0.400 m) x Φ B = BA = (0.400 m) x ε = d Φ B /dt = B d [(0.400 m) x]/dt = B(0.400 m) dx /dt = B (0.400 m)v ε = (1.25 T)(0.400 m)(0.0200 m/s) = 0.0100 V 29.25 EVALUATE: 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 IDENTIFY: ε = vBL SET UP: L = 5.00 × 10−2 m mph = 0.4470 m/s 29.26 ε 1.50 V = 46.2 m/s = 103 mph (0.650 T)(5.00 × 10−2 m) EVALUATE: This is a large speed and not practical It is also difficult to produce a 5.00-cm wide region of 0.650 T magnetic field IDENTIFY: ε = vBL EXECUTE: v = BL = SET UP: mph = 0.4470 m/s G = 10−4 T ⎛ 0.4470 m/s ⎞ −4 EXECUTE: (a) ε = (180 mph) ⎜ ⎟ (0.50 × 10 T)(1.5 m) = 6.0 mV This is much too small to be mph ⎝ ⎠ noticeable ⎛ 0.4470 m/s ⎞ −4 (b) ε = (565 mph) ⎜ ⎟ (0.50 × 10 T)(64.4 m) = 0.813 V This is too small to be noticeable ⎝ mph ⎠ 29.27 EVALUATE: Even 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 IDENTIFY and SET UP: ε = vBL Use Lenz’s law to determine the direction of the induced current The force Fext required to maintain constant speed is equal and opposite to the force FI that the magnetic field exerts on the rod because of the current in the rod EXECUTE: (a) ε = vBL = (7.50 m/s)(0.800 T)(0.500 m) = 3.00 V G (b) 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 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction (c) I = ε R = 29-7 G 3.00 V = 2.00 A FI = ILB sin φ = (2.00 A)(0.500 m)(0.800 T)sin 90° = 0.800 N FI is to the 1.50 Ω left To keep the bar moving to the right at constant speed an external force with magnitude Fext = 0.800 N and directed to the right must be applied to the bar (d) The rate at which work is done by the force Fext is Fext v = (0.800 N)(7.50 m/s) = 6.00 W The rate at which thermal energy is developed in the circuit is I R = (2.00 A)2 (1.50 Ω) = 6.00 W These two rates are 29.28 equal, as is required by conservation of energy EVALUATE: 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 IDENTIFY: Use the results of Example 29.5 Use the three approaches specified in the problem for determining the direction of the induced current I = ε /R G SET UP: Let A be directed into the figure, so a clockwise emf is positive EXECUTE: (a) ε = vBl = (5.0 m/s)(0.750 T)(1.50 m) = 5.6 V (b) (i) Let q be a positive charge in the moving bar, as shown in Figure 29.28a The magnetic force on this G G G charge is F = qv × B, which points upward This force pushes the current in a counterclockwise direction through the circuit (ii) Φ B is positive and is increasing in magnitude, so d Φ B /dt > Then by Faraday’s law ε < 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 29.28b ε 5.6 V = 0.22 A (c) ε = RI I = = R 25 Ω EVALUATE: All three methods agree on the direction of the induced current Figure 29.28 29.29 IDENTIFY: The motion of the bar due to the applied force causes a motional emf to be induced across the ends of the bar, which induces a current through the bar The magnetic field exerts a force on the bar due to this current ε BvL SET UP: The applied force is to the left and equal to Fapplied = FB = ILB ε = BvL and I = = R R G EXECUTE: (a) B out of page and Φ B decreasing, so the field of the induced current is out of the page inside the loop and the induced current is counterclockwise (b) Combining Fapplied = FB = ILB and ε = BvL, we have I = ε R = BvL vB L2 Fapplied = The rate at R R (vBL) [(5.90 m/s)(0.650 T)(0.360 m)]2 = = 0.0424 W which this force does work is Papplied = Fapplied v = R 45.0 Ω EVALUATE: The power is small because the magnetic force is usually small compared to everyday forces © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-8 Chapter 29 29.30 IDENTIFY: The motion of the bar due to the applied force causes a motional emf to be induced across the ends of the bar, which induces a current through the bar and through the resistor This current dissipates energy in the resistor SET UP: PR = I R, ε = BvL = IR G EXECUTE: (a) B is out of the page and Φ B is increasing, so the field of the induced current is into the page inside the loop and the induced current is clockwise P 0.840 W emf BvL (b) PR = I R so I = R = = = 0.1366 A I = R R 45.0 Ω R IR (0.1366 A)(45.0 Ω) = = 26.3 m/s BL (0.650 T)(0.360 m) EXECUTE: This speed is around 60 mph, so it would not be very practical to generate energy this way IDENTIFY: The motion of the bar causes an emf to be induced across its ends, which induces a current in the circuit SET UP: ε = BvL, I = ε /R G G BvL EXECUTE: FB on the bar is to the left so v is to the right Using ε = BvL and I = ε /R, we have I = R IR (1.75 A)(6.00 Ω) = = 35.0 m/s v= BL (1.20 T)(0.250 m) EVALUATE: This speed is greater