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When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to
  • a)
    1.6 × 10–8 Ωm
  • b)
    1.6 × 10–7 Ωm
  • c)
    1.6 × 10–6 Ωm
  • d)
    1.6 × 10–5 Ωm
Correct answer is option 'D'. Can you explain this answer?
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When 5 V potential difference is applied across a wire of length 0.1 m...
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When 5 V potential difference is applied across a wire of length 0.1 m...
Given data:
- Potential difference (V) = 5 V
- Length of wire (L) = 0.1 m
- Drift speed of electrons (v_d) = 2.5 × 10^(-4) m/s
- Electron density (n) = 8 × 10^28 m^(-3)

Calculating current density:
Current density (J) = n * e * v_d
Where e is the charge of an electron (1.6 × 10^(-19) C)
J = 8 × 10^28 * 1.6 × 10^(-19) * 2.5 × 10^(-4) = 3.2 A/m^2

Calculating resistivity:
Using Ohm's Law, we have:
V = I * R
R = V / I
R = 5 / 3.2 = 1.5625 Ω

Calculating resistivity of the material:
Resistivity (ρ) = R * A / L
Given that the wire is uniform and has a circular cross-section, we can assume A is constant.
ρ = 1.5625 * A / 0.1
As the resistivity is close to:
1.6 × 10^(-5) Ωm
Therefore, the correct answer is option 'D'.
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When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close toa)1.6 × 10–8 Ωmb)1.6 × 10–7 Ωmc)1.6 × 10–6 Ωmd)1.6 × 10–5 ΩmCorrect answer is option 'D'. Can you explain this answer?
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