THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 108 Assignment # 9 SOLUTIONS:
 
INDUCTANCE & CIRCUITS
 
Wed. 9 Mar. 2005 - finish by Wed. 16 Mar.

1. Solenoid as an $RL$ Circuit: A long wire with net resistance $R = 120$ $\Omega$ is wound onto a nonmagnetic spindle to make a solenoid whose cross-sectional area is $A = 0.02$ m$^2$ and whose effective length is $\ell = 0.5$ m. (Treat the coil as an ideal, long solenoid.) Using a battery with a 1 M$\Omega$ internal resistance, a magnetic field of $B_0 = 0.6$ T has been built up inside the solenoid. At $t=0$ the battery is shorted out and then disconnected so that the current begins to be dissipated by the coil's resistance $R$. We find that after 3.6 ms the field in the coil has fallen to 0.1 T.
   1. How many joules of energy are stored in the coil at $t=0$?

      ANSWER: The energy density in a magnetic field is given in general by ${\displaystyle {U \over V} = {1\over2\mu_0} B^2 }$, so the total stored energy is ${\displaystyle U = {\ell A \over 2\mu_0} B^2 }$, in this case ${\displaystyle U_0 = {0.5\times 0.02 \times (0.6)^2 \over 2 \times 4 \pi \times 10^{-7} } }$, or  .

   2. How long does it take for the stored energy to fall to half its initial value?

      ANSWER: Since $U \propto B^2$, $U \to {1\over2} U_0$ when $B \to {1\over\sqrt{2}} B_0$ and since $B \propto I$, this occurs when $I \to {1\over\sqrt{2}} I_0$. We know that the current in an $RL$ circuit decays exponentially, so $I(t) = I_0 \exp(-t/\tau)$. We are therefore looking for the time when $I/I_0 = 1/\sqrt{2} = \exp(-t_f/\tau)$, or $t_f = \tau \ln(\sqrt{2})$. We can determine the time constant $\tau$ from the given fact that $B/B_0 = 0.1/0.6= \exp(-3.6\times 10^{-3}/\tau)$ or $\tau = 3.6\times 10^{-3} / \ln 6= 2.01\times 10^{-3}$ s. This then gives  .

   3. What is the total number of turns in the coil?

      ANSWER: The time constant is $\tau = L/R$. With $R = 120\; \Omega$ this gives $L = 0.241$ H (Henries). It is also true that a solenoid of this form has $L = \mu_0 n^2 A \ell = \mu_0 N^2 A/\ell$, giving ${\displaystyle N^2 = {\ell L \over \mu_0 A} = {0.5\times 0.241\over 4 \pi \times 10^{-7} \times 0.02} = 4.80\times 10^6 }$ or  .

2. $LC$ Circuit Time-Dependence: In an $LC$ circuit with $C = 90$ $\mu$F the current is given as a function of time by $I = 3.4\cos ( 1800t + 1.25)$, where $t$ is in seconds and $I$ is in amperes.
   1. How soon after $t=0$ will the current reach its maximum value?

      ANSWER: The phase of the oscillation is $\theta = \omega t + \phi$. By inspection, $\omega = 1800$ s$^{-1}$ and $\phi = 1.25$ rad. Thus $\cos \theta$ will reach its maximum amplitude at $\theta = \pi$, for which $t = (\pi - 1.25)/1800$ or  . However, since this gives a maximum negative current, one might argue that the maximum (positive) current will first occur when $\theta = 2\pi$ or for $t = (2\pi - 1.25)/1800$ or  . Either answer is acceptable.

   2. Calculate the inductance.

      ANSWER: Since $\omega = 1/\sqrt{LC} = 1800$ s$^{-1}$, $L \times (90\times 10^{-6}$ F $) = 1/(1800)^2$, giving

   3. Find the total energy in the circuit.

      ANSWER: When $i = i_{\rm max} = 3.4$ A, all the energy is in the inductance $L$. Later on this gets shared back and forth with the capacitance, but the total energy never changes. Thus $U = {1\over2}Li_{\rm max}^2 = {1\over2} \times 3.43\times 10^{-3} \times (3.4)^2$ or  .

3. Build Your Own Circuit: You are given a 12 mH inductor and two capacitors of 7.0 and 3.0 $\mu$F capacitance. List all the resonant frequencies that can be produced by connecting these circuit elements in various combinations.
   
   ANSWER: Generally only the loops with both an $L$ and a $C$ can resonate. Any "external" $C$ is just a "spectator" (consider $\sum \Delta {\cal E} = 0$ on the outermost loop in C or D). Thus

   $\omega_A$	=	${\displaystyle \left[ L \left( {1 \over C_1} + {1 \over C_2} \right)^{-1} \right]^{-{1\over2}} }$	=	6299 s$^{-1}$;	$\omega_C$	=	${\displaystyle \left[ L C_1 \right]^{-{1\over2}} }$	=	3450 s$^{-1}$
   $\omega_B$	=	${\displaystyle \left[ L \left( C_1 + C_2 \right) \right]^{-{1\over2}} }$	=	2887 s$^{-1}$;	$\omega_D$	=	${\displaystyle \left[ L C_2 \right]^{-{1\over2}} }$	=	5270 s$^{-1}$

4. $LRR$ Circuit Time-Dependence: In the circuit shown, the ${\cal E} = 12$ V battery has negligible internal resistance, the inductance of the coil is $L = 0.12$ H and the resistances are $R_1 = 120$ $\Omega$ and $R_2 = 70$ $\Omega$. The switch S is closed for several seconds, then opened. Make a quantitatively labelled graph with an abscissa of time (in milliseconds) showing the potential of point A with respect to ground, just before and then for 10 ms after the opening of the switch. Show also the variation of the potential at point B over the same time period.
   
   ANSWER: After $S$ has been closed for a long time, $dI/dt = 0$ and $L$ acts like a plain wire. Then both resistors have the same potential drop as the battery: ${\cal E} = I_1 R_1 = I_2 R_2$, giving $I_1 = 12$ V $/120\; \Omega = 0.1$ A and $I_2 = 12$ V $/70\; \Omega = 0.1714$ A. Also at that time ${\cal E}_A = {\cal E}_B = {\cal E} = 12$ V. When the switch opens at $t=0$, the isolated right loop is just an $LR$ circuit with $L = 0.12$ H and $R = R_1 + R_2 = 190\; \Omega$. The current $I_2 = 0.1714$ A flowing through $L$ cannot change suddenly but that through $R_1$ immediately reverses direction and is thereafter equal to $I_2 = I$. Subsequently $I(t) = [0.1714$ A$] e^{-t/\tau}$ where $\tau = L/R = 0.12/190 = 6.316\times 10^{-4}$ s. Point $A$ is at a voltage ${\cal E}_A = -IR_1$ with respect to ground and point $B$ is at a voltage ${\cal E}_B = +IR_2$ with respect to ground. These have initial values ${\cal E}_A(0) = -20.57$ V and ${\cal E}_B(0) = +12$ V and both decay exponentially toward zero with time constant $\tau$.