than 60 mph! IDENTIFY: A motional emf is induced across the blood vessel SET UP and SOLVE: (a) Each slab of flowing blood has maximum width d and is moving perpendicular to the field with speed v ε = vBL becomes ε = vBd v= 29.31 29.32 (b) B = ε vd = 1.0 × 10−3 V (0.15 m/s)(5.0 × 10−3 m) = 1.3 T (c) The blood vessel has cross-sectional area A = π d /4 The volume of blood that flows past a cross section of the vessel in time t is π (d /4)vt The volume flow rate is volume/time = R = π d 2v/4 v = so R = 29.33 29.34 ε Bd π d ⎛ ε ⎞ πε d ⎜ ⎟= ⎝ Bd ⎠ B EVALUATE: A very strong magnetic field (1.3 T) is required to produce a small potential difference of only mV IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends SET UP: The induced potential is ε = vBL sin φ EXECUTE: Note that φ = 90° in all these cases because the bar moved perpendicular to the magnetic field But the effective length of the bar, L sin θ, is different in each case (a) ε = vBL sin θ = (2.50 m/s)(1.20 T)(1.41 m) sin (37.0°) = 2.55 V, with a at the higher potential because positive charges are pushed toward that end (b) Same as (a) except θ = 53.0°, 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 4.23 V For Vb > Va, it must move upward and to the left (toward the second quadrant) perpendicular to its length EVALUATE: The orientation of the bar affects the potential induced across its ends IDENTIFY: While the circuit is entering and leaving the region of the magnetic field, the flux through it will be changing This change will induce an emf in the circuit SET UP: When the loop is entering or leaving the region of magnetic field the flux through it is changing and there is an induced emf The magnitude of this induced emf is ε = BLv The length L is 0.750 m When the loop is totally within the field the flux through the loop is not changing so there is no induced emf The induced current has magnitude I = ε and direction given by Lenz’s law R © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction EXECUTE: (a) I = 29-9 ε BLv (1.25 V)(0.750 m)(3.0 m/s) = = = 0.225 A The magnetic field through the loop 12.5 Ω R R is directed out of the page and is increasing, so the magnetic field of the induced current is into the page inside the loop and the induced current is clockwise (b) The flux is not changing so ε and I are zero (c) I = ε = 0.225 A The magnetic field through the loop is directed out of the page and is decreasing, so R the magnetic field of the induced current is out of the page inside the loop and the induced current is counterclockwise (d) Let clockwise currents be positive At t = the loop is entering the field It is totally in the field at time ta and beginning to move out of the field at time tb The graph of the induced current as a function of time is sketched in Figure 29.34 Figure 29.34 29.35 EVALUATE: Even though the circuit is moving throughout all parts of this problem, an emf is induced in it only when the flux through it is changing While the coil is entirely within the field, the flux is constant, so no emf is induced IDENTIFY: Apply Eqs (29.9) and (29.10) SET UP: Evaluate the integral if Eq (29.10) 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 B (r ) = for r > R EXECUTE: (a) (b) E = dΦB dB dB =A = π r12 dt dt dt G d Φ B π r12 dB r1 dB = = The direction of E is shown in Figure 29.35a 2π r1 dt 2π r1 dt dt (c) All the flux is within r < R, so outside the solenoid E = d Φ B π R dB R dB = = 2π r2 dt 2π r2 dt 2r2 dt (d) The graph is sketched in Figure 29.35b dΦB dB π R dB (e) At r = R/2, ε = = π ( R /2) = dt dt dt dΦB dB (f) At r = R, ε = = π R2 dt dt dΦB dB = π R2 (g) At r = R, ε = dt dt EVALUATE: The emf is independent of the distance from the center of the cylinder at all points outside it Even though the magnetic field is zero for r > R, the induced electric field is nonzero outside the solenoid and a nonzero emf is induced in a circular turn that has r > R © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-10 Chapter 29 Figure 29.35 29.36 IDENTIFY: Use Eq (29.10) to calculate the induced electric field E at a distance r from the center of the solenoid Away from the ends of the solenoid, B = μ0nI inside and B = outside (a) SET UP: The end view of the solenoid is sketched in Figure 29.36 Let R be the radius of the solenoid Figure 29.36 G G dΦB to an integration path that is a circle of radius r, where r < R We need to dt calculate just the magnitude of E so we can take absolute values G G EXECUTE: v∫ E ⋅ dl = E (2π r ) Apply v∫ E ⋅ dl = − Φ B = Bπ r , − G G v∫ E ⋅ dl E = 12 r = − dΦB dB = π r2 dt dt dΦB dB implies E (2π r ) = π r dt dt dB dt dB dI = μ0 n dt dt dI Thus E = 12 r μ0n = 12 (0.00500 m)(4π × 10−7 T ⋅ m/A)(900 m −1 )(60.0 A/s) = 1.70 × 10−4 V/m dt (b) r = 0.0100 cm is still inside the solenoid so the expression in part (a) applies B = μ0nI , so dI = (0.0100 m)(4π × 10−7 T ⋅ m/A)(900 m −1 )(60.0 A/s) = 3.39 × 10−4 V/m dt EVALUATE: Inside the solenoid E is proportional to r, so E doubles when r doubles IDENTIFY: Apply Eq (29.11) with Φ B = μ0niA E = 12 r μ0n 29.37 SET UP: EXECUTE: A = π r , where r = 0.0110 m In Eq (29.11), r = 0.0350 m ε = dΦB d d di di E 2π r = ( BA) = ( μ0niA) = μ0 nA and ε = E (2π r ) Therefore, = dt μ0 nA dt dt dt dt di (8.00 × 10−6 V/m)2π (0.0350 m) = = 9.21 A/s dt μ0 (400 m −1 )π ( 0.0110 m ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-16 Chapter 29 d ΦB μ0b di di ε = ln(1 + a/c) = − e−t/RC and dt dt dt 2π R C d ΦB N μ0b ε Nμ0b ε induced = N = ln(1 + a/c) e− t/RC = ln(1 + a/c) i The resistance of the small loop 2π 2π dt RC R C is (25)(0.600 m)(1.0 Ω/m) = 15.0 Ω EXECUTE: ε induced = (25)(2.00 × 10−7 T ⋅ m/A)(0.200 m)ln(1 + 10.0/5.0) (10 Ω)(20 × 10−6 F) (5.00 A) ε induced 0.02747 V = = 1.83 × 10−3 A = 1.83 mA The 15.0 Ω R current in the large loop is counterclockwise The magnetic field through the small loop is into the page and the flux is increasing, so the flux due to the induced current in the small loop is out of the page and the induced current in the small loop is counterclockwise EVALUATE: The answer is actually independent of N because the emf induced in the small coil is proportional to N and the resistance of that coil is also proportional to N Since I = ε /R, the N will cancel out IDENTIFY: The changing current in the solenoid will cause a changing magnetic field (and hence changing flux) through the secondary winding, which will induce an emf in the secondary coil dΦB SET UP: The magnetic field of the solenoid is B = μ0ni, and the induced emf is ε = N dt ε induced = 0.02747 V The induced current is 29.51 EXECUTE: B = μ0ni = (4π × 10−7 T ⋅ m/A)(90.0 × 102 m −1 )(0.160 A/s2 )t = (1.810 × 10−3 T/s )t The total flux through secondary winding is (5.0) B (2.00 × 10−4 m ) = (1.810 × 10−6 Wb/s )t ε =N dΦB = (3.619 × 10−6 V/s)t i = 3.20 A says 3.20 A = (0.160 A/s )t and t = 4.472 s This gives dt ε = (3.619 × 106 V/s)(4.472 s) = 1.62 × 10−5 V EVALUATE: This a very small voltage, about 16 μ V 29.52 IDENTIFY: A changing magnetic field causes a changing flux through a coil and therefore induces an emf in the coil dΦB and the magnetic flux through a coil is SET UP: Faraday’s law says that the induced emf is ε = − dt defined as Φ B = BA cos φ EXECUTE: In this case, Φ B = BA, where A is constant So the emf is proportional to the negative slope of the magnetic field The result is shown in Figure 29.52 EVALUATE: 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 29.52 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29.53 29-17 dΦB The flux is changing because the magnitude of the magnetic field of the dt 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 SET UP: (a) IDENTIFY: (i) ε = Consider a narrow strip of width dx and a distance x from the long wire, as shown in Figure 29.53a The magnetic field of the wire at the strip is B = μ0 I/2π x The flux through the strip is d Φ B = Bb dx = ( μ0 Ib /2π )(dx /x) Figure 29.53a ⎛ μ Ib ⎞ r + a dx EXECUTE: The total flux through the loop is ΦB = ∫ d Φ B = ⎜ ⎟ ∫ x ⎝ 2π ⎠ r ⎛ μ Ib ⎞ ⎛ r + a ⎞ ΦB = ⎜ ⎟ ln ⎜ ⎟ ⎝ 2π ⎠ ⎝ r ⎠ d ΦB d Φ B dr μ0 Ib ⎛ a ⎞ = = ⎜− ⎟v dt dr dt 2π ⎝ r ( r + a ) ⎠ ε = μ0 Iabv 2π r ( r + a ) (ii) IDENTIFY: ε = Bvl for a bar of length l moving at speed v perpendicular to a magnetic field B Calculate the induced emf in each side of the loop, and combine the emfs according to their polarity SET UP: The four segments of the loop are shown in Figure 29.53b EXECUTE: The emf in each side ⎛μ I⎞ of the loop is ε1 = ⎜ ⎟ vb, ⎝ 2π r ⎠ μ0 I ⎞ ⎟ vb, ε = ε = π ( r + a) ⎠ ⎝ ⎛ ε2 = ⎜ Figure 29.53b Both emfs ε1 and ε are directed toward the top of the loop so oppose each other The net emf is ε = ε1 − ε = μ0 Ivb ⎛ 1 ⎞ μ0 Iabv ⎜ − ⎟= 2π ⎝ r r + a ⎠ 2π r (r + a ) This expression agrees with what was obtained in (i) using Faraday’s law (b) (i) IDENTIFY and SET UP: The flux of the induced current opposes the change in flux © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-18 Chapter 29 G EXECUTE: B is ⊗ ΦB is decreasing, so the flux Φind of the induced current is ⊗ and the current is clockwise (ii) IDENTIFY and SET 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 and of the loop are shown in Figure 29.53c Figure 29.53c EXECUTE: B is larger at