Exercises — Self-inductance L, mutual inductance M

Hum kya use karenge: definition $L=N\Phi /I$ (flux linkage $N\Phi$, sirf $\Phi$ nahi). Kyun: inductance hota hai "flux linkage per amp" — ek pure geometry number. $L=\frac{N\Phi }{I}=\frac{200\times 3\times 10^{-4}}{5}=\frac{6\times 10^{-2}}{5}=1.2\times 10^{-2}\text{H}=12\text{mH}.$ Answer: $L=12\text{mH}$.

Hum kya use karenge: EMF form $\epsilon =-LdI/dt$, toh $L=\epsilon /(dI/dt)$. Kyun: kisi quantity ki units uski defining equation se match karni chahiye. $[L]=\frac{[\epsilon ]}{[dI/dt]}=\frac{\text{V}}{\text{A/s}}=\frac{\text{V⋅s}}{\text{A}}.$ Answer: $1\text{H}=1\text{V⋅s/A}$ (equivalently $\text{Wb/A}$).

Hum kya use karenge: $L={\mu }_0{N}^{2}A/\ell$. Kyun: yeh ek long solenoid ke liye derived box hai; seedha plug-in. $L=\frac{(4\pi \times 10^{-7})(800{)}^2(5\times 10^{-4})}{0.40}.$ Pieces mein build karo: $N^2=6.4\times 10^5$; $N^{2}A=6.4\times 10^5\times 5\times 10^{-4}=320$; $\ell$ se divide karo: $320/0.40=800$; ${\mu }_0$ se multiply karo: $L=(4\pi \times 10^{-7})(800)=3200\pi \times 10^{-7}\approx 1.005\times 10^{-3}\text{H}.$ Answer: $L\approx 1.0\text{mH}$.

Hum kya use karenge: $\epsilon =-LdI/dt$. Kyun: current constant rate se change ho raha hai, toh $dI/dt$ ek single number hai. $\frac{dI}{dt}=\frac{4-0}{2\times 10^{-3}}=2000\text{A/s}.$ $\epsilon =-(1.0\times 10^{-3})(2000)=-2.0\text{V}.$ Sign ka matlab: current badh raha hai, toh Lenz's law ke hisaab se induced EMF rise ka oppose karta hai (current ko rokne ki koshish karta hai). Magnitude $=2.0\text{V}$.

Hum kya use karenge: $\epsilon =-LdI/dt$ aur $\frac{d}{dt}\sin (\omega t)=\omega \cos (\omega t)$. Derivative kyun: EMF change ko track karta hai, isliye hum current ko differentiate karte hain. $\frac{dI}{dt}=2\cdot 400\cos (400t)=800\cos (400t)\text{A/s}.$ $\epsilon =-(50\times 10^{-3})(800\cos 400t)=-40\cos (400t)\text{V}.$ Peak $|\epsilon |=40\text{V}$. Figure dekho: EMF ($\cos$) tab peak karta hai jab current ($\sin$) zero cross karta hai — wahan current sabse tezi se change ho raha hota hai. Current ke peaks par slope zero hota hai, toh $\epsilon =0$.

Hum kya use karenge: $M={\mu }_0\frac{N_1{N}_2}{\ell }A$ (parent note mein derive kiya gaya hai). Kyun: sirf inner solenoid field produce karta hai; uska flux $B_{1}A$ outer ke har $N_2$ turns mein se guzarta hai. Figure dekho. $M=(4\pi \times 10^{-7})\frac{(1000)(15)}{0.50}(8\times 10^{-4}).$ Pieces: $N_1{N}_2=1.5\times 10^4$; $/\ell =3\times 10^4$; $\times A=3\times 10^4\times 8\times 10^{-4}=24$; $\times {\mu }_0$: $M=(4\pi \times 10^{-7})(24)\approx 3.02\times 10^{-5}\text{H}\approx 30\mu \text{H}.$ Answer: $M\approx 30\mu \text{H}$.

