Exercises — Self-inductance L, mutual inductance M

What we use: the definition $L=N\Phi /I$ (flux linkage $N\Phi$, not just $\Phi$). Why: inductance is "flux linkage per amp" — a 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: $L=12\text{mH}$.

What we use: the EMF form $\epsilon =-LdI/dt$, so $L=\epsilon /(dI/dt)$. Why: the units of a quantity must match its defining equation. $[L]=\frac{[\epsilon ]}{[dI/dt]}=\frac{\text{V}}{\text{A/s}}=\frac{\text{V⋅s}}{\text{A}}.$ Answer: $1\text{H}=1\text{V⋅s/A}$ (equivalently $\text{Wb/A}$).

What we use: $L={\mu }_0{N}^{2}A/\ell$. Why: this is the derived box for a long solenoid; pure plug-in. $L=\frac{(4\pi \times 10^{-7})(800{)}^2(5\times 10^{-4})}{0.40}.$ Build it in pieces: $N^2=6.4\times 10^5$; $N^{2}A=6.4\times 10^5\times 5\times 10^{-4}=320$; divide by $\ell$: $320/0.40=800$; times ${\mu }_0$: $L=(4\pi \times 10^{-7})(800)=3200\pi \times 10^{-7}\approx 1.005\times 10^{-3}\text{H}.$ Answer: $L\approx 1.0\text{mH}$.

What we use: $\epsilon =-LdI/dt$. Why: the current changes at a constant rate, so $dI/dt$ is a single number. $\frac{dI}{dt}=\frac{4-0}{2\times 10^{-3}}=2000\text{A/s}.$ $\epsilon =-(1.0\times 10^{-3})(2000)=-2.0\text{V}.$ Sign meaning: current is rising, so by Lenz's law the induced EMF opposes the rise (tries to hold current back). Magnitude $=2.0\text{V}$.

What we use: $\epsilon =-LdI/dt$ and $\frac{d}{dt}\sin (\omega t)=\omega \cos (\omega t)$. Why the derivative: EMF tracks change, so we differentiate the current. $\frac{dI}{dt}=2\cdot 400\cos (400t)=800\cos (400t)\text{A/s}.$ $\epsilon =-(50\times 10^{-3})(800\cos 400t)=-40\cos (400t)\text{V}.$ Peak $|\epsilon |=40\text{V}$. Look at the figure: EMF ($\cos$) peaks when the current ($\sin$) crosses zero — that is where the current is changing fastest. At the current peaks the slope is zero, so $\epsilon =0$.

What we use: $M={\mu }_0\frac{N_1{N}_2}{\ell }A$ (derived in the parent note). Why: only the inner solenoid produces field; its flux $B_{1}A$ threads each of the $N_2$ outer turns. See the figure. $M=(4\pi \times 10^{-7})\frac{(1000)(15)}{0.50}(8\times 10^{-4}).$ 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: $M\approx 30\mu \text{H}$.

What we use: $M=k\sqrt{L_1{L}_2}\Rightarrow k=\frac{M}{\sqrt{L_1{L}_2}}$. Why: $k$ is the fraction of one coil's flux that reaches the other; it can never exceed $1$. $\sqrt{L_1{L}_2}=\sqrt{8\times 2}\text{mH}=\sqrt{16}\text{mH}=4\text{mH}.$ $k=\frac{3}{4}=\mathrm{0.75.}$ Since $0\le 0.75\le 1$, it is physically allowed ($75\%$ flux coupling). Answer: $k=0.75$, allowed.

What we use: $U=\frac{1}{2}L{I}^2$; energy added $=U_f-U_i$. Why: stored magnetic energy depends on the square of current, so the change is a difference of squares — not $\frac{1}{2}L(\Delta I{)}^2$. $\Delta U=\frac{1}{2}L\left(I_f^2-I_i^2\right)=\frac{1}{2}(1.0\times 10^{-3})(4^2-1^2).$ $=\frac{1}{2}(1.0\times 10^{-3})(16-1)=\frac{1}{2}(1.0\times 10^{-3})(15)=7.5\times 10^{-3}\text{J}.$ Answer: $\Delta U=7.5\text{mJ}$.

What we use: solve $L={\mu }_0{N}^{2}A/\ell$ for $N$. Why: $L$ grows as $N^2$, so we invert with a square root. $N=\sqrt{\frac{L\ell }{{\mu }_{0}A}}=\sqrt{\frac{(2.0\times 10^{-3})(0.20)}{(4\pi \times 10^{-7})(4\times 10^{-4})}}.$ 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 — find $M$. With $k=1$: $M=\sqrt{L_1{L}_2}=\sqrt{4\times 9}\text{mH}=6\text{mH}$. Step 2 — differentiate the current. $\frac{d{I}_1}{dt}=0.5\cdot 2000\cos (2000t)=1000\cos (2000t)\text{A/s}.$ 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 $=6\text{V}$.

What we use: $U=\frac{1}{2}L{I}^2$. $U=\frac{1}{2}(2.0\times 10^{-3})(3{)}^2=\frac{1}{2}(2.0\times 10^{-3})(9)=9.0\times 10^{-3}\text{J}=9.0\text{mJ}.$ Check the scaling: doubling $I$ to $6\text{A}$ would give $\frac{1}{2}(2.0\times 10^{-3})(36)=36\text{mJ}$ — exactly $4\times$, confirming $U\propto I^2$. Answer: $U=9.0\text{mJ}$.

Defining relation for $L$ from flux? ::: $L=N\Phi /I$ (flux linkage per amp). Why is L1·Q1 not $L=\Phi /I$? ::: You must use total linkage $N\Phi$; the $N$ turns each carry $\Phi$. In L3·Q1, when is $|\epsilon |$ maximal? ::: When $I=0$ (current crossing zero, changing fastest). Why square root in L5·Q1? ::: Because $L\propto N^2$, so $N=\sqrt{L\ell /({\mu }_{0}A)}$. Energy change formula used in L4·Q2? ::: $\Delta U=\frac{1}{2}L(I_f^2-I_i^2)$, not $\frac{1}{2}L(\Delta I{)}^2$.