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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 27 of 35
Marks: +1, -0
Define current density and relaxation time. Derive an expression for resistivity of a conductor in terms of number density of charge carriers in the conductor and relaxation time.
Current density: Current density is the current per unit area of cross-section of a conductor.
J=IAJ = \frac{I}{A}
Relaxation time: Relaxation time is the time interval between two successive collisions of electrons in a conductor.
Expression for resistivity:
vd=eEmτv_d = \frac{e E}{m} \tau
or, vd=eVmlτ  (since, E=Vl)v_d = \frac{e V}{m l} \tau \; \text{(since, } E = \frac{V}{l} \text{)}
or, V=vdmleτV = \frac{v_d m l}{e \tau}
or, IR=vdmleτI R = \frac{v_d m l}{e \tau} (since from Ohm's law, V=IRV = I R )
or, R=vdmleτIR = \frac{v_d m l}{e \tau I}
ρlA=vdmleτI(since R=ρlA)\frac{\rho l}{A} = \frac{v_d m l}{e \tau I} \quad \text{(since } R = \frac{\rho l}{A}\text{)}
ρ=vdmAeτI\rho = \frac{v_d m A}{e \tau I}
  \;
 or, ρ=vdmAeτ(neAvd)\text{ or, } \rho = \frac{v_d m A}{e \tau (n e A v_d)}  (since, I=neAvd ) \quad \text{ (since, } I = n e A v_d \text{ ) }
ρ=mne2τ\therefore \rho = \frac{m}{n e^2 \tau}
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