This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Ili kukadiria msongamano wa mkondo na kasi ya mkengeuko ya elektroni za upitishaji katika waya wa shaba, tutatumia fomula zifuatazo: Data Uliyopewa: Mkondo (I) = 0.5 A Eneo la msalaba (A) = 0.0000002 m^2 = 2 × 10^-7 m^2 Msongamano wa shaba (_Cu) = 9000 kg/m^3 Masi ya atomiki ya shaba (M_Cu) = 63.5 g/mol = 0.0635 kg/mol Kila atomu ya shaba huchangia elektroni moja ya upitishaji. Chaji ya elektroni (e) = 1.602 × 10^-19 C Namba ya Avogadro (N_A) = 6.022 × 10^23 mol^-1 Hatua 1: Hesabu Msongamano wa Mkondo (Current Density, J). Msongamano wa mkondo ni mkondo kwa kila eneo la msalaba. J = (I)/(A) Badilisha maadili: J = 0.5 A2 × 10^-7 m^2 J = 0.25 × 10^7 A/m^2 J = 2.5 × 10^6 A/m^2 Hatua 2: Hesabu Msongamano wa Namba ya Elektroni za Upitishaji (n). Kwa kuwa kila atomu ya shaba huchangia elektroni moja ya upitishaji, msongamano wa namba ya elektroni ni sawa na msongamano wa namba ya atomu za shaba. n = _CuM_Cu × N_A Badilisha maadili: n = 9000 kg/m^30.0635 kg/mol × 6.022 × 10^23 mol^-1 n = (141732.28 mol/m^3) × (6.022 × 10^23 mol^-1) n ≈ 8.53 × 10^28 m^-3 Hatua 3: Hesabu Kasi ya Mkengeuko (Drift Velocity, v_d). Uhusiano kati ya mkondo na kasi ya mkengeuko hutolewa na fomula: I = n A v_d e Panga upya fomula ili kupata v_d: v_d = (I)/(n A e) Badilisha maadili: v_d = 0.5 A(8.53 × 10^28 m^-3) × (2 × 10^-7 m^2) × (1.602 × 10^-19 C) v_d = (0.5)/((8.53 × 2 × 1.602) × 10^28 - 7 - 19) v_d = (0.5)/(27.32 × 10^2) v_d = (0.5)/(2732) v_d ≈ 0.000183 m/s v_d ≈ 1.8 × 10^-4 m/s Majibu: Msongamano wa mkondo ni 2.5 × 10^6 A/m^2 Kasi ya mkengeuko ni 1.8 × 10^-4 m/s Tuma swali linalofuata 📸