Hum kya use karenge: $M=k\sqrt{L_1{L}_2}\Rightarrow k=\frac{M}{\sqrt{L_1{L}_2}}$. Kyun: $k$ ek coil ke flux ka wo fraction hai jo doosre tak pahunchta hai; yeh kabhi $1$ se zyada nahi ho sakta. $\sqrt{L_1{L}_2}=\sqrt{8\times 2}\text{mH}=\sqrt{16}\text{mH}=4\text{mH}.$ $k=\frac{3}{4}=\mathrm{0.75.}$ Kyunki $0\le 0.75\le 1$ hai, yeh physically allowed hai ($75\%$ flux coupling). Answer: $k=0.75$, allowed.

Hum kya use karenge: $U=\frac{1}{2}L{I}^2$; energy added $=U_f-U_i$. Kyun: stored magnetic energy current ke square par depend karti hai, toh change squares ka difference hota hai — $\frac{1}{2}L(\Delta I{)}^2$ nahi. $\Delta U=\frac{1}{2}L\left(I_f^2-I_i^2\right)=\frac{1}{2}(1.0\times 10^{-3})(4^2-1^2).$ $=\frac{1}{2}(1.0\times 10^{-3})(16-1)=\frac{1}{2}(1.0\times 10^{-3})(15)=7.5\times 10^{-3}\text{J}.$ Answer: $\Delta U=7.5\text{mJ}$.

Hum kya use karenge: $L={\mu }_0{N}^{2}A/\ell$ ko $N$ ke liye solve karo. Kyun: $L$ $N^2$ ke saath badhta hai, toh hum square root se invert karte hain. $N=\sqrt{\frac{L\ell }{{\mu }_{0}A}}=\sqrt{\frac{(2.0\times 10^{-3})(0.20)}{(4\pi \times 10^{-7})(4\times 10^{-4})}}.$ Numerator: $2.0\times 10^{-3}\times 0.20=4.0\times 10^{-4}$. Denominator: $4\pi \times 10^{-7}\times 4\times 10^{-4}=16\pi \times 10^{-11}\approx 5.027\times 10^{-10}$. $N=\sqrt{\frac{4.0\times 10^{-4}}{5.027\times 10^{-10}}}=\sqrt{7.958\times 10^5}\approx 892.$ Answer: $N\approx 892$ turns.

Step 1 — $M$ nikalo. $k=1$ ke saath: $M=\sqrt{L_1{L}_2}=\sqrt{4\times 9}\text{mH}=6\text{mH}$. Step 2 — current differentiate karo. $\frac{d{I}_1}{dt}=0.5\cdot 2000\cos (2000t)=1000\cos (2000t)\text{A/s}.$ Step 3 — mutual EMF. ${\epsilon }_2=-M\frac{d{I}_1}{dt}=-(6\times 10^{-3})(1000\cos 2000t)=-6\cos (2000t)\text{V}.$ Peak $|{\epsilon }_2|=6\text{V}$. Answer: $M=6\text{mH}$, peak EMF $=6\text{V}$.

Hum kya use karenge: $U=\frac{1}{2}L{I}^2$. $U=\frac{1}{2}(2.0\times 10^{-3})(3{)}^2=\frac{1}{2}(2.0\times 10^{-3})(9)=9.0\times 10^{-3}\text{J}=9.0\text{mJ}.$ Scaling check karo: $I$ ko double karke $6\text{A}$ karne par $\frac{1}{2}(2.0\times 10^{-3})(36)=36\text{mJ}$ milega — exactly $4\times$, jo confirm karta hai $U\propto I^2$. Answer: $U=9.0\text{mJ}$.

Flux se $L$ ke liye defining relation kya hai? ::: $L=N\Phi /I$ (flux linkage per amp). L1·Q1 mein $L=\Phi /I$ kyun nahi? ::: Total linkage $N\Phi$ use karni chahiye; $N$ turns har ek $\Phi$ carry karta hai. L3·Q1 mein $|\epsilon |$ kab maximum hota hai? ::: Jab $I=0$ ho (current zero cross kar raha ho, sabse tezi se change ho raha ho). L5·Q1 mein square root kyun? ::: Kyunki $L\propto N^2$, toh $N=\sqrt{L\ell /({\mu }_{0}A)}$. L4·Q2 mein use ki gayi energy change formula? ::: $\Delta U=\frac{1}{2}L(I_f^2-I_i^2)$, $\frac{1}{2}L(\Delta I{)}^2$ nahi.