A cell of emf 5 V and negligible internal resistance is connected across a conducting wire of cross-sectional area 2 ~mm ^2 . If the electron density in the wire is 5 10²⁸ and the drift velocity of the electrons is 0.125 mms ⁻¹ then the resistance offered by the wire is:

Step-by-step solution

The current I in a conductor is given by the relation I = nAev_d , where n is the electron density, A is the cross-sectional area, e is the elementary charge, and v_d is the drift velocity. Given values are: n = 5 10²⁸ m ⁻³ A = 2 mm ^2 = 2 10⁻⁶ m ^2 v_d = 0.125 mm s ⁻¹ = 0.125 10⁻³ m s ⁻¹ = 1.25 10⁻⁴ m s ⁻¹ e = 1.6 10⁻¹⁹ C Substituting these values into the formula for current: I = (5 10²⁸) (2 10⁻⁶) (1.6 10⁻¹⁹) (1.25 10⁻⁴) I = (5 2 1.6 1.25) 10^ 28 - 6 - 19 - 4 I = 20 10⁻¹ = 2 A Using Ohm's law, V = IR , where V =