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.
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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.
Let the current \( I = 1 \, \text{A} \) pass through the copper rod. The electric field \( E \) and drift velocity \( v_d \) are related to the current by: \[ I = n A e v_d \] where: - \( n \) is the number of free electrons per unit volume, - \( A \) is the cross-sectional area, - \( e \) is the charge of an electron, - \( v_d \) is the drift velocity. Also, the electric field \( E \) is related to the drift velocity by: \[ E = \rho J \] where \( \rho \) is the resistivity of the material and \( J \) is the current density. Now, the drift velocity and electric field are inversely proportional to the area of cross-section, meaning the electric field at point A and point B can be compared as: \[ \frac{E_A}{E_B} = \frac{A_B}{A_A} \] Substituting the areas: \[ \frac{E_A}{E_B} = \frac{2.0 \times 10^{-7}}{1.0 \times 10^{-7}} = 2 \] Thus, the ratio of electric fields at points A and B is: \[ \frac{E_A}{E_B} = 2 \] Now, to calculate the drift velocity at point B, we can use the equation: \[ v_{dB} = \frac{I}{n A_B e} \] Substituting the given values: \[ v_{dB} = \frac{1}{8.5 \times 10^{28} \times 2.0 \times 10^{-7} \times 1.6 \times 10^{-19}} \approx 3.7 \times 10^{-4} \, \text{m/s} \] Thus, the drift velocity at point B is: \[ v_{dB} \approx 3.7 \times 10^{-4} \, \text{m/s} \]
