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If n, e, tau, m, are representing electr...

If n, e, `tau`, m, are representing electron density charge, relaxation time and mass of an electron respectively then the resistance of wire of length 1 and cross sectional area A is given by

A

`(ml)/(Ne^(2) A^(2) tau)`

B

`(2 m tau A)/(Ne^(2) l )`

C

`(Ne^(2) tau A)/(2 ml)`

D

`(Ne^(2) A)/(2 m tau l)`

Text Solution

Verified by Experts

The correct Answer is:
A
Since average drift velocity `= (1)/(2) (e E)/( m) xx tau`
Now `I = Ne A xx ("avg. drift velocity")`
`= (Ne^(2) AE)/( 2 ml ) xx tau = ( Ne^(2) AV)/( 2ml) xx tau`
`R= ( V)/( I) = (m l) /( Ne^(2) tau A^(2) )`, were N is electron density.
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