The cross-sectional area of a wire is 1 . 0 × 10 - 7 m 2 and the density of free electrons is 2 . 0 × 10 28 m - 3 . What will be the drift velocity of the free electrons for a current of 3 . 2 A in the wire?

Formula used : I = neAv d , Where I is electric current, n is the no. of electrons, A is the area of cross-section, and V d is the drift velocity of electron. ⇒ v d = I neA = 3 . 2 2 × 10 28 × 1 . 6 × 10 − 19 × 10 − 7 = 1 10 2 = 0 . 01 ms - 1

90 C charge flows for 75 minutes in silver wire of diameter 1 . 0 mm . The free-electron density in the wire is 5 . 8 × 10 28 m - 3 . Find the current and the drift velocity of free electrons in the wire.

Formula used : Q = I . T Where Q is electric charge I is electric current T is time . I = 90 75 × 60 = 1 50 = 0 . 02 A Formula : v d = I neA Where A is area of cross-section. v d = 0 .02 5 .8 × 10 28 × 1 .6 × 10 − 19 × 3 .14 × 10 − 6 2 .74 × 10 − 6 m / s

A very small current is sent through a wire of length 15 m and cross-sectional area 6 . 0 × 10 - 7 m - 2 Its resistance is measured to be 5 . 0 Ω . Find the specific resistance of the experiment. What will be the specific conductance?

Formula used : R = ρ 1 A , σ = 1 / ρ Where R is the resistance A is the area. l is the length σ is the conductivity ρ is the resistivity. ρ = R ⋅ A l = 5 × 6 × 10 − 7 15 = 2 × 10 − 7 Ω m σ = 1 ρ = 1 2 × 10 − 7 = 5 × 10 6 ( Ω m ) - 1 ρ = 2 × 10 − 7 Ωm Conductivity , σ = 5 × 10 − 6 ( Ω m ) - 1

What length of a wire of diameter 0 . 46 mm and specific resistance 50 × 10 - 6 ohm - cm will be required to prepare a coil of 10 ohm ?

Formula used : resistance R = ρ l A Where l is length, A is area, And ρ is resistivity Step 1 Given : R = 10 Ω A = 3 .14 × ( 0 . 23 ) 2 × 10 6 ρ = 50 × 10 − 6 ohm. cm = 50 × 10 − 8 ohm. m Step 2 : l = R ⋅ A ρ l = 10 × 3 .14 × ( 0 .23 ) 2 × 10 − 6 50 × 10 − 8 l = 0 .033231 × 10 + 2 m = 3 .3231 m l = 332 .31 m