segment since it is closer to the long wire, so FB is larger in segment and the 29.54 induced current in the loop is clockwise This agrees with the direction deduced in (i) using Lenz’s law (c) EVALUATE: 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 IDENTIFY: Apply Faraday’s law SET UP: For rotation about the y-axis the situation is the same as in Examples 29.3 and 29.4 and we can apply the results from those examples EXECUTE: (a) Rotating about the y-axis: the flux is given by Φ B = BA cosφ and ε max = ω BA = (35.0 rad/s)(0.450 T)(6.00 × 10−2 m) = 0.945 V dΦB = and ε = dt (c) Rotating about the z-axis: the flux is given by Φ B = BA cosφ and (b) Rotating about the x-axis: ε max = ω BA = (35.0 rad/s)(0.450 T)(6.00 × 10−2 m) = 0.945 V 29.55 EVALUATE: 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 IDENTIFY: Apply the results of Example 29.3, so ε max = Nω BA for N loops SET UP: For the minimum ω , let the rotating loop have an area equal to the area of the uniform magnetic field, so A = (0.100 m)2 N = 400, B = 1.5 T, A = (0.100 m)2 and ε max = 120 V gives EXECUTE: ω = ε max /NBA = (20 rad/s)(1 rev/2π rad)(60 s/1 min) = 190 rpm EVALUATE: In ε max = ω BA, ω is in rad/s 29.56 IDENTIFY: Apply the results of Example 29.3, generalized to N loops: ε max = N ω BA v = rω SET UP: In the expression for ε max , ω must be in rad/s 30 rpm = 3.14 rad/s ε max 9.0 V = = 18 m ω NB (3.14 rad/s)(2000 turns)(8.0 × 10−5 T) (b) Assuming a point on the coil at maximum distance from the axis of rotation we have EXECUTE: (a) Solving for A we obtain A = v = rω = A π ω= 18 m π (3.14 rad/s) = 7.5 m/s EVALUATE: The device is not very feasible The coil would need a rigid frame and the effects of air resistance would be appreciable © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29.57 IDENTIFY: Apply Faraday’s law in the form ε av = − N 29-19 ΔΦ B to calculate the average emf Apply Lenz’s Δt law to calculate the direction of the induced current SET UP: Φ B = BA The flux changes because the area of the loop changes ΔΦ B ΔA π r2 π (0.0650/2 m) =B =B = (1.35 T) = 0.0179 V = 17.9 mV Δt Δt Δt 0.250 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 G EVALUATE: Faraday’s law can be used to find the direction of the induced current Let A be into the page Then ΦB is positive and decreasing in magnitude, so d ΦB /dt < Therefore ε > and the induced EXECUTE: (a) ε av = 29.58 current is clockwise around the loop IDENTIFY: The movement of the rod causes an emf to be induced across its ends, which causes a current to flow through the circuit The magnetic field exerts a force on this current SET UP: The magnetic force is Fmag = ILB, the induced emf is ε = vBL ∑ F = ma applies to the rod, and a = dv/dt vBL vB L2 vB L2 dv F− = ma F − =m R R dt R 2 v F t dv′ Ft FR ⎛ vB L ⎞ , which gives = − 2 ln ⎜1 − Integrating to find the time gives ∫ dt ′ = ∫ ⎟ 2 0 m m FR ⎟⎠ B L ⎜⎝ v′ B L 1− FR Solving for t and putting in the numbers gives ⎛ ⎞ Rm ⎛ vB L2 ⎞ 25.0 m/s t = − 2 ln ⎜1 − ⎟⎟ = −(0.120 kg)(888.9 s/kg)ln ⎜1 − ⎟ = 1.59 s ⎜ (1.90 N)(888.9 s/kg) FR B L ⎝ ⎝ ⎠ ⎠ EVALUATE: We cannot use the constant-acceleration kinematics formulas because as the speed v of the rod changes, the magnetic force on it also changes Therefore the acceleration of the rod is not constant IDENTIFY: Use Faraday’s law to calculate the induced emf and Ohm’s law to find the induced current Use Eq (27.19) to calculate the magnetic force FI on the induced current Use the net force F − FI in EXECUTE: The net force on the rod is F − iLB = ma i = 29.59 Newton’s second law to calculate the acceleration of the rod and use that to describe its motion (a) SET UP: The forces in the rod are shown in Figure 29.59a EXECUTE: I= ε = dΦB = BLv dt BLv R Figure 29.59a G d ΦB to find the direction of I: Let A be into the page Then ΦB > The area of the circuit is dt G d ΦB > Then ε < and with our direction for A this means that ε and I are increasing, so dt counterclockwise, as shown in the sketch The force FI on the rod due to the induced current is given by G G G G G FI = Il × B This gives FI to the left with magnitude FI = ILB = ( BLv/R ) LB = B L2v/R Note that FI is Use ε = − directed to oppose the motion of the rod, as required by Lenz’s law © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-20 Chapter 29 EVALUATE: The net force on the rod is F − FI , so its acceleration is a = ( F − FI )/m = ( F − B L2v/R )/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 vt The graph of v versus t is sketched in Figure 29.59b (Recall that a is the slope of the tangent to the v versus t curve.) Figure 29.59b F − B L2vt /R RF = and vt = 2 m B L EVALUATE: A large F produces a large vt If B is larger, or R is smaller, the induced current is larger at a (b) EXECUTE: v = vt when a = so given v so FI is larger and the terminal speed is less 29.60 IDENTIFY: Apply Newton’s second law to the bar The bar will experience a magnetic force due to the induced current in the loop Use a = dv/dt to solve for v At the terminal speed, a = SET UP: The induced emf in the loop has a magnitude BLv The induced emf is counterclockwise, so it opposes the voltage of the battery, ε ε − BLv EXECUTE: (a) The net current in the loop is I = The acceleration of the bar is R F ILB sin(90°) (ε − BLv ) LB (ε − BLv) LB and solve for v using the To find v(t ), set dv = a = a= = = dt mR m m mR method of separation of variables: v t LB dv ε − B L2t/mR ) = (22 m/s)(1 − e−t/15 s ) The graph of v versus t is sketched ∫ (ε − BLv) =∫ mR dt → v = BL (1 − e in Figure 29.60 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 = ε /R = 2.4 A, F = ILB = 1.296 N, and a = F/m = 1.4 m/s [12 V − (1.5 T)(0.36 m)(2.0 m/s)](0.36 m)(1.5 T) = 1.3 m/s (0.90 kg)(5.0 Ω) (d) Note that as the speed increases, the acceleration decreases The speed will asymptotically approach the 12 V terminal speed ε = = 22 m/s, which makes the acceleration zero BL (1.5 T)(0.36 m) (c) When v = 2.0 m/s, a = EVALUATE: 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 29.60 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29.61 29-21 G G IDENTIFY: Apply ε = BvL Use ∑ F = ma applied to the satellite motion to find the speed v of the satellite mm SET UP: The gravitational force on the satellite is Fg = G E , where m is the mass of the satellite r and r is the radius of its orbit mm v2 GmE EXECUTE: B = 8.0 × 10−5 T, L = 2.0 m G E = m and r = 400 × 103 m + RE gives v = = r r r 7.665 × 103 m/s Using this v in ε = vBL gives ε = (8.0 × 10−5 T)(7.665 × 103 m/s)(2.0 m) = 1.2 V 29.62 29.63 EVALUATE: The induced emf is large enough to be measured easily IDENTIFY: The induced emf is ε = BvL, where L is measured in a direction that is perpendicular to both the magnetic field and the velocity of the bar SET UP: The magnetic force pushed positive charge toward the high potential end of the bullet EXECUTE: (a) ε = BLv = (8 × 10−5 T)(0.004 m)(300 m/s) = 96 μ V Since a positive charge moving to the east would be deflected upward, the top of the bullet will be at a higher potential G G (b) For a bullet that travels south, v and B are along the same line, there is no magnetic force and the induced emf is zero G (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 EVALUATE: 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 IDENTIFY: Find the magnetic field at a distance r from the center of the wire Divide the rectangle into narrow strips of width dr, find the flux through each strip and integrate to find the total flux SET UP: Example 28.8 uses Ampere’s law to show that the magnetic field inside the wire, a distance r from the axis, is B (r ) = μ0 Ir/2π R EXECUTE: Consider a small strip of length W and width dr that is a distance r from the axis of the wire, as μ IW shown in Figure 29.63 The flux through the strip is d Φ B = B (r )W dr = r dr The total flux through 2π R R μ IW ⎛ μ IW ⎞ the rectangle is Φ B = ∫ d Φ B = ⎜ ⎟ ∫ r dr = 4π ⎝ 2π R ⎠ EVALUATE: Note that the result is independent of the radius R of the wire Figure 29.63 29.64 IDENTIFY: Apply Faraday’s law to calculate the magnitude and direction of the induced emf G SET UP: Let A be directed out of the page in Figure P29.64 in the textbook This means that counterclockwise emf is positive EXECUTE: (a) ΦB = BA = B0π r02 (1 − 3(t /t0 ) + 2(t /t0 )3 ) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-22 Chapter 29 (b) ε = − d ΦB d B π r2 = − B0π r02 (1 − 3(t /t0 )2 + 2(t /t0 )3 ) = − 0 ( −6(t /t0 ) + 6(t /t0 ) ) dt dt t0 B0π r02 ⎛⎜ ⎛ t ⎞ ⎛ t ⎞ ⎞⎟ −3 ⎜ ⎟ − ⎜ ⎟ At t = 5.0 × 10 s, t0 ⎜ ⎝ t0 ⎠ ⎝ t0 ⎠ ⎟ ⎝ ⎠ ⎛ ⎞ B π (0.0420 m)2 ⎜ ⎛ 5.0 × 10−3 s ⎞ ⎛ 5.0 × 10−3 s ⎞ ⎟ ε =− ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ = 0.0665 V ε is positive so it is ⎜ 0.010 s ⎟ 0.010 s ⎠ ⎝ 0.010 s ⎠ ⎠ ⎝⎝ counterclockwise ε ε 0.0665 V (c) I = ⇒ Rtotal = r + R = ⇒ r = − 12 Ω = 10.2 Ω Rtotal I 3.0 × 10−3 A ε =− (d) Evaluating the emf at t = 1.21 × 10−2 s and using the equations of part (b), ε = −0.0676 V, and the current flows clockwise, from b to a through the resistor ⎛ ⎛ t ⎞2 ⎛ t ⎞ ⎞ t (e) ε = when = ⎜ ⎜ ⎟ − ⎜ ⎟ ⎟ = and t = t0 = 0.010 s ⎜ ⎝ t ⎠ ⎝ t0 ⎠ ⎟ t0 ⎝ ⎠ G EVALUATE: At t = t0 , B = At t = 5.00 × 10−3 s, B is in the + kˆ direction and is decreasing in G magnitude