Question

Current I (= 1 A) is passing through a copper rod (n = \(8.5 \times 10^{28} \, \text{m}^{-3}\)) of varying cross-sections as shown in the figure. The areas of cross-section at points A and B along its length are \(1.0 \times 10^{-7} \, \text{m}^2\) and \(2.0 \times 10^{-7} \, \text{m}^2\) respectively. Calculate:

(I) the ratio of electric fields at points A and B.
(II) the drift velocity of free electrons at point B.

Hint:

In problems involving current through varying cross-sections, use the relationship between current, drift velocity, and electric field to solve for quantities of interest. Remember that the electric field is inversely proportional to the cross-sectional area.

Answer 1

Given a current \( I = 1 \, \text{A} \) flowing through a copper rod, the relationship between current, free electron density \( n \), cross-sectional area \( A \), electron charge \( e \), and drift velocity \( v_d \) is defined as: \[ I = n A e v_d \]
The electric field \( E \) is related to the current density \( J \) and resistivity \( \rho \) by: \[ E = \rho J \]
Since current density \( J = I/A \), both the electric field \( E \) and drift velocity \( v_d \) are inversely proportional to the cross-sectional area \( A \). Therefore, the ratio of electric fields at points A and B is: \[ \frac{E_A}{E_B} = \frac{A_B}{A_A} \] With \( A_A = 1.0 \times 10^{-7} \, \text{m}^2 \) and \( A_B = 2.0 \times 10^{-7} \, \text{m}^2 \): \[ \frac{E_A}{E_B} = \frac{2.0 \times 10^{-7}}{1.0 \times 10^{-7}} = 2 \] The ratio of electric fields at points A and B is thus \( \frac{E_A}{E_B} = 2 \).
To calculate the drift velocity at point B, we use the equation: \[ v_{dB} = \frac{I}{n A_B e} \] Substituting the values \( I = 1 \, \text{A} \), \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \), \( A_B = 2.0 \times 10^{-7} \, \text{m}^2 \), and \( e = 1.6 \times 10^{-19} \, \text{C} \): \[ v_{dB} = \frac{1}{8.5 \times 10^{28} \times 2.0 \times 10^{-7} \times 1.6 \times 10^{-19}} \approx 3.7 \times 10^{-4} \, \text{m/s} \] The drift velocity at point B is approximately \( 3.7 \times 10^{-4} \, \text{m/s} \).