 Phyllis Fleming Physics Physics 108
 1 ΦB = B . A = BA cos 0o = BA. ε = - ΔΦB/ Δt   = - ( ΔB/ Δt)A   = - [(0.040 - 0.080)N/A-m/(2.0 s)]4.0 m2   = 0.080 N-m/C   = 0.080 V. I = ε/R = 0.080 V/0.04 ω = 2 A. With the magnetic field into the page and decreasing, the sense of the current will be such to oppose the change that produced it, or clockwise. A clockwise-induced current produces a magnetic field into the page and opposes the decreasing external magnetic field into the page.
 2 dΦB = B . dA = 0.0120 N/A-m2(x sin ωt j + y cos ωt k) . dA k                      = 0.0120 N/A-m2 y cos ωt (dx dy),  where ω = 103 s-1. ΦB = 0.0120 N/A-m2 cos ωt o∫b dx o∫a y dy       = 0.0120 N/A-m2 cos ωt (ba2/2) ε = - dΦB/dt = 0.0060 N/A-m2 ωsin ωt (ba2)                     = 0.0060 N/A-m2 (103 s-1)sin 103 s-1t(0.0015 m3)                     = 9.0 x 10-3 V sin 103 s-1t                     = 9.0 mV sin 103 s-1t.
 3 For a toroid, all the flux is confined to the inside of the toroid. B = NµoI/2 πr. ΦB = ∫ B . dA = NµoI/2 π o∫a dz R∫R + b dr/r      = (NµoI/2 π)a ln (R + b)/R      = (NaµoIo sin ωt/2 π)) ln (R + b)/R. ε = - N’ dΦB/dt    = - ωNN’[a(µo/2 π)Io cos ωt)]ln (R + b)/R    = 12 π x 105 s-1(0.02m)(2 x 10-7N/A2)(50 A){ln(6 + 3)/6} cos 120 π s-1 t    = 0.306 V cos 120 π s-1 t.
 4 B = µonI. ΦB = ∫ B . dA = µonI( πR2)      = µon[30 A (1 - e -1.6t/s)]( πR2). ε = - N dΦB/dt = µoNn(1.6/s)30 A(e -1.6t/s)( πR2)    = (4 x 10-7 N/A2)(250)(400/m)(1.6/s)(30 A)(e -1.6t/s) π(0.06 m)2    = 2.17 x 10-2 (e -1.6t/s)V.
 5 The initial magnetic flux (ΦB)i = B(area) = B(xL). The later magnetic flux = B(x + Δx)L. The change in flux = ΔΦ= BL{(x + Δx) -x} = BL Δx. The electromotive force = ε = ΔΦ/ Δt = BL( Δx/ Δt) = BLv. I haven't used the minus sign because we find the sense of the current from Lenz's law. The sense of the induced current is such to oppose the change that produces it. As the rod moves to the right, the magnetic flux out of the page increases. To oppose this, the induced current must be clockwise. A current clockwise will produce a field into the page. If you curl the fingers of your right hand clockwise, your thumb points into the page. The current I = ε/R = BLv/R. Power dissipated in the resistance R = I2R = (BLv/R)2R = B2L2v2/R, since I = ε/R = BLv/R. In Fig. 4 above, the induced current is clockwise, which means positive charge moves down through the bar of length L. The force on the current carrying bar due to the magnetic field B is Fm = I( L x B).  L is taken in the direction of the current or down. B is out of the page, so Fm can be neither in or out nor up or down. It must be to the right or the left. If you rotate vector L into vector B, you find Fm is to the left. The external agent must exert a force F equal in magnitude to Fm to move the rod with a constant velocity.  F = Fm = ILB sin 90o = ILB = (BLv/R)LB = B2L2v/R. Power = work/time = (force x distance)/time = force x (distance/time) = force x v. Power generated by external agent = (B2L2v/R)v = B2L2v2/R. Notice that the answers to (f) and (h) are the same, as, of course, they should be if you believe in conservation of energy and all those good things. The external agent must do work in order to produce the induced emf and current. Lenz's law is just a statement of conservation of energy.
 6 As the bar slides down, the magnetic flux into the page increases. To oppose this, the induced current must be to the right in the bar. The induced current then produces a magnetic field out of the page inside the loop. The "motional emf" = BLv.  I = ε/R = BLv/R.  The magnetic force on the bar of length L is Fm = (I)LB sin 90o = (BLv/R)LB(1) = B2L2v/R. For a current to the right in a field into the page the force on the bar is up. For constant velocity, Fnet = B2L2v/R - mg = 0  and  v = mgR/B2L2 = 0.50 kg(9.8 m/s2)(0.01 N-m/A-C)/(0.20 N/A-m)2(0.5 m)2 = 4.90 m/s.
