1. The initial magnetic flux (F_B)_i = B(area) = B(xL).
   
   
2. The later magnetic flux = B(x + Dx)L.
   The change in flux = DF= BL{(x + Dx) -x} = BLDx.
   
   
3. The electromotive force = e = DF/Dt = BL(Dx/Dt) = 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 = e/R = BLv/R.
   
   
6. Power dissipated in the resistance R = I²R = (BLv/R)²R = B²L²v²/R,
   since I = e/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 = e/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. dF = B ^. dA = (B_o N/A-m – C N/A-m³-s² r²t²)k • 2pr dr
                  = 2p(B_or N/A-m – C N/A-m³-s² r³t²)dr
   
   
2.  F = _o∫^R 2p(B_or N/A-m – C N/A-m³-s² r³t²)dr
       = B_o p R² N/A-m - pCR⁴t² N/A-m³-s² /2
   
   
3. e = - dF/dt = pC 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 Q = [(BL)²v cos Q/R]cos Q= (BL)² v cos² Q/R.

1. The magnetic field inside the solenoid is B = µ_onI. The number of loops in a length l is nl; each loop has a flux pR²B, where R is the radius of the solenoid. The flux through all the loops in a length l is F_B = pR²µ_on²Il. The inductance L = F_B/I = pR²µ_on²l. The inductance per unit length is pR²µ_on².
   
   
2. L = pR²µ_on²l= p(2 x 10^-2 m)²(4px 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 2p r = -dB/dt pr² = - d(0.05 s^-2 t + 0.4)/dt T pr².
E2pr = (- 0.1 s ^–2 t N/A/m)pr²  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/2pr.
   The flux over the toroid cross section is found from integration.
   
   F =B (adr)
   
                        =(Nµ_oI / 2pr)(a dr)
   
                        = (Nµ_oIa)[ln (R + a)/R]
   
   Then  L = NF / I = µ_oN²a/2p [ln (R + a)/R]
   
   
2. The magnetic energy density u_B = (1/2µ_o)B² = (1/2µ_o)( µ_oNI/2pr)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 = 2pra dr.
   
   The magnetic energy in the ring = u_B dV = [(µ_oNI/2pr)²/2µ_o][2pra dr].
   
   To find the total energy in the toroid,
   we integrate from  r = R  to  r = R + a:
   
   U_B = [µ_oN²I²a/4p] _R∫^{R + a} dr/r = [µ_oN²I²a/4p] ln [(R + a)/R].
   
   
4. U_B = 1/2 LI² = 1/2 (µ_oN²I²a/2p)ln[(R + a)/R]
                        = (µ_oN²I²a/4p)ln[(R + a)/R].
   
   

1. - d/dtB • dA = E • d
   Changing magnetic flux produces an electric field.
   
   
2. µ_oe_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_oe_o)^1/2.
   e_oc² = 1/µ_o.
   
   Thus S = e_oc² (E x B).
   
   
2. u_E = 1/2 e_oE²  and  u_B = B²/2µ_o.
   B = E/c.
   u_B = E²/2c²µ_o = 1/2 e_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 4p 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/2pr.
   
   
2. S = 1/µ_o (EB) = 1/µ_o(µ_oI²R/2prL) = (I²R/2prL).
   
   
3. Power = Energy/second = S (surface area) = (I²R/2prL)(2prL) = I²R.
   
   

1. Average Intensity = Energy/area =
   7.0 W/(p)(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 4p 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 = s/e_o = q/Ae_o = q_o sin wt/pR²e_o.
   
   
2. F_E = E(area).
   Choose for the area a circle between the plates with r < R and area pr².
   F_E = (q_o sin wt/e_opR²)(pr²) = (q_o sin wt) (r²/e_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_oe_odF_e / dt

   For r < R,

   B2pr = µ_oe_o d(q_o sin wt r²/e_oR²)/dt
        B = µ_owq_o cos wt r/2pR²

   
   
4. S = 1/µ_o E x B.  For sin wt > 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_owq_o cos wt/2pR   and
   S = 1/µ_o EB sin 90^o = wq_o² sin wt cos wt/2p²e_oR³.

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

   = ( wq_o² sin wt cos wt/2p²e_oR³) (2pRd)
   = wq_o² (sin wt cos wt)d/pe_oR².

   Rate of increase of electrical energy

   = d(q²/2C)/dt
   = [d{(q_o sin wt)²}/dt]/(2e_opR²/d)
   = wq_o² (sin wt cos wt)d/pe_oR²

   = energy flow into cylindrical area between the plates.
   
   

1. The resonance frequency equals

   w₀ = 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 W = 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 w = 0.95w₀,  X = X_L - X_c = wL - 1/wC =
   (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 w = 10.8 W.
   
   The maximum current I_p = V_p/Z = 1.0V/10.8 W = 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 F, where F is the phase angle with tan F
   = X/R = (-10.3 w)/3.3 W = 1.26 and  F = - 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 w= 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 W = 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 + (wL) ²]^1/2.

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

1. For Fig. 13a,  V_{p out} = I_p (1/wc) = {V_{p in} / [R² + (1/wC) ²]^1/2}(1/wC).
   
   Multiplying numerator and denominator by (wC),
   V_{p out} = {V_{p in} / [(RwC)² + 1]^1/2   or
   V_{p out} / V_{p in} = 1/[(RwC)² + 1]^1/2.
   
   For small values of w,
   (RwC)² is very small and V_{p out} / V_{p in} approaches 1.
   
   For large values of w,
   (RwL)² is very large and V_{p out} / V_{p in} approaches 0.
   
   The sketch of V_{p out} / V_{p in} vs. w 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/wC) ²]^1/2}(R).
   
   Dividing numerator and denominator by (R),
   V_{p out} = {V_{p in} / [1 +(1/RwC)² ]^1/2    or
   V_{p out} / V_{p in} = 1/[1 + [(1/RwC)² ]^1/2.
   
   For small values of w,
   (1/RwC)² is very large and V_{p out} / V_{p in} approaches 0.
   
   For large values of w,
   (1/RwL)² is very small and V_{p out} / V_p in approaches 1.
   
   The sketch of V_{p out} / V_{p in} vs. w 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 + (wL – 1/wC)²).
   
   Since w_o² = 1/LC,   (wL – 1/wC)² = (L²/w²)(w² - w_o²)².
   
   P_av  = V²_rmsR/[R² + (L²/w²)(w² - w_o²)²]
          = V²_rmsRw²/[(Rw)² + (L²)(w² - w_o²)²].
   
   
2. At resonance,  w = w_o,   P_av at resonance = V²_rms / R.
   
   
3. A plot of P_av as a function of w is shown below in Fig. for #25:
   
   
   
   
4. V²_rmsRw² / [(Rw)² + (L²)(w² - w_o²)²] = V²_rms / 2R
   2R²w² = (Rw)² + (L²)(w² - w_o²)²
   R²w² = (L²)(w² - w_o²)²
   ±Rw = L(w² - w_o²) w² ± Rw/L - w_o² = 0
   w = ±R/2L ± [(R/2L)² + w_o²]^1/2
   w_high = R/2L + [(R/2L)² + w_o²]^1/2
   w_low = - R/2L + [(R/2L)² + w_o²]^1/2
   w_high - w_low = Dw = R/L
   
   
5. Q_o = w_o/Dw= w_oL/R