1. **Problem statement:** We have a cylindrical coil (Zylinderspule) with length $l = 20$ cm, number of turns $N = 500$, and diameter $d = 5.2$ cm. A time-varying current $I(t)$ flows through it, and we want to find the self-induced voltage $U_L(t)$ as a function of time. 2. **Relevant formula:** The self-inductance $L$ of a solenoid is given by: $$L = \mu_0 \frac{N^2 A}{l}$$ where $\mu_0 = 4\pi \times 10^{-7}$ H/m is the permeability of free space, $A$ is the cross-sectional area, and $l$ is the length. The induced voltage (self-induction voltage) is: $$U_L = -L \frac{dI}{dt}$$ 3. **Calculate parameters:** - Convert length to meters: $l = 20\text{ cm} = 0.20$ m - Radius $r = \frac{d}{2} = \frac{5.2}{2} = 2.6$ cm = 0.026 m - Cross-sectional area: $$A = \pi r^2 = \pi (0.026)^2 = \pi \times 0.000676 = 0.002124\, \text{m}^2$$ 4. **Calculate inductance $L$:** $$L = 4\pi \times 10^{-7} \times \frac{500^2 \times 0.002124}{0.20}$$ $$= 4\pi \times 10^{-7} \times \frac{250000 \times 0.002124}{0.20}$$ $$= 4\pi \times 10^{-7} \times \frac{531}{0.20}$$ $$= 4\pi \times 10^{-7} \times 2655$$ $$= (4 \times 3.1416) \times 10^{-7} \times 2655 = 12.5664 \times 10^{-7} \times 2655$$ $$= 3.335 \times 10^{-3} \text{ H} = 3.335 \text{ mH}$$ 5. **Current $I(t)$ description:** - Interval I (0 to 10 ms): $I = 100$ mA (constant) - Interval II (10 to 20 ms): $I$ decreases linearly from 100 mA to -150 mA - Interval III (20 to 30 ms): $I = -150$ mA (constant) - Interval IV (30 to 35 ms): $I$ increases linearly from -150 mA to 50 mA - Interval V (35 to 50 ms): $I = 50$ mA (constant) 6. **Calculate $\frac{dI}{dt}$ for each interval:** - I, III, V: $\frac{dI}{dt} = 0$ (constant current) - II: Linear decrease from 100 mA to -150 mA over 10 ms $$\frac{dI}{dt} = \frac{-150 - 100}{20 - 10} = \frac{-250}{10} = -25\, \text{mA/ms} = -25 \times 10^{-3} / 10^{-3} = -25 \text{ A/s}$$ - IV: Linear increase from -150 mA to 50 mA over 5 ms $$\frac{dI}{dt} = \frac{50 - (-150)}{35 - 30} = \frac{200}{5} = 40\, \text{mA/ms} = 40 \text{ A/s}$$ 7. **Calculate induced voltage $U_L$ for each interval:** Using $U_L = -L \frac{dI}{dt}$ and $L = 3.335 \times 10^{-3}$ H: - I, III, V: $U_L = 0$ V - II: $$U_L = -3.335 \times 10^{-3} \times (-25) = 0.0834 \text{ V} = 83.4 \text{ mV}$$ - IV: $$U_L = -3.335 \times 10^{-3} \times 40 = -0.1334 \text{ V} = -133.4 \text{ mV}$$ **Final result:** $$U_L(t) = \begin{cases} 0 & 0 \leq t < 10 \text{ ms} \\ 83.4 \text{ mV} & 10 \leq t < 20 \text{ ms} \\ 0 & 20 \leq t < 30 \text{ ms} \\ -133.4 \text{ mV} & 30 \leq t < 35 \text{ ms} \\ 0 & 35 \leq t \leq 50 \text{ ms} \end{cases}$$ This shows the self-induction voltage as a piecewise function depending on the slope of the current.