## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$7.2\cdot10^{-4}$ H
At position $a$, we can assume that the switch has been closed for a long time which means that there is a steady current in the circuit (a current that is not changing). We can find the current at this instant by using Kirchoff's law: $\varepsilon - IR - \displaystyle L\frac{dI}{dt} = 0$ $3 - I(50) - L(0) = 0\qquad\rightarrow\qquad3-I(50)=0$ $\displaystyle I = \frac{\varepsilon}{R} = \frac{3}{50}$ Amps As soon as the switch is moved from position $a$ to position $b$, the steady current in the system at position $a$ is the initial current as soon as we move to position $a$. $I_0 = \displaystyle \frac{\varepsilon}{R}= \frac{3}{50}$ Amps Given that after $5$ microseconds the energy in the inductor is half of the initial energy, and knowing that $U_L = \displaystyle \frac{1}{2}LI^2$ and for $LR$ circuits $I(t) = I_0e^{-t/(L/R)}$ we can solve for the inductance of the inductor. $\displaystyle \frac{1}{2} = \frac{U_f}{U_0} = \frac{\frac{1}{2}LI_f^2}{\frac{1}{2}LI_0^2} = \frac{I_f^2}{I_0^2} = \frac{(I_0e^{-t/(L/R)})^2}{(\varepsilon/R)^2}$ $\displaystyle \frac{(\varepsilon/R)^2}{2} = (I_0e^{-t/(L/R)})^2$ $\displaystyle \sqrt{\frac{(\varepsilon/R)^2}{2}} = I_0e^{-t/(L/R)}$ $\displaystyle \frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0} = e^{-t/(L/R)}$ $\displaystyle \ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0}) = -t/(L/R) = \frac{-t}{L/R} = \frac{-Rt}{L}$ $\displaystyle \frac{\ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0})}{-Rt} = \frac{1}{L}$ $\displaystyle \frac{-Rt}{\ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0})} = L$ After finding this awful looking equation, we see that our answer is $L = \displaystyle \frac{-Rt}{\ln(\frac{\sqrt{\frac{(\varepsilon/R)^2}{2}}}{I_0})} = \displaystyle \frac{-(50)(5\cdot10^{-6})}{\ln(\frac{\sqrt{\frac{(3/50)^2}{2}}}{(3/50)})} = 7.2\cdot10^{-4}$ H