Lenz’s law therefore says ε is counterclockwise At t = 0.0121 s, B is in the + kˆ direction and is increasing in magnitude Lenz’s law therefore says ε is clockwise These results for the direction of ε agree with the results we obtained from Faraday’s law 29.65 (a) and (b) IDENTIFY and SET UP: The magnetic field of the wire is μ I given by B = and varies along 2π r the length of the bar At every point along G the bar B has direction into the page Divide the bar up into thin slices, as shown in Figure 29.65a Figure 29.65a G G G G G EXECUTE: The emf d ε induced in each slice is given by d ε = v × B ⋅ dl v × B is directed toward the ⎛μ I⎞ wire, so d ε = −vB dr = −v ⎜ ⎟ dr The total emf induced in the bar is ⎝ 2π r ⎠ μ0 Iv ⎞ μ0 Iv d + L dr μ Iv d +L = − [ ln(r ) ]d ⎜ ⎟ dr = − ∫ d 2π 2π r ⎝ 2π r ⎠ b d +L⎛ a d Vba = ∫ d ε = − ∫ μ0 Iv μ Iv (ln( d + L) − ln( d )) = − ln(1 + L/d ) 2π 2π EVALUATE: The minus sign means that Vba is negative, point a is at higher potential than point b G G G (The force F = qv × B 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) IDENTIFY: Use Faraday’s law to calculate the induced emf SET UP: The wire and loop are sketched in Figure 29.65b Vba = − © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29-23 EXECUTE: As the loop moves to the right the magnetic flux through it doesn’t change dΦ Thus ε = − B = and I = dt Figure 29.65b 29.66 29.67 EVALUATE: 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 IDENTIFY: ε = vBL, where v is the component of velocity perpendicular to the field direction and perpendicular to the bar SET UP: Wires A and C have a length of 0.500 m and wire D has a length of 2(0.500 m) = 0.707 m G G EXECUTE: Wire A: v is parallel to B, so the induced emf is zero G G G Wire C: v is perpendicular to B The component of v perpendicular to the bar is v cos 45° ε = (0.350 m/s)(cos 45°)(0.120 T)(0.500 m) = 0.0148 V G G G Wire D: v is perpendicular to B The component of v perpendicular to the bar is v cos 45° ε = (0.350 m/s)(cos 45°)(0.120 T)(0.707 m) = 0.0210 V G G EVALUATE: The induced emf depends on the angle between v and B and also on the angle between G v and the bar (a) IDENTIFY: Use the expression for motional emf to calculate the emf induced in the rod SET UP: The rotating rod is shown in Figure 29.67a The emf induced in a thin G G G slice is d ε = v × B ⋅ dl Figure 29.67a G G G EXECUTE: Assume that B is directed out of the page Then v × B is directed radially outward and G G G dl = dr , so v × B ⋅ dl = vB dr v = rω so d ε = ω Br dr The d ε for all the thin slices that make up the rod are in series so they add: L ε = ∫ d ε = ∫ ω Br dr = 12 ω BL2 = 12 (8.80 rad/s)(0.650 T)(0.240 m) = 0.165 V EVALUATE: ε increases with ω , B or L2 (b) No current flows so there is no IR drop in potential Thus the potential difference between the ends equals the emf of 0.165 V calculated in part (a) (c) SET UP: The rotating rod is shown in Figure 29.67b Figure 29.67b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-24 Chapter 29 EXECUTE: The emf between the center of the rod and each end is ε = 12 ω B( L/2)2 = 14 (0.165 V) = 0.0412 V, 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 EVALUATE: ω and B are the same as in part (a) but L of each half is 12 L for the whole rod ε is proportional to L2 , so is smaller by a factor of 29.68 IDENTIFY: The power applied by the person in moving the bar equals the rate at which the electrical energy is dissipated in the resistance SET UP: From Example 29.6, the power required to keep the bar moving at a constant velocity is P= ( BLv ) R EXECUTE: (a) R = ( BLv) [(0.25 T)(3.0 m)(2.0 m/s)]2 = = 0.090 Ω P 25 W (b) For a 50-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 0.20 m, ( BLv) [(0.25 T)(0.20 m)(2.0 m/s)]2 = = 0.11 W R 0.090 Ω EVALUATE: 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 (a) IDENTIFY: Use Faraday’s law to calculate the induced emf, Ohm’s law to calculate I, and Eq (27.19) to calculate the force on the rod due to the induced current SET UP: The force on the wire is shown in Figure 29.69 P= 29.69 EXECUTE: When the wire has speed v the induced emf is ε = BvL and the BvL induced current is I = ε /R = R Figure 29.69 G G G The induced current flows upward in the wire as shown, so the force F = Il × B exerted by the magnetic G 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 = ILB = B L2v /R G G (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 = 0, so x0 = ∑ Fx = ma x says − F = ma x F B L2v =− m mR Use this expression to solve for v (t ) : ax = − ax = dv B L2v dv B L2 =− =− dt and dt mR v mR dv′ B L2 t = − dt ′ ∫v0 v′ mR ∫ v © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction ln(v ) − ln(v0 ) = − 29-25 B L2t mR 2 ⎛ v ⎞ B L2t ln ⎜ ⎟ = − and v = v0e− B L t /mR v mR ⎝ 0⎠ Note: At t = 0, v = v0 and v → when t → ∞ Now solve for x(t ): v= x 2 2 dx = v0e − B L t / mR so dx = v0e − B L t /mR dt dt t ∫ dx′ =∫ v0e − B L2t /mR dt ′ t 2 2 mRv ⎛ mR ⎞ x = v0 ⎜ − 2 ⎟ ⎡⎢ e− B L t ′ /mR ⎤⎥ = 20 (1 − e− B L t /mR ) ⎦0 B L ⎝ B L ⎠⎣ Comes to rest implies v = This happens when t → ∞ t → ∞ gives x = mRv0 B L2 Thus this is the distance the wire travels before coming to rest EVALUATE: 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 = − B L2v0 /mR, then the stopping distance would be x = −v02 /2a x = mRv0 /2 B L2 The actual stopping distance is twice this 29.70 G G G IDENTIFY: Since the bar is straight and the magnetic field is uniform, integrating d ε = v × B ⋅ dl along G G G the length of the bar gives ε = (v × B ) ⋅ L G G SET UP: v = (6.80 m/s)iˆ L = (0.250 m)(cos36.9°iˆ + sin 36.9° ˆj ) G G G G EXECUTE: (a) ε = (v × B ) ⋅ L = (6.80 m/s) iˆ × ((0.120 T) iˆ − (0.220 T) ˆj − (0.0900 T) kˆ ) ⋅ L ε = ((0.612 V/m) ˆj − (1.496 V/m)kˆ ) ⋅ ((0.250 m)(cos36.9°iˆ + sin 36.9° ˆj )) ε = (0.612 V/m)(0.250 m)sin 36.9° = 0.0919 V = 91.9 mV (b) The higher potential end is the end to which positive charges in the rod are pushed by the magnetic G G force v × B has a positive y-component, so the end of the rod marked + in Figure 29.70 is at higher potential G G G EVALUATE: Since v × B has nonzero ˆj and kˆ components, and L has nonzero iˆ and ˆj components, only G the kˆ component of B contributes to ε In fact, | ε |=| vx Bz Ly | (6.80 m/s)(0.0900 T)(0.250 m)sin 36.9° = 0.0919 V = 91.9 mV Figure 29.70 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-26 29.71 Chapter 29 G G IDENTIFY: Use Eq (29.10) to calculate the induced electric field at each point and then use F = qE SET UP: G G dΦB to a dt concentric circle of radius r, as shown G in Figure 29.71a Take A to be into the G page, in the direction of B Apply v∫ E ⋅ dl = − Figure 29.71a G G dΦ B > 0, so v∫ E ⋅ dl is negative This means that E is tangent to the dt circle in the counterclockwise direction, as shown in Figure 29.71b EXECUTE: B increasing then gives G G v∫ E ⋅ dl = − E (2π r ) dΦB dB = π r2 dt dt Figure 29.71b dB dB so E = 12 r dt dt point a The induced electric field and the force on q are shown in Figure 29.71c − E (2π r ) = −π r dB F = qE = 12 qr dt G G F is to the left ( F is in the same G direction as E since q is positive) Figure 29.71c point b The induced electric field and the force on q are shown in Figure 29.71d dB F = qE = 12 qr dt G F is toward the top of the page Figure 29.71d point c r = here, so E = and F = 29.72 EVALUATE: 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 IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends The propeller acts as such a bar SET 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 ε = vBdx, where B is uniform © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29-27 EXECUTE: (a) Call x the distance from the center to an element of length dx, and L the length of the propeller The speed of dx is xω, giving d ε = vBdx = xω Bdx ε = ∫ L/2 xω Bdx = ω BL2 /8 (b) The potential difference is zero since the potential is the same at both ends of the propeller ⎛ 220 rev ⎞ (2.0 m) −4 (c) ε = (2π rad/rev) ⎜ = 5.8 × 10−4 V = 0.58 mV ⎟ (0.50 × 10 T) 60 s ⎝ ⎠ EVALUATE: A potential difference of about 29.73 mV is not large enough to be concerned about in a propeller IDENTIFY: Apply Eq (29.14) SET UP: ⑀ = 3.5 × 10−11 F/m EXECUTE: iD = ⑀ dΦE = (3.5 × 10−11 F/m)(24.0 × 103 V ⋅ m/s3 )t iD = 21 × 10−6 A gives t = 5.0 s dt EVALUATE: iD depends on the rate at which Φ E is changing 29.74 IDENTIFY and SET UP: Apply Ohm’s law to the dielectric to relate the current in the dielectric to the charge on the plates Use Eq (25.1) for the current and obtain a differential equation for q (t ) Integrate this equation to obtain q (t ) and i (t ) Use E = q/⑀ A and Eq (29.16) to calculate jD EXECUTE: (a) Apply Ohm’s law to the dielectric: The capacitor is sketched in Figure 29.74 v (t ) R q (t ) ⑀ A v(t ) = and C = K C d i (t ) = Figure 29.74 ⎛ d ⎞ v(t ) = ⎜ ⎟ q(t ) ⎝ K ⑀0 A ⎠ v(t ) ⎛ q(t )d ⎞ ⎛ A ⎞ q(t ) =⎜ But the ⎟⎜ ⎟= R ⎝ K ⑀0 A ⎠ ⎝ ρ d ⎠ K ⑀0 ρ current i (t ) in the dielectric is related to the rate of change dq/dt of the charge q (t ) on the plates by i (t ) = −dq/dt (a positive i in the direction from the + to the – plate of the capacitor corresponds to a The resistance R of the dielectric slab is R = ρ d/A Thus i (t ) = dq dt dq ⎛ ⎞ =⎜ =− Integrate both ⎟ q (t ) q K ρ⑀0 dt ⎝ K ρ⑀0 ⎠ sides of this equation from t = 0, where q = Q0 , to a later time t when the charge is q (t ) decrease in the charge) Using this in the above gives − ⎛ ⎞ t ⎛ q ⎞ dq t dq ⎛ Q0 ⎞ −t /K ρ⑀0 = −⎜ and q (t ) = Q0e−t/K ρ⑀0 Then i (t ) = − =⎜ ⎟ ∫ dt ln ⎜ ⎟=− ⎟e q K ρ⑀ dt ⎝ K ρ⑀ ⎠ ⎝ K ρ⑀0 ⎠ ⎝ Q0 ⎠ i (t ) ⎛ Q0 ⎞ −t /K ρ⑀0 =⎜ The conduction current flows from the positive to the negative plate and jC = ⎟e A ⎝ AK ρ⑀0 ⎠ of the capacitor q (t ) q (t ) (b) E (t ) = = ⑀ A K ⑀0 A q ∫Q © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 29-28 Chapter 29 jD (t ) = ⑀ dE dE dq(t )/dt i (t ) = K ⑀0 = K ⑀0 = − C = − jC (t ) dt dt K ⑀0 A A G The minus sign means that jD (t ) is directed from the negative to the positive plate E is from + to – but dE/dt is negative (E decreases) so jD (t ) is from – to + EVALUATE: There is no conduction current to and from the plates so the concept of displacement current, G G with jD = − jC in the dielectric, allows the current to be continuous at the capacitor 29.75 IDENTIFY: The conduction current density is related to the electric field by Ohm's law The displacement current density is related to the rate of change of the electric field by Eq (29.16) SET UP: dE/dt = ω E0 cos ωt EXECUTE: (a) jC (max) = E0 ρ = 0.450 V/m = 1.96 × 10−4 A/m 2300 Ω ⋅ m ⎛ dE ⎞ (b) jD (max) = ⑀ ⎜ = ⑀ 0ω E0 = 2π ⑀0 fE0 = 2π ⑀0 (120 Hz)(0.450 V/m) = 3.00 × 10−9 A/m ⎟ ⎝ dt ⎠max E (c) If jC = jD then = ω⑀0 E0 and ω = = 4.91 × 107 rad/s ρ ρ⑀0 f = 29.76 ω 4.91 × 107 rad/s = = 7.82 × 106 Hz 2π 2π EVALUATE: (d) The two current densities are out of phase by 90° because one has a sine function and the other has a cosine, so the displacement current leads the conduction current by 90° IDENTIFY: A current is induced in the loop because of its motion and because of this current the magnetic field exerts a torque on the loop SET UP: Each side of the loop has mass m/4 and the center of mass of each side is at the center of each side The flux through the loop is ΦB = BA cos φ G G G EXECUTE: (a) τ g = ∑ rcm × mg summed over each leg ⎛ L ⎞⎛ m ⎞ ⎛ L ⎞⎛ m ⎞ ⎛m⎞ τ g = ⎜ ⎟⎜ ⎟ g sin(90° − φ ) + ⎜ ⎟⎜ ⎟ g sin(90° − φ ) + ( L) ⎜ ⎟ g sin(90° − φ ) ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝4⎠ mgL cos φ (clockwise) G G τ B = τ × B = IAB sin φ (counterclockwise) τg = I= ε R =− BA d BA dφ BAω cos φ = sin φ = sin φ The current is going counterclockwise looking to the − kˆ R dt R dt R B A2ω B L4ω mgL B L4ω sin φ = sin φ The net torque is τ = cos φ − sin φ , R R R opposite to the direction of the rotation (b) τ = Iα (I being the moment of inertia) About this axis I = mL2 Therefore, 12 12 ⎡ mgL B L4ω ⎤ g 12 B L2ω α= φ − φ = φ − cos sin cos sin φ ⎢ ⎥ mL2 ⎣⎢ 5mR R ⎦⎥ 5L EVALUATE: (c) The magnetic torque slows down the fall (since it opposes the gravitational torque) (d) Some energy is lost through heat from the resistance of the loop IDENTIFY: The motion of the bar produces an induced current and that results in a magnetic force on the bar G G SET UP: FB is perpendicular to B, so is horizontal The vertical component of the normal force equals mg cos φ , so the horizontal component of the normal force equals mg tan φ direction Therefore, τ B = 29.77 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Electromagnetic Induction 29-29 EXECUTE: (a) As the bar starts to slide, the flux is decreasing, so the current flows to increase the flux, LB LB d Φ B LB dA LB vL2 B which means it flows from a to b FB = iLB = ε= (vL cos φ ) = cos φ B = = R R dt R dt R R (b) At the terminal speed the horizontal forces balance, so mg tan φ = (c) i = ε R = vt L2 B Rmg tan φ cos φ and vt = 2 R L B cos φ d Φ B dA B v LB cos φ mg tan φ = B = (vt L cos φ ) = t = R dt R dt R R LB (d) P = i R = Rm g tan φ L2 B ⎛ Rmg tan φ ⎞ Rm g tan φ (e) Pg = Fvt cos(90° − φ ) = mg ⎜⎜ 2 ⎟⎟ sin φ and Pg = L2 B ⎝ L B cos φ ⎠ EVALUATE: The power in part (e) equals that in part (d), as is required by conservation of energy © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... the induced current is clockwise P 0.840 W emf BvL (b) PR = I R so I = R = = = 0 .136 6 A I = R R 45.0 Ω R IR (0 .136 6 A)(45.0 Ω) = = 26.3 m/s BL (0.650 T)(0.360 m) EXECUTE: This speed is around... critical field, so inside the superconductor B = and G G G G (0 .130 T)iˆ B0 =− = −(1.03 × 105 A/m) iˆ Outside the superconductor, B = B0 = (0 .130 T) iˆ and M =− G M = μ0 μ0 G G (b) The field is greater... 1.33 T = (0.380 T/s3 )t , which gives t = 1.518 s At this t, I = (5.739 × 10−3 A/s )(1.518 s) = 0. 0132 A 29.22 EVALUATE: As the field changes, the current will also change IDENTIFY: The magnetic