1. The initial magnetic flux (Φ_B)_i = B(area) = B(xL).
   
   
2. The later magnetic flux = B(x + Δx)L.
   The change in flux = ΔΦ= BL{(x + Δx) -x} = BL Δx.
   
   
3. The electromotive force = ε = ΔΦ/ Δt = BL( Δx/ Δt) = BLv.
   
   
4. 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.
   
   
5. The current I = ε/R = BLv/R.
   
   
6. Power dissipated in the resistance R = I²R = (BLv/R)²R = B²L²v²/R,
   since I = ε/R = BLv/R.
   
   
7. 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 F_m = I( L x B).  L is taken in the direction of the current or down. B is out of the page, so F_m 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 F_m is to the left. The external agent must exert a force F equal in magnitude to F_m to move the rod with a constant velocity.  F = F_m = ILB sin 90^o = ILB = (BLv/R)LB = B²L²v/R.
   
   
8. Power = work/time = (force x distance)/time = force x (distance/time) = force x v. Power generated by external agent = (B²L²v/R)v = B²L²v²/R.
   
   
9. 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.
   
   

1. 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.
   
   
2. The "motional emf" = BLv.  I = ε/R = BLv/R.  The magnetic force on the bar of length L is F_m = (I)LB sin 90^o = (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, F_net = 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.
   
   

1. dΦ = B ^. dA = (B_o N/A-m – C N/A-m³-s² r²t²)k • 2 πr dr
                  = 2 π(B_or N/A-m – C N/A-m³-s² r³t²)dr
   
   
2.  Φ = _o∫^R 2 π(B_or N/A-m – C N/A-m³-s² r³t²)dr
       = B_o π R² N/A-m - πCR⁴t² N/A-m³-s² /2
   
   
3. ε = - dΦ/dt = πC N/A-m³-s² R⁴t
   
   
4. 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.
   
   

F_m cos Θ = [(BL)²v cos Θ/R]cos Θ= (BL)² v cos² Θ/R.

1. The magnetic field inside the solenoid is B = µ_onI. 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 Φ_B = πR²µ_on²I λ. The inductance L = Φ_B/I = πR²µ_on² λ. The inductance per unit length is πR²µ_on².
   
   
2. L = πR²µ_on² 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.
   
   
3. Emf = LdI/dt = 6.32 x 10^-3 V/(A/s) (3.0 x 10² A/s) = 1.9V.
   
   

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

1. The magnetic field inside of a toroid is B = µ_oNI/2 πr.
   The flux over the toroid cross section is found from integration.
   
   Φ =B (adr)
   
                        =(Nµ_oI / 2 πr)(a dr)
   
                        = (Nµ_oIa)[ln (R + a)/R]
   
   Then  L = NΦ / I = µ_oN²a/2 π[ln (R + a)/R]
   
   
2. The magnetic energy density u_B = (1/2µ_o)B² = (1/2µ_o)( µ_oNI/2 πr)².
   
   
3. 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 = u_B dV = [(µ_oNI/2 πr)²/2µ_o][2 πra dr].
   
   To find the total energy in the toroid,
   we integrate from  r = R  to  r = R + a:
   
   U_B = [µ_oN²I²a/4 π] _R∫^{R + a} dr/r = [µ_oN²I²a/4 π] ln [(R + a)/R].
   
   
4. U_B = 1/2 LI² = 1/2 (µ_oN²I²a/2 π)ln[(R + a)/R]
                        = (µ_oN²I²a/4 π)ln[(R + a)/R].
   
   

1. - d/dtB • dA = E • d
   Changing magnetic flux produces an electric field.
   
   
2. µ_o ε_o d/dtE • dA = B • d
   Changing electric flux produces a magnetic field without a conduction current.

1. S = 1/µ_o (E x B).
   c = 1/(µ_oε_o)^1/2.
   ε_oc² = 1/µ_o.
   
   Thus S = ε_oc² (E x B).
   