 7  After 1 s, one-half of the loop is in the magnetic field so the magnetic flux = BA = (1T)(0.5m2) = 0.5 Wb. After 2 s, the entire loop is in B:  ΦB = 1 Wb. From  t = 2s  to  t = 10 s,  the entire loop is in the field. At  t =10 s, the front edge of the loop reaches the edge of the field boundary. At  t = 11 s, only half the loop is in the field, and at  t = 12 s  none of the loop is in the field. From  t = 12 s on the flux is zero. Since the emf = - dΦB/dt,  you find the emf from the negative slope of the ΦB vs t  graph, as shown in the figures above. As the loop enters the magnetic field, the sense of the induced current is clockwise in the loop producing a magnetic field into the page to oppose the change that produced it.  As it leaves the field, the current is counterclockwise. From  t = 0  to  t = 2 s, and from  t = 10 s  to  t = 12 s, I = ε/R = 0.50 V/0.01ω = 50 A. From  t = 2s  to  t = 10 s,  I = 0.
 8 dΦ = B . dA = (Bo N/A-m – C N/A-m3-s2 r2t2)k • 2 πr dr                = 2 π(Bor N/A-m – C N/A-m3-s2 r3t2)dr  Φ = o∫R 2 π(Bor N/A-m – C N/A-m3-s2 r3t2)dr     = Bo π R2 N/A-m - πCR4t2 N/A-m3-s2 /2 ε = - dΦ/dt = πC N/A-m3-s2 R4t C is a positive constant. As time increases, the second term in B increases. Since this is a negative term, the magnetic field, which is up, is decreasing with time, the induced current must produce a magnetic field up. If you point the thumb of your right hand up and view your curled fingers from above, you see the current is counterclockwise.
 9 dΦB = B . dA = B xL cos Θ ε = - dΦB/dt = - B (dx/dt)L cos Θ = - BvL cos Θ I = ε/R = BvLcos Θ/R Fm = I(L x B) Fm = ILB sin 90o = ILB = [BvL cos Θ/R](LB) = (BL)2v cos Θ/R The direction of the magnetic force is to the left. The component of the magnetic force up the plane equals Fm cos Θ = [(BL)2v cos Θ/R]cos Θ= (BL)2 v cos2 Θ/R. The component of the gravitational force down the plane = mg sin Θ. For a constant velocity,  Fnet = mg sin Θ - (BL)2 v cos2 Θ/R = 0   or v = mgR sin Θ/(BL cos Θ)2.
 10 Consider the element of area dA = dr’ L. The magnetic flux through this area  dΦB = B dr’ L, where B at the position of the area = µoi/2 πr’. So  dΦB = µoILdr’/2 πr’ and ΦB = ∫ µoILdr’/2 πr’ from r’ = r to r’ = (r + w).  Emf = -dΦB/dt = -µoIL/2 π{r/(r + w)} {r(1) - (r+w)(1)}/r2(dr/dt), where dr/dt = v. Emf = (µoIL/2 π) {wv/r(r + w)}. Iinduced = Emf/R = (µoIL/2 πR) {wv/r(r + w)}. The magnetic field lines due to the very long current carrying wire are into the page in the region of the loop. As the loop moves away from the wire the magnetic flux inside the loop decreases. To oppose the change that produced it, the induced current in the loop is in a clockwise sense producing magnetic field lines into the page.
 11 The magnetic field inside the solenoid is B = µonI. The number of loops in a length λ is n λ; each loop has a flux πR2B, where R is the radius of the solenoid. The flux through all the loops in a length λ is ΦB = πR2µon2I λ. The inductance L = ΦB/I = πR2µon2 λ. The inductance per unit length is πR2µon2. L = πR2µon2 λ= π(2 x 10-2 m)2(4 πx 10-7 N/A2)(2 x 103 m-1)2(1 m)    = 6.32 x 10-3 V/(A/s) = 6.32 x 10-3 V/(A/s) = 6.32 x 10-3 H. Emf = LdI/dt = 6.32 x 10-3 V/(A/s) (3.0 x 102 A/s) = 1.9V.
