PHYS 208 Lecture 20

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Inductance

An inductor is any device that produces a magnetic field.

Mutual Inductance

The current in one circuit affects emf in the other through magnetic fields

${\mathrm{\Phi }}_2=M_{21}{i}_1$, where $M$ is the mutual inductance of loop 2 due to loop 1

$M$ has units of Henry [V / (A / s) = V · s / A = H]

$M=\frac{N\mathrm{\Phi }}{I_{other}}=-\frac{{ℰ}}{\frac{\mathrm{d}{I}_{other}}{\mathrm{d}t}}$

Example

(a) Calculate the inductance of a very long solenoid that has $n$ turns of wire per meter of radius $a$ carrying a current of $i$. there is another coil with $N$ turns of radius $R$ around a section of the solenoid.

$\begin{array}{rlr}{B}_{solenoid}& ={\mu }_{0}i(t)n& \text{along axis by RHR}\\ {\mathrm{\Phi }}_R& =N{\mathrm{\Phi }}_{oneturn}=N\int \overrightarrow{B}\cdot \mathrm{b}\overrightarrow{A}\\ & =NB(t)(\pi a^2)& \overrightarrow{B}\text{ from solenoid is only flux through large loop}\\ & =N{\mu }_{0}ni(t)(\pi a^2)\\ M& =\frac{Ph{i}_R}{i(t)}\\ & =N{\mu }_{0}Nn(\pi a^2)\end{array}$

For N = 100 turns, n = 1000 turns/meter, a = 10 cm:

M = 4 π² × 10⁻⁶ Henrys

(b) What is the emf induced in the R-loop if the current in the solenoid changes at a rate of $\frac{\mathrm{d}i}{\mathrm{d}t}$?

${ℰ}=-\frac{M}{\mathrm{d}i/\mathrm{d}t}$

Another Example

What is the inductance between a toroid with $N_T$ turns, an inner radius of $a$, an outer radius of $b$, and a height (thickness) of $c$ and a loop of radius $R$ around the toroid?

Recall that the magnetic field of a solenoid at a distance $r$ from the center (hole) of the donut is given by $B(r)=\frac{{\mu }_{0}iN}{2\pi r}$ (B-field is non-uniform)

$\begin{array}{rl}{\mathrm{\Phi }}_{1loop}& =\int \mathrm{B}\cdot \mathrm{d}\overrightarrow{A}=\underset{a}{\overset{b}{\int }}\left(\frac{{\mu }_{0}i{N}_T}{2\pi r}\right)\left(c\mathrm{d}r\right)\\ & =\frac{{\mu }_{0}i{N}_{T}c}{2\pi }\underset{a}{\overset{b}{\int }}\frac{\mathrm{d}r}{r}\\ & =\frac{{\mu }_{0}i{N}_{T}c}{2\pi }\ln \left(\frac{b}{a}\right)\\ M& =\frac{{\mathrm{\Phi }}_{tot}}{i(t)}=\frac{{\mu }_0{N}_{T}Nc}{2\pi }\ln \left(\frac{b}{a}\right)\end{array}$

Self Inductance

a circuit that produces a magnetic field affects its own magnetic flux and thus its own emf. The induced emf counters the emf of the system and is called "back emf"

${ℰ}=-L\frac{\mathrm{d}I}{\mathrm{d}t}$

for $N$ turns,

$L=\frac{N{\mathrm{\Phi }}_{oneturn}}{I}$

Example

Self-inductance of toroid in #Another Example

$\begin{array}{rl}L& =\frac{{\mathrm{\Phi }}_{tot}}{I}=\frac{N{\mathrm{\Phi }}_{1turn}}{I}\\ & =\frac{{\mu }_0{N}^{2}c}{2\pi }\ln \left(\frac{b}{a}\right)\end{array}$

Magnetic Energy

Suppose we start with an inductor that initially has no current. As we add current, the back emf counters the addition, so work is being done to produce the magnetic field. Therefore, the magnetic field of an inductor stores energy.

Work done by power supply:

$\frac{\mathrm{d}W}{\mathrm{d}t}=Power=Vi=Li\frac{\mathrm{d}i}{\mathrm{d}t}$

So the work done is

$\begin{array}{rl}dW& =Li\mathrm{d}i\\ W& =\int dW=\frac{1}{2}L{I}^2\end{array}$

Example

Find the energy stored in a finite solenoid of length $L$ with a radius $R$, a current $I$, and $N$ turns

$\begin{array}{rl}B& ={\mu }_{0}i\frac{N}{L}\\ {\mathrm{\Phi }}_{1loop}& =\frac{{\mu }_{0}iN}{L}(\pi R^2)\\ L& =\frac{N^2{\mu }_0\pi R^2}{L}\\ U_B& =\frac{1}{2}\left(\frac{N^2{\mu }_0\pi R^2}{L}\right)I^2\\ & =\frac{1}{2}{B}^2\frac{L\pi R^2}{{\mu }_0}\\ \frac{U_B}{\pi R^{2}L}=u_B& =\frac{1}{2}\frac{B^2}{{\mu }_0}\end{array}$

Recall from our study of electric fields that the potential energy per unit volume of an electric field is

$u_E=\frac{1}{2}{ϵ}_0{E}^2$