Drift velocity $\vec{v}_d$

$\vec{J}=-n e \vec{v}_\alpha$

$\left|v_a\right|$ is small $\sim a$ few $\mathrm{mm} / \mathrm{s}$.

$\vec{J}, \vec{E}$ Ohm's Law

$\vec{J}=\sigma \vec{E}$

$\vec{E}=\rho \vec{J}$

$\sigma=\text { Conductivily }$

$\rho=\text { Resistivily }$

$R=\frac{\Delta V}{I} \quad \text { Ohms}$

$R \propto$ Length

$\propto \frac{1}{\text { Area of cross Section}}$

6. Microscopic view of Ohm's Law.

$\tau$: Relaxation time

$\sigma=\frac{\eta{e}^2 \tau}{m}$

$\tau \sim 10^{-14}$

$v_d=\frac{e E \tau}{m}$

$R \propto {L}$

$\propto \frac {1}{A}$

Ease with which a charge carrier moves inside a solid when subjected to an electric field.

$\mu = \frac{\left|v_d\right|}{|E|}=\frac{m / s}{v / m} =\frac{m^2}{v-s}$

$v_d=\frac{e E \tau}{m}$

$\mu = \frac{e \tau}{m} \sim \frac{10^{-19}+10^{-14}}{10^{-30}}$

$\sim 10^{-3}$ to $10^{-4} \mathrm{~m^2}/vs$

$\sigma=\frac{n e^2 \tau}{m}=e n \mu \quad$

$\mu=\frac{e \tau}{m}$

$\sigma=e\left[n \mu_e+p \mu_n\right]$

Cu:

$n =8.5 \times 10^{28} / \mathrm{m}^2$

$\sigma =5.8 \times 10^7 \mathrm{~s} / \mathrm{m}$

$\mu=\frac{\sigma}{n e} =\frac{5.8 \times 10^7}{8.5 \times 10^{28} \times 1.6 \times 10^{-19}}=0.0042 \mathrm{m}^2 / \mathrm{v}-\mathrm{s} =42 \mathrm{cm}^2 / \mathrm{v}-\mathrm{s}$

$v_d =\mu E$

$E=10 \mathrm{V}$

$=4.3 \times 10^{-3} \times 10$.

$= 4.2 \mathrm{~cm} / \mathrm{s}$

Deviation from linearity - ohm's law.

Does not depend on sign of $v$.

I : Non Linear region

II: Negative resistance

1. Wire bound

Alloys of manganin. constantine, Nichrome wires

fraction of $\Omega$ to several hundred ohms.

2. Carbon Resistance.

Color coding

Significant Figures

multiplier

(Missing)

230 = $23 \times 10^{1}$ ROB

4 Bands

B. B. Roy of Great Britain has a very Good Wife

$47 \times 10^2 \pm$ 10 %

$= 4.7 \mathrm{k} \Omega \pm$ 10 %

Resistance of a sample depends on temperature

Thermal Expansion $\Delta L=\alpha L \cdot \Delta t$

• $\rho-\rho_0= \rho_0 \alpha\left(T-T_0\right)$

• $\rho=\rho_0\left[1+\alpha\left(T-T_0\right)\right]$

$\alpha$ : Temperature coefficient

$\alpha = \frac{1}{\rho_0} \frac{(\rho-\rho_0)}{T-T_0}$

$=\frac{1}{\rho_0} \frac{\Delta \rho}{\Delta t}$

Al. 3 valence electrons.

at $0^{\circ} \mathrm{C}$

$\rho = 2.7 \times 10^{-8} \Omega \mathrm{~m}$.

$\alpha=$ temperature coefficient = $4.3 \times 10^{-3} \mathrm{~K}^{-1}$

calculate $\rho$ at Room Temp $25^{0} \mathrm{C}$

$\rho_{25} =\rho_0(1+\alpha . \Delta T)$

$=\rho_0\left(1+4.3 \times 10^{-3} \times 25\right) =\rho_0(1.1)$

$=2.7 \times 1.1=2.97 \times 10^{-8} \Omega / \mathrm{m}$

Al: Atomic mass 27.

mass density $2700 \mathrm{~kg} / \mathrm{m}^3$

$\frac{2700}{27 \times 10^{-3}} \times 6 \times 10^{23} \approx 6 \times 10^{28}$

$n=1.8 \times 10^{29} / \mathrm{m}^3$

electron density, $\sigma=\frac{n e^2 \tau}{m}$

$\tau \sim 7 \times 10^{-15} \mathrm{s}$, $\mu=\frac{\sigma}{n e}=12 \mathrm{cm}^2 / \mathrm{v}-\mathrm{s}$

$\lambda=14.4 \mathrm{~nm}$

Pt-Resistance thermometer

$T=0^{\circ} \mathrm{C} \quad R=5 \Omega$

$T=100^{\circ} \mathrm{C} \quad R=5.4 \Omega$

Unknown temperature Heat bath R=6 $\Omega$

$\rho=\rho_0[1+\alpha \Delta T] 100^{\circ} \mathrm{C}$

$R=R_0[1+\alpha \Delta T]$

$5.4 \Omega=5 \Omega[1+\alpha \cdot 100]$

$\alpha =8 \times 10^{-4}/^{\circ} C$

$6=5\left[1+8 \times 10^{-4} \Delta {T}\right]$

$\Delta T=250^{\circ} \mathrm{C}$

$\rho_0=1.7 \times 10^{-8} \Omega \mathrm{m}$

? $T$ at which lts resistivity will double?

$\rho_T=\rho_0(1+\alpha T)$