There is a current of 4.8 ampere in a conductor. The number of electrons per second that cross any section normal to the direction of flow is

Question

There is a current of 4.8 ampere in a conductor. The number of electrons per second that cross any section normal to the direction of flow is (A) 1019 (B) 2 × 1019 (C) 3 × 1019 (D) 7 × 1020

Tardigrade Admin · Accepted Answer

Let N electrons are flowing through a conductor in time t. ∴ Current , I=(N e/t) or (N/t)=(I/e)=(4.8/1.6 × 10-19)=3 × 1019 electrons per second

Q. There is a current of 4.8 ampere in a conductor. The number of electrons per second that cross any section normal to the direction of flow is

$1 0^{19}$

$2 \times 1 0^{19}$

$3 \times 1 0^{19}$

$7 \times 1 0^{20}$

Let $N$ electrons are flowing through a conductor in time $t$ .
$∴$ Current $, I = \frac{N e}{t}$
or $\frac{N}{t} = \frac{I}{e} = \frac{4.8}{1.6 \times 1 0 ^{- 19}} = 3 \times 1 0^{19}$ electrons per second

Q. There is a current of 4.8 ampere in a conductor. The number of electrons per second that cross any section normal to the direction of flow is

1019

2×1019

3×1019

7×1020

Let N electrons are flowing through a conductor in time t. ∴ Current ,I