Question

A potential drop per unit length along a wire is 5 × 10-3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.

Answer 1

Data: K = 5 × 10^-3 \(\frac{\mathrm{V}}{\mathrm{m}}\), L = 216 cm = 216 × 10^-2 m
E = KL
∴ E = 5 × 10^-3 × 216 × 10^-2
= 1080 × 10^-5
= 0.01080V
The emf of the cell is 0.01080 volt.
(Note: For K = 0.5 V/m, we get, E = 1.08V (reasonable value)]