An ideal inductor of inductance \(L\) is connected to an ac power supply, which provides an emf \(\mathscr{E}(t)=\mathscr{E}_{\mathrm{m}}\) sin \(\omega t\) (a) Write an expression for the current in the inductor as a function of time. [Hint: See Eq. \((20-7) .]\) (b) What is the ratio of the maximum emf to the maximum current? This ratio is called the reactance. (c) Do the maximum emf and maximum current occur at the same time? If not, how much time separates them?

According to Faraday's law of electromagnetic induction, the induced emf in an inductor is related to the rate of change of the current through the inductor. Mathematically, it is expressed as: \[\mathscr{E}(t) = -L \frac{dI(t)}{dt}\] In this exercise, we are given: \[\mathscr{E}(t) = \mathscr{E}_m \sin(\omega t)\]

Plugging the given emf equation into Faraday's law, we get: \[-L \frac{dI(t)}{dt} = \mathscr{E}_m \sin(\omega t)\] Now, we need to solve this differential equation to find the expression for the current \(I(t)\). To do this, we can integrate both sides with respect to time: \[-L \int \frac{dI(t)}{dt} dt = \mathscr{E}_m \int \sin(\omega t) dt\] After integrating and rearranging, we get: \[I(t) = -\frac{\mathscr{E}_m}{\omega L} \cos(\omega t) + C\] To find the constant \(C\), we can use the initial condition that the current is zero when \(t=0\): \[I(0) = -\frac{\mathscr{E}_m}{\omega L} \cos(0) + C = 0\] Solving for \(C\), we find that \(C=0\). Thus, our final expression for the current is: \[I(t) = -\frac{\mathscr{E}_m}{\omega L} \cos(\omega t)\]

Reactance is the ratio of maximum emf to maximum current: \[X_L = \frac{\mathscr{E}_m}{I_m}\] We can find the maximum current \(I_m\) by noting that the maximum value of the cosine function is 1: \[I_m = \frac{\mathscr{E}_m}{\omega L}\] Now, we can calculate the reactance: \[X_L=\frac{\mathscr{E}_m}{I_m} = \omega L\]

To find the time difference, we can look at the time when the maximum values of the emf and current occur. For the given emf function, the maximum occurs when the sine function equals 1, which is when: \[\omega t = \frac{\pi}{2}\] For the current function, the maximum occurs when the cosine function equals -1, which is when: \[\omega t = \pi\] The time difference between these two maximums is: \[\Delta t = \frac{\pi}{\omega} - \frac{\pi}{2 \omega} = \frac{\pi}{2\omega}\] In conclusion: (a) The current in the inductor as a function of time is: \[I(t) = -\frac{\mathscr{E}_m}{\omega L} \cos(\omega t)\] (b) The reactance is: \[X_L = \omega L\] (c) The maximum emf and maximum current do not occur at the same time. The time difference between them is: \[\Delta t = \frac{\pi}{2\omega}\]