As per the Drude model of metals, the electrical resistance of a metallic wire of
length L and cross-section area A is
(Consider 𝜏 as the relaxation time , m as electron mass, n as carrier concentration
and e as electronic charge)
Options:
(A) mL
ne2A𝜏
(B) 2mL
ne2A𝜏
(C) mL
2ne2A𝜏
(D) mL
4ne2A𝜏
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.17)**
**1. Drude Model of Electrical Conduction**:
- Under an applied electric field $E$, conduction electrons experience drift acceleration and collisions with relaxation time $\tau$.
- Drift velocity: $v_d = \frac{e E \tau}{m}$
- Current density: $J = n e v_d = n e \left(\frac{e E \tau}{m}\right) = \left(\frac{n e^2 \tau}{m}\right) E$
**2. Derivation of Wire Resistance**:
- Electrical conductivity:
$\sigma = \frac{J}{E} = \frac{n e^2 \tau}{m}$
- Electrical resistivity:
$\rho = \frac{1}{\sigma} = \frac{m}{n e^2 \tau}$
- Resistance of a wire with length $L$ and uniform cross-sectional area $A$:
$R = \rho \frac{L}{A} = \left(\frac{m}{n e^2 \tau}\right) \frac{L}{A} = \frac{m L}{n e^2 A \tau}$
**3. Official Verification**:
The verified official IIT answer is **Option (A)** (`mL / (ne^2 A 𝜏)`).
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