   
2. u_E = 1/2 ε_oE²  and  u_B = B²/2µ_o.
   B = E/c.
   u_B = E²/2c²µ_o = 1/2 ε_oE² = u_E
   
   
3. S_av = 1/µ_o (E_mB_m/2)
          = 1/µ_o (E_m²/2c)
          = (1/2cµ_o)(E_m²)
          = [1/(2 x 3 x 10⁸ m/s x 4 π x 10^-7 N/A²)](E_m²)
          = [1.33 x 10^-3 A²-s/N-m](E_m²).

1. V_R = IR = EL  or  E = IR/L.   B = µ_oI/2 πr.
   
   
2. S = 1/µ_o (EB) = 1/µ_o(µ_oI²R/2 πrL) = (I²R/2 πrL).
   
   
3. Power = Energy/second = S (surface area) = (I²R/2 πrL)(2 πrL) = I²R.
   
   

1. Average Intensity = Energy/area =
   7.0 W/( π)(5 x 10^-4 m)² = 8.9 x 10⁶ W/m² = 8.9 MW/m².
   
   
2. S_av = 1/2cµ_o (E_m²).
   
   E_m = (2µ_ocS_av)^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.
   
   

1. E = σ/ ε_o = q/A ε_o = q_o sin ωt/ πR² ε_o.
   
   
2. Φ_E = E(area).
   Choose for the area a circle between the plates with r < R and area  πr².
   F_E = (q_o sin ωt/ ε_oπR²)( πr²) = (q_o sin ωt) (r²/ ε_oR²).
   
   
3. 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 = µ_o ε_odΦ_e / dt

   For r < R,

   B2 πr = µ_oε_o d(q_o sin ωt r²/ ε_oR²)/dt
        B = µ_oωq_o cos ωt r/2 πR²

   
   
4. S = 1/µ_o 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.
   
   
5. For r = R,

   B = µ_oωq_o cos ωt/2 πR   and
   S = 1/µ_o EB sin 90^o = ωq_o² sin ωt cos ωt/2 π² ε_oR³.

   
   
6. The energy flows into the cylindrical area between the plates.
   
   This surface area = 2 πRd, and
   
   S(area)

   = ( ωq_o² sin ωt cos ωt/2 π² ε_oR³) (2 πRd)
   = ωq_o² (sin ωt cos ωt)d/πε_oR².

   Rate of increase of electrical energy

   = d(q²/2C)/dt
   = [d{(q_o sin ωt)²}/dt]/(2 ε_o πR²/d)
   = ωq_o² (sin ωt cos ωt)d/πε_oR²

   = energy flow into cylindrical area between the plates.
   
   

1. 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

2. At resonance, maximum current I_p = V_p/R = 1.O V/3.3 ω= 0.30 A
   
   
3. At resonance, average power dissipated

   P_av = 1/2 (I_p)²R
          = [1/2R](V_p)²
          = [1/2(3.3 V/A)](1.0 V)²
          = 0.15 W
   

   
   
4. When ω = 0.95ω₀,  X = X_L - X_c = 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 I_p = V_p/Z = 1.0V/10.8 ω = 0.093 A,
   approximately 1/3 of the maximum current at resonance.
   
   Average power dissipated =1/2(I_p)²R = (1/2)(V_p²/Z)R/Z
   = (1/2)(V_p²/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 (72.4^o) = 0.014 W,
   which is about 10% of the average power at resonance.

1. I₁ = 10V/5 ω= 2 A.
   
   
2. Since the current has not started to build up,  I₂ = 0.
   
   
3. I = (2 A + 0) = 2 A.  Since I₂ = 0,
   the potential difference across R₂ = 0.
   
   
4. Since V_ab = 10 V and V_R the potential difference across R₂ = 0,
   the potential difference V_L across the inductor = 10 V.
   
   
5. L dI₂/dt = V_L   or   dI₂/dt = V_L/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) V_R = I₂ R₂ = V_ab = 10 V.
   (e) L dI₂/dt = V_L = 0,  because
   

6. dI₂/dt = 0.
   
   

I_p = V _{p in}/(R² + X_L ²)^1/2
    = V_{p in} /[R² + (ωL) ²]^1/2.

1. For Fig. 12a above,  V_{p out} = I_p (ωL) = {V_{p in}/[R² + (ωL) ²]^1/2}(ωL).
   
   Dividing numerator and denominator by (ωL),
   V_{p out} = {V_{p in}/[(R/ωL)² + 1]^1/2   or
   V_{p out} / V_{p in} = 1/[(R/ωL)² + 1]^1/2.
   
   For small values of ω,
   (R/ωL)² is very large and V_{p out} / V_{p in} is very small.
   
   For large values of ω,
   (R/ωL)² is very small and V_{p out} / V_{p in} approaches 1.
   
   The sketch of V_{p out} / V_{p 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.
   
   
2. For Fig. 12b above,  V_{p out} = I_p (R) = {V_{p in}/[R² + (ωL) ²]^1/2}(R).
   
   Dividing numerator and denominator by (R),
   V_{p out} = {V_{p in}/[1 + (ωL/R)² ]^1/2   or
   V_{p out} / V_{p in} = 1/[1 + (ωL/R)² ]^1/2.
   
   For small values of ω,
   (ωL/R)² is very small and V_{p out} / V_{p in} approaches 1.
   
   For large values of ω,
   (ωL/R)² is very large and V_{p out} / V_{p in} approaches 0.
   
   The sketch of V_{p out} / V_{pi 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.
   
   

I_p = V_{p in} / (R² + X_C²)^1/2
     = V_{p out} / [R² + (1/ωC) ²]^1/2.

1. For Fig. 13a,  V_{p out} = I_p (1/ωc) = {V_{p in} / [R² + (1/ωC) ²]^1/2}(1/ωC).
   
   Multiplying numerator and denominator by (ωC),
   V_{p out} = {V_{p in} / [(RωC)² + 1]^1/2   or
   V_{p out} / V_{p in} = 1/[(RωC)² + 1]^1/2.
   
   For small values of ω,
   (RωC)² is very small and V_{p out} / V_{p in} approaches 1.
   
   For large values of ω,
   (RωL)² is very large and V_{p out} / V_{p in} approaches 0.
   
   The sketch of V_{p out} / V_{p 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.
   
   
2. For Fig. 13b,  V_{p out} = I_p (R) = {V_{p in} / [R² + (1/ωC) ²]^1/2}(R).
   
   Dividing numerator and denominator by (R),
   V_{p out} = {V_{p in} / [1 +(1/RωC)² ]^1/2    or
   V_{p out} / V_{p in} = 1/[1 + [(1/RωC)² ]^1/2.
   
   For small values of ω,
   (1/RωC)² is very large and V_{p out} / V_{p in} approaches 0.
   
   For large values of ω,
   (1/RωL)² is very small and V_{p out} / V_p in approaches 1.
   
   The sketch of V_{p out} / V_{p 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.
   
   

1. P_av = I²_rms R = V²_rmsR/(R² + (ωL – 1/ωC)²).
   
   Since ω_o² = 1/LC,   (ωL – 1/ωC)² = (L²/ω²)(ω² - ω_o²)².
   
   P_av  = V²_rmsR/[R² + (L²/ω²)(ω² - ω_o²)²]
          = V²_rmsRω²/[(Rω)² + (L²)(ω² - ω_o²)²].
   
   
2. At resonance,  ω = ω_o,   P_av at resonance = V²_rms / R.
   
   
3. A plot of P_av as a function of ω is shown below in Fig. for #25:
   
   
   
   
4. V²_rmsRω² / [(Rω)² + (L²)(ω² - ω_o²)²] = V²_rms / 2R
   2R²ω² = (Rω)² + (L²)(ω² - ω_o²)²
   R²ω² = (L²)(ω² - ω_o²)²
   ±Rω = L(ω² - ω_o²) ω² ± Rω/L - ω_o² = 0
   ω = ±R/2L ± [(R/2L)² + ω_o²]^1/2
   ω_high = R/2L + [(R/2L)² + ω_o²]^1/2
   ω_low = - R/2L + [(R/2L)² + ω_o²]^1/2
   ω_high - ω_low = Δω = R/L
   
   
5. Q_o = ω_o/Δω= ω_oL/R