Numerical Problems Based on Drift Velocity for Class 12 Physics

Given parameters: Area $A = 10^{-6}\text{ m}^2$, Drift velocity $v_d = 10^{-4}\text{ ms}^{-1}$, Electron density $n = 8.5 \times 10^{28}\text{ m}^{-3}$, and elementary charge $e = 1.6 \times 10^{-19}\text{ C}$.

The relation between electric current and drift velocity is given by:

Substitute the given values into the formula:

Current $I = 2.7\text{ A}$, Area $A = 2.5 \times 10^{-7}\text{ m}^2$, Electron density $n = 9 \times 10^{28}\text{ m}^{-3}$, Charge $e = 1.6 \times 10^{-19}\text{ C}$.

Using the formula for current $I = nAev_d$, rearrange to solve for $v_d$:

Converting to mm/s:

Current $I = 1.8\text{ A}$, Area $A = 0.5\text{ mm}^2 = 0.5 \times 10^{-6}\text{ m}^2$, $n = 8.8 \times 10^{28}\text{ m}^{-3}$.

1. Calculate current density ($J$):

2. Calculate drift speed ($v_d$):

Using the relation $J = nev_d$:

Potential difference $V = 5\text{ V}$, Length $l = 0.1\text{ m}$, Drift speed $v_d = 2.5 \times 10^{-4}\text{ ms}^{-1}$, $n = 8 \times 10^{28}\text{ m}^{-3}$.

Current density is given by $J = nev_d$ and also by Ohm’s law in microscopic form, $J = \frac{E}{\rho}$, where $E$ is the electric field and $\rho$ is resistivity.

Equating the two expressions for current density:

Rounding to standard figures:

Diameter $d = 1.0\text{ mm} = 10^{-3}\text{ m}$. Current $I = 0.2\text{ A}$. Since 1 charge carrier is associated with each atom, the number density of electrons $n = 8.4 \times 10^{28}\text{ m}^{-3}$.

First, calculate the cross-sectional area ($A$):

Calculate the drift velocity ($v_d$):

Current $I = 2\text{ A}$, Length $l = 4\text{ m}$, Area $A = 1\text{ mm}^2 = 10^{-6}\text{ m}^2$, $n = 10^{29}\text{ m}^{-3}$.

The time taken ($t$) is given by distance divided by velocity: $t = \frac{l}{v_d}$.

From the current formula, $v_d = \frac{I}{nAe}$. Substitute this into the time equation:

Charge $q = 10\text{ C}$, Time $t = 5\text{ mins} = 300\text{ s}$, Radius $r = 1\text{ mm} = 10^{-3}\text{ m}$.
Density $n = 5 \times 10^{22}\text{ cm}^{-3} = 5 \times 10^{22} \times 10^6\text{ m}^{-3} = 5 \times 10^{28}\text{ m}^{-3}$.

1. Calculate Current ($I$):

2. Calculate Drift Velocity ($v_d$):

First, find the cross-sectional area $A = \pi r^2 = \pi(10^{-3})^2 = 3.14 \times 10^{-6}\text{ m}^2$.

Current $I = 10\text{ A}$ (same in series).

(i) Current density in Cu wire:

Diameter of Cu wire $d_{Cu} = 0.16\text{ cm} = 1.6 \times 10^{-3}\text{ m}$.
Area $A_{Cu} = \frac{\pi (1.6 \times 10^{-3})^2}{4} = 2.01 \times 10^{-6}\text{ m}^2$.

(ii) Drift velocity in Al wire:

Diameter of Al wire $d_{Al} = 0.25\text{ cm} = 2.5 \times 10^{-3}\text{ m}$.
Area $A_{Al} = \frac{\pi (2.5 \times 10^{-3})^2}{4} = 4.908 \times 10^{-6}\text{ m}^2$.

Current $I = 30\text{ A}$, Area $A = 2 \times 10^{-6}\text{ m}^2$.

1. Find number density ($n$):

Density $d = 8.9\text{ g cm}^{-3} = 8.9 \times 10^3\text{ kg m}^{-3}$.
Atomic mass $M = 63\text{ g} = 63 \times 10^{-3}\text{ kg}$.

2. Calculate Drift Velocity ($v_d$):

3. Calculate RMS Velocity ($v_{rms}$):

Temperature $T = 27^\circ\text{C} = 300\text{ K}$, Mass of electron $m_e = 9.1 \times 10^{-31}\text{ kg}$.

The RMS velocity of electrons due to thermal motion is significantly larger than the drift velocity.

Current $I = 10\text{ A}$, Area $A = 3.14 \times 10^{-6}\text{ m}^2$.
Assuming 1 free electron per atom for silver, find number density $n$:

Now, compute the drift velocity ($v_d$):