 12  Because the field is increasing and it is into the page, the electric field line will be counterclockwise.  Take as a path a circle going counterclockwise with the radius of the circle = r. From symmetry, E, d = 0 and E is constant everywhere around the circle. ∫ E . d = E 2 πr. The magnetic field varies with time but is constant with distance over a circle of radius R. The path of the electric field surrounds a circle of radius r. -d/dt ∫ B . dA = - dB/dt ( πr2) = E 2 π r = -dB/dt πr2 = - d(0.05 s-2 t + 0.4)/dt T πr2. E2 πr = (- 0.1 s –2 t N/A/m) πr2  or E(r,t) =- 0.05 s-2 t N/A-m r E (0.04 m, 4 s) = -0.05(4 )(0.04) N/C = 8 x 10-3 N/C
 13 For build up of the current  I(t) = (Emf/R) (1 - e-Rt/L). As  t  approaches a very large number,  e-Rt/L  approaches zero and I = Imax = (Emf)/R   or   R = (Emf)/Imax = 3.0V/24 A= (1/8) ω. For decay of the current,  I(t) = Imax e-Rt/L. I(0.22s) = 12A = 24 A e-(1/8 Ω)(0.22s)/L. (1/8 ω)(0.22 s) =2.75 x 10-2 (V/A)s. 12A = 24 A e-0.0275 (V/A)s/L or  12/24 = 1/2 = e-0.0275(V/A)s/L. Taking reciprocals of both sides of the equation, 2 = e+0.0275(V/A)s/L. Taking the ln of both sides of the equation gives ln 2 = 0.0275(V/A)s/L. L = 0.0275 (V/A)s/ln 2 = 40 x 10-3 V/(A/s) = 40 mH.
 14 (a) UE = 1/2 q2/C = 1/2 (1.0 x10-4 C)2/(5.0 x 10-6C/V) = 1.0 x 10-3 J. (b) and (c) q/C + L dI/dt = 0  or  L d2q/dt2 + q/C = 0  and  d2q/dt2 + (1/LC)q = 0. Compare with d2x/dt2 + (k/m)x = 0 and see that for q = maximum q = qm at t = 0, q(t) = qm cos 2 πt/T with T = 2 π (1/LC)1/2                                          = 2 π [1/(5.0 x 10-2 V-s/A)(1.0 x 10-4 C/V)]1/2                                          = 2.8 x 103 s. I(t) = Im sin 2 π/T. The electric energy UE = q2/2C is a maximum at t = 0, t = T/2, t = T, and so forth. The magnetic energy UB = 1/2 LI2 is a maximum at t = T/4, t = 3T/4, t = 5T/4, and so forth.
 15 The magnetic field inside of a toroid is B = µoNI/2 πr. The flux over the toroid cross section is found from integration. Φ = B (adr)                      = (NµoI / 2 πr)(a dr)                      = (NµoIa)[ln (R + a)/R] Then  L = NΦ / I = µoN2a/2 π[ln (R + a)/R] The magnetic energy density uB = (1/2µo)B2 = (1/2µo)( µoNI/2 πr)2. The volume of a thin ring of thickness dr and height a at a distance r from the axis of the toroid is  dV = 2 πra dr. The magnetic energy in the ring = uB dV = [(µoNI/2 πr)2/2µo][2 πra dr]. To find the total energy in the toroid, we integrate from  r = R  to  r = R + a: UB = [µoN2I2a/4 π] R∫R + a dr/r = [µoN2I2a/4 π] ln [(R + a)/R]. UB = 1/2 LI2 = 1/2 (µoN2I2a/2 π)ln[(R + a)/R]                      = (µoN2I2a/4 π)ln[(R + a)/R].
 16 - d/dt B • dA = E • d Changing magnetic flux produces an electric field. µo εo d/dt E • dA = B • d Changing electric flux produces a magnetic field without a conduction current.
 17 S = 1/µo (E x B). c = 1/(µoεo)1/2. εoc2 = 1/µo. Thus S = εoc2 (E x B). uE = 1/2 εoE2  and  uB = B2/2µo. B = E/c. uB = E2/2c2µo = 1/2 εoE2 = uE Sav = 1/µo (EmBm/2)        = 1/µo (Em2/2c)        = (1/2cµo)(Em2)        = [1/(2 x 3 x 108 m/s x 4 π x 10-7 N/A2)](Em2)        = [1.33 x 10-3 A2-s/N-m](Em2).
 18 VR = IR = EL  or  E = IR/L.   B = µoI/2 πr. S = 1/µo (EB) = 1/µo(µoI2R/2 πrL) = (I2R/2 πrL). Power = Energy/second = S (surface area) = (I2R/2 πrL)(2 πrL) = I2R.
