Does the number of electrons colliding in a wire get doubled when length of wire is made twice with its area of cross-section remain constant.
My calculations for this are:-
For wire of length $L$ and area of cross-section:
Let the average distance after travelling which electron will collide $=d$
The number of free electrons in wire $=n $
Time taken between two successive collisions of electron $=τ$
Electric field across it $=E_1$
And Voltage apllied across wire $=V$
Average velocity of electron $= E_1eτ/2m$ The number of collision of electron in time $t = E_1eτtn/2md$
When using a wire of length 2L but with same area of cross-section and same voltage apllied across it:
Number of free electrons $ =2n$
Electric field inside wire $= E_2=V/2L=E_1/2$
Average velocity of electrons $= E_2eτ/2m=E_1eτ/2×2m$
And at last, the number of collisions $= E_2eτt×2n/2md=E_1eτtn/2md$
We can see here that number of collision remains same when length of wire is changed.
*Edit: There are lot of errors above. So I am writing correct version down( correct in my sense)
For wire of length $L$:
The number of free electrons in wire $=n$
Time taken between two successive collisions of electron (it is constant for a material for reason I don't know)$=τ$
Electric field across it $=E_1$
And Voltage apllied across wire $=V$
The number of collision of electron in time $t = tn/ \tau $
When using a wire of length 2L and area of cross section A but with same area of cross-section and same voltage apllied across it:
Number of free electrons $ =2n $
Average velocity of electrons $= t2n/τ$
Finally, why is the $\tau$ uneffected by Electric field, $E = V/L$? ($L$is the length of wire).
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