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

$\Phi _{2}=M_{21}i_{1}$, 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]

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

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.

${\begin{aligned}B_{solenoid}&=\mu _{0}i(t)n&{\mbox{along axis by RHR}}\\\Phi _{R}&=N\Phi _{one\ turn}=N\int {\vec {B}}\cdot \mathrm {b} {\vec {A}}\\&=NB(t)(\pi a^{2})&{\vec {B}}{\mbox{ from solenoid is only flux through large loop}}\\&=N\mu _{0}ni(t)(\pi a^{2})\\M&={\frac {Phi_{R}}{i(t)}}\\&=N\mu _{0}Nn(\pi a^{2})\end{aligned}}$

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 ${\frac {\mathrm {d} i}{\mathrm {d} t}}$?

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

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} and a loop of radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} around the toroid?

Recall that the magnetic field of a solenoid at a distance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} from the center (hole) of the donut is given by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B(r) = \frac{\mu_0 i N}{2 \pi r}} (B-field is non-uniform)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Phi_{1\ loop} &= \int \mathrm{B} \cdot \mathrm{d}\vec{A} = \int_a^b \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} \int_a^b \frac{\mathrm{d}r}{r} \\ &= \frac{\mu_0 i N_T c}{2 \pi} \ln \left( \frac{b}{a} \right) \\ M &= \frac{\Phi_{tot}}{i(t)} = \frac{\mu_0 N_T N c}{2 \pi} \ln \left( \frac{b}{a} \right) \end{align}}

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"

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{E} = -L \frac{\mathrm{d}I}{\mathrm{d}t}}

for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} turns,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L = \frac{N \Phi_{one\ turn}}{I}}

Example

Self-inductance of toroid in #Another Example

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} L &= \frac{\Phi_{tot}}{I} = \frac{N \Phi_{1\ turn}}{I} \\ &= \frac{\mu_0 N^2 c}{2 \pi} \ln \left( \frac{b}{a} \right) \end{align}}

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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d}W}{\mathrm{d}t} = Power = Vi = Li \frac{\mathrm{d}i}{\mathrm{d}t}}

So the work done is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}dW &= Li \mathrm{d}i \\ W &= \int dW = \frac{1}{2} LI^2\end{align}}

Example

Find the energy stored in a finite solenoid of length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} with a radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} , a current Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} turns

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} B &= \mu_0 i \frac{N}{L} \\ \Phi_{1\ loop} &= \frac{\mu_0 i N}{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{align}}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_E = \frac{1}{2} \epsilon_0 E^2}