There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is

Question

There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is (A) 1019 (B) 2× 1019 (C) 3× 1019 (D) 3.2× 1019

Tardigrade Admin · Accepted Answer

I = (q/t) = (ne/I) . Here I = 3.2 A, e = 1.6 × 10-19 C Hence n = 2 × 1019

Q. There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is

$1 0^{19}$

$2 \times 1 0^{19}$

$3 \times 1 0^{19}$

$3.2 \times 1 0^{19}$

$I = \frac{q}{t} = \frac{n e}{I}$ . Here $I$ = 3.2 A, $e = 1.6 \times 1 0^{- 19}$ C Hence $n = 2 \times 1 0^{19}$

Q. There is a current of 3.2 A in a conductor. The number of electrons that cross any section normal to the direction of flow per second is

1019

2×1019

3×1019

3.2×1019

I=tq​=Ine​ . Here I = 3.2 A, e=1.6×10−19 C Hence n=2×1019

$1 0^{19}$

$2 \times 1 0^{19}$

$3 \times 1 0^{19}$

$3.2 \times 1 0^{19}$

$I = \frac{q}{t} = \frac{n e}{I}$ . Here $I$ = 3.2 A, $e = 1.6 \times 1 0^{- 19}$ C Hence $n = 2 \times 1 0^{19}$