 19 Average Intensity = Energy/area = 7.0 W/( π)(5 x 10-4 m)2 = 8.9 x 106 W/m2 = 8.9 MW/m2. Sav = 1/2cµo (Em2). Em = (2µocSav)1/2       = (2 x 4 π x 10-7 N/A2 x 3 x 108 m/s x 8.9 x 106 N/m-s)1/2       = 8.2 x 104 N/C.
 20 E = σ/ εo = q/A εo = qo sin ωt/ πR2 εo. ΦE = E(area). Choose for the area a circle between the plates with r < R and area  πr2. FE = (qo sin ωt/ εoπR2)( πr2) = (qo sin ωt) (r2/ εoR2). By symmetry, along a circle whose plane is parallel to the parallel plates, B, ds = 0 and B is constant everywhere around the circle, B • ds = µo εodΦe / dt For r < R, B2 πr = µoεo d(qo sin ωt r2/ εoR2)/dt      B = µoωqo cos ωt r/2 πR2 S = 1/µo E x B.  For sin ωt > 0, the direction of B is as shown in the figure above, and S is radially in toward the axis of the capacitor. For r = R, B = µoωqo cos ωt/2 πR   and S = 1/µo EB sin 90o = ωqo2 sin ωt cos ωt/2 π2 εoR3. The energy flows into the cylindrical area between the plates. This surface area = 2 πRd, and S(area) = ( ωqo2 sin ωt cos ωt/2 π2 εoR3) (2 πRd) = ωqo2 (sin ωt cos ωt)d/πεoR2. Rate of increase of electrical energy = d(q2/2C)/dt = [d{(qo sin ωt)2}/dt]/(2 εo πR2/d) = ωqo2 (sin ωt cos ωt)d/πεoR2 = energy flow into cylindrical area between the plates.
 21 The resonance frequency equals ω0 = 1/(LC)1/2      = 1/[(0.010 V-s/A)(1.0 x 10-6 C/V)]1/2      = 104 Hz      = 104 s-1 At resonance, maximum current Ip = Vp/R = 1.O V/3.3 ω= 0.30 A At resonance, average power dissipated Pav = 1/2 (Ip)2R        = [1/2R](Vp)2        = [1/2(3.3 V/A)](1.0 V)2        = 0.15 W When ω = 0.95ω0,  X = XL - Xc = ωL - 1/ωC = (9500 s-1)(0.010 V-s/A) - [1/(9500 s-1)(1.0 x 10-6 C/V) = - 10.3 W The impedance Z = [R2 + X2]1/2 = [3.22 + (-10.3)2]1/2 ω = 10.8 ω. The maximum current Ip = Vp/Z = 1.0V/10.8 ω = 0.093 A, approximately 1/3 of the maximum current at resonance. Average power dissipated =1/2(Ip)2R = (1/2)(Vp2/Z)R/Z = (1/2)(Vp2/Z)cos Φ, where Φis the phase angle with tan Φ = X/R = (-10.3 ω)/3.3 ω = 1.26 and  Φ= - 1.26 rad = - 72.40. Average Power = 1/2(1.0 V2/10.8 V/A)cos (72.4o) = 0.014 W, which is about 10% of the average power at resonance.
 22 Notice that the potential difference Vab always equals 10 V. While the switch is closed, the potential difference across R1 and the series circuit of R2 and L will always be 10 V. When the switch S is just closed, I1 = 10V/5 ω= 2 A. Since the current has not started to build up,  I2 = 0. I = (2 A + 0) = 2 A.  Since I2 = 0, the potential difference across R2 = 0. Since Vab = 10 V and VR the potential difference across R2 = 0, the potential difference VL across the inductor = 10 V. L dI2/dt = VL   or   dI2/dt = VL/L = 10 V/ 5 V-s/A = 2 A/s. Note 1 H = V-s/A. After a very long time, I2 has reached a constant maximum of 10 V/R2 = 10 V/10 ω = 1 A,  and dI2/dt = 0.  So the answers are now: (a) As always, I1 = 2 A. (b) I2 = 1 A. (c) I = 3A. (d) VR = I2 R2 = Vab = 10 V. (e) L dI2/dt = VL = 0,  because dI2/dt = 0.
 23 In general,  Ip = Vp in/[(R2 + (XL – XC)2]1/2. When there is no capacitor in the circuit, Ip = V p in/(R2 + XL 2)1/2     = Vp in /[R2 + (ωL) 2]1/2. For Fig. 12a above,  Vp out = Ip (ωL) = {Vp in/[R2 + (ωL) 2]1/2}(ωL). Dividing numerator and denominator by (ωL), Vp out = {Vp in/[(R/ωL)2 + 1]1/2   or Vp out / Vp in = 1/[(R/ωL)2 + 1]1/2. For small values of ω, (R/ωL)2 is very large and Vp out / Vp in is very small. For large values of ω, (R/ωL)2 is very small and Vp out / Vp in approaches 1. The sketch of Vp out / Vp in vs. ω in Fig. b below describes this case. Since the effect of small frequencies is small, this is called a high frequency pass filter. For Fig. 12b above,  Vp out = Ip (R) = {Vp in/[R2 + (ωL) 2]1/2}(R). Dividing numerator and denominator by (R), Vp out = {Vp in/[1 + (ωL/R)2 ]1/2   or Vp out / Vp in = 1/[1 + (ωL/R)2 ]1/2. For small values of ω, (ωL/R)2 is very small and Vp out / Vp in approaches 1. For large values of ω, (ωL/R)2 is very large and Vp out / Vp in approaches 0. The sketch of Vp out / Vpi in vs. ω in Fig. a below describes this case. Since the effect of large frequencies is small, this is called a low frequency pass filter.
 24 In general, Ip = Vp in/[(R2 + (XL – XC)2]1/2. When there is no inductor in the circuit, Ip = Vp in / (R2 + XC2)1/2      = Vp out / [R2 + (1/ωC) 2]1/2. For Fig. 13a,  Vp out = Ip (1/ωc) = {Vp in / [R2 + (1/ωC) 2]1/2}(1/ωC). Multiplying numerator and denominator by (ωC), Vp out = {Vp in / [(RωC)2 + 1]1/2   or Vp out / Vp in = 1/[(RωC)2 + 1]1/2. For small values of ω, (RωC)2 is very small and Vp out / Vp in approaches 1. For large values of ω, (RωL)2 is very large and Vp out / Vp in approaches 0. The sketch of Vp out / Vp in vs. ω in Fig. a below describes this case. Since the effect of large frequencies is small, this is called a low frequency pass filter. For Fig. 13b,  Vp out = Ip (R) = {Vp in / [R2 + (1/ωC) 2]1/2}(R). Dividing numerator and denominator by (R), Vp out = {Vp in / [1 +(1/RωC)2 ]1/2    or Vp out / Vp in = 1/[1 + [(1/RωC)2 ]1/2. For small values of ω, (1/RωC)2 is very large and Vp out / Vp in approaches 0. For large values of ω, (1/RωL)2 is very small and Vp out / Vp in approaches 1. The sketch of Vp out / Vp in vs. ω in Fig. b below describes this case. Since the effect of small frequencies is small, this is called a high frequency pass filter.
 25 Pav = I2rms R = V2rmsR/(R2 + (ωL – 1/ωC)2). Since ωo2 = 1/LC,   (ωL – 1/ωC)2 = (L2/ω2)(ω2 - ωo2)2. Pav  = V2rmsR/[R2 + (L2/ω2)(ω2 - ωo2)2]        = V2rmsRω2/[(Rω)2 + (L2)(ω2 - ωo2)2]. At resonance,  ω = ωo,   Pav at resonance = V2rms / R. A plot of Pav as a function of ω is shown below in Fig. for #25: V2rmsRω2 / [(Rω)2 + (L2)(ω2 - ωo2)2] = V2rms / 2R 2R2ω2 = (Rω)2 + (L2)(ω2 - ωo2)2 R2ω2 = (L2)(ω2 - ωo2)2 ±Rω = L(ω2 - ωo2) ω2 ± Rω/L - ωo2 = 0 ω = ±R/2L ± [(R/2L)2 + ωo2]1/2 ωhigh = R/2L + [(R/2L)2 + ωo2]1/2 ωlow = - R/2L + [(R/2L)2 + ωo2]1/2 ωhigh - ωlow = Δω = R/L Qo = ωo/Δω= ωoL/R
 26 Ip = Vp/[(R2 + (XL – XC)2]1/2.    At resonance, Ip = Vp/R. VpL = Ip (ωoL) = (Vp/R)(ωoL) = (10 V/10 ω)(ωoL). ωo = (1/LC)1/2. (ωoL) = (L/C)1/2 = [(1.0 V-s/A/1.0 x 10-6 C/V)]1/2 = 103 ω. VpL = (Vp/R)(ωoL) = (10 V/10 ω)103 ω= 103 V.

 Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming August 8, 2002 May 3, 2003