Question

A conductor of length \( l \) is connected across an ideal cell of emf \( E \). Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If \( R \) and \( R' \) are the initial and final values of resistance, and \( v_d \) and \( v'_d \) are the initial and final values of drift velocity, find the relation between (i) \( R' \) and \( R \) and (ii) \( v'_d \) and \( v_d \).

Hint:

Stretching a conductor doubles its length and halves its cross-sectional area, leading to a fourfold increase in resistance and a doubling of the drift velocity to maintain the current.

Answer 1

The resistance \( R \) of a conductor is given by: \[ R = \rho \frac{l}{A} \] where \( \rho \) is the resistivity of the material, \( l \) is the length of the conductor, and \( A \) is the cross-sectional area. When the length is increased to \( 2l \), the new resistance \( R' \) is given by: \[ R' = \rho \frac{2l}{A'} \] Since the volume of the conductor remains constant while stretching, the volume before and after stretching must be equal: \[ lA = 2lA' \] \[ A' = \frac{A}{2} \] Thus, the final resistance \( R' \) becomes: \[ R' = \rho \frac{2l}{A/2} = 4 \times \rho \frac{l}{A} = 4R \] Therefore, the relation between \( R' \) and \( R \) is: \[ R' = 4R \] For drift velocity, the drift velocity \( v_d \) is related to the current and the electric field by: \[ J = nq v_d \] where \( n \) is the number of charge carriers per unit volume, \( q \) is the charge of an electron, and \( v_d \) is the drift velocity. When the length of the conductor is stretched, the electric field \( E \) remains the same, but the cross-sectional area decreases. Since the current is conserved, the drift velocity must increase, as the current density remains the same. Therefore, the final drift velocity \( v'_d \) is related to the initial drift velocity \( v_d \) by: \[ v'_d = 2v_d \]