A wire of length l and area A is connected to an ideal battery . The drift speed of electrons is v. If the wire is twisted to half its length, what will be the new drift speed ?

Explanation

Let `R_(1)` be the initial resistance of the metal wire having length `l_(1)` and cross section `A_(1)`. Then we can write `R_(1) = rho (l_(1))/(A_(1))`, where `rho` is the electrical resistivity of the material of wire. Let V be the potential difference applied. Current flowing through the wire can be written as:
`I_(1) = (V)/(R_(1)) rArr I_(1) = (VA_(1))/(rho l_(1))`
We know that `I = "ne "Av_(d)`
`I_(1) = "ne"A_(1)v_(d1)= (VA_(1))/(rho l_(1)) rArr v_(d1) = (V)/(rho n e l_(1))` ...(ii)
Similarly we can find `v_(d2) = (V)/(rho n e l_(2))` ....(iii)
`rArr (v_(d2))/(v_(d1)) = (l_(1))/(l_(2)) =(1)/(2)`