Question

A potential difference of 30 V is applied between the ends of a conductor of length 100 m and resistance 0.5 W and uniform area of cross-section. The total linear momentum of free electrons is -

• A. 3.4 × 10^–6 kg/s
• B. 4.3 × 10^–6 kg/s
• C. 3.4 × 10^–8 kg/s
• D. 4.3 × 10^–8 kg/s

Answer 1

Answer: (C)

3.4 × 10^–8 kg/s

Answer 2

Current, I = $\frac{V}{R}$ = $\frac{30}{0.5}$ = 60 A

Total no. of free e^– s, N = nAl

and linear momentum of each e¯s, P = mv_a

\ Total momentum of all free e¯s,

P = (nAl) (mu_a)

But I = neAu_a , so nAu_a = $\frac{I}{e}$

\ P = $\frac{I\mathcal{l}m}{e}$ = $\frac{60 \times 100 \times 9.1 \times 10^{- 31}}{1.6 \times 1{0^{–}}^{19}}$

3.4 ×10^–8 kg/s