Electrical resistivity $\rho$ of a material is given microscopically by:

• Here, $n$ is the number density of free electrons (which depends solely on the nature of the material), and $\tau$ is the relaxation time between collisions (which depends strongly on temperature due to lattice vibrations).
• Resistivity is an intensive property and does not depend on the geometrical dimensions (length, area, or shape) of the conductor.

• In the absence of an electric field, no net force acts on the free electrons between collisions. Hence, by Newton's first law, they travel at constant velocity along straight line paths between successive collisions with the lattice ions.
• In the presence of a uniform electric field $\vec{E}$, each electron experiences a continuous electrostatic acceleration $\vec{a} = -\frac{e\vec{E}}{m}$. Under this constant transverse acceleration combined with their random thermal velocity, the resulting trajectory between successive collisions is curved (parabolic in general).

Thermal energy $H$ dissipated in a resistor of resistance $R$ under potential difference $V$ for time $t$ is:

Rearranging for resistance $R$:

Given values:

• $V = 20\,\text{V}$
• $t = 20\,\text{s}$
• $H = 800\,\text{J}$
  Substituting:
  $$R = \frac{(20)^2 \times 20}{800} = \frac{400 \times 20}{800} = \frac{8000}{800} = 10\,\Omega$$

• Radius of the wire:
  $$r = \frac{D}{2} = \frac{1.4\,\text{mm}}{2} = 0.7\,\text{mm} = 7 \times 10^{-4}\,\text{m}$$
• Area of cross-section of wire:
  $$A = \pi r^2 = \frac{22}{7} \times (7 \times 10^{-4}\,\text{m})^2 = \frac{22}{7} \times 49 \times 10^{-8} = 154 \times 10^{-8}\,\text{m}^2$$
• Total length $L$ of wire from resistance formula $R = \rho \frac{L}{A}$:
  $$L = \frac{R A}{\rho} = \frac{4\,\Omega \times 154 \times 10^{-8}\,\text{m}^2}{2 \times 10^{-7}\,\Omega\cdot\text{m}} = 2 \times 15.4 = 30.8\,\text{m}$$
• Circumference of one circular turn of coil of radius $R_c = 7\,\text{cm} = 0.07\,\text{m}$:
  $$C = 2\pi R_c = 2 \times \frac{22}{7} \times 0.07\,\text{m} = 0.44\,\text{m}$$
• Total number of turns $N$:
  $$N = \frac{L}{C} = \frac{30.8\,\text{m}}{0.44\,\text{m}} = 70$$