ANSWER TO CHAPTER 30 HOMEWORK

1.  The magnetic field inside the outer solenoid is

                    M =m_oN₁N₂A/l.

                        = (2000)(4p x 10^-7T.m/A)(300turns)p(0.0200m)²/(2.44m)

                        = 3.88 x 10^-2H = 38.8mH.

2.  We use the reulg for the inductance from Ex.30-4:

                    L/l = (m_ol/2p)ln(r₂/r₁)

                         = (2 x 10^-7T.m/A)ln(3.0mm/r₁) < 40 x 10^-9H/m, which gives  r₁ > 2.5mm.

3.  The magnetic field at the center of the loop is

                   B = m_oI/2R.

         The energy density of the magnetic field is

                   u_B = 1/2B²/m_o = m_oI²/8R² = (4p x 10^-7T.m/A)(30A)²/8(0.280m)² =  1.8 x10^-3J/m³.

4.   (a)  For an LR circuit, we have

                   I = I_max(1- e^-t/t);

                   1/2 = 1 -e^-t/t,     or -t/t = -(-2.56ms)/t = ln1/2, which gives t =  3.69ms.

          (b) We find the resistance from

                    t = L/R;

                   3.69 x 10^-3s = (310H)/R, which gives R = 8.39 x 10⁴W = 83.9kW.

5.  (a)  The resonant frequency is

                    f_o = (1/2p)(1/LC)^1/2 = (1/2p)[1/(175 x10^-3H)(760 x10^-12F)]^1/2 = 1.38 x10⁴Hz =  13.8kHz.

         (b)  The maximum charge on the capacitor is

                    Q_o = CV.

               so the peak value of the current is

                     I_o = Q_ow = CVw = (760 x 10^-12F)(135V)2p(1.38 x 10⁴Hz) = 8.90 x 10^-3A = 8.90mA.

          (c)  The maximum energy stored in the inductor is 

                     U_Lmax =1/2LI_o² = 1/2(175 x10^-3H)(8.90 x10^-3A)² =  6.93 x 10^-6J.  

6.  We assume underdamping, with

                      w' = w_o = 1/(LC)^1/2, and T = 2p/w_o = 2p(LC)^1/2

         The charge on the capacitor is

                       Q= Q_oe^-Rt/2Lcos (w't + f).

         The energy stored in the capacitor and inductor can be expressed in terms of the amplitude of the cosine function:

                       U = Q²/2C = Q_o²e^-Rt/L/2C = U_oe^-RT/L.

          In one period the energy is reduced by 5.5 percent, so we have

                       0.945U_o = U_oe^-Rt/L, which gives ln(1/0.945) = RT/L = 2pR(C/L)^1/2;

                       ln(1/0.945) = 2pR[(1.00 x10^-6F)/(65 x 10^-3H)]^1/2, which gives  R = 2.30W

           We can check to see if we have underdamping:

                       R² = (2.30W)² = 5.3W²;

                       4L/C = 4(65 x 10^-3H)/(1.00 x 10^-6F) = 2.6 x 10⁵W².

            Thus R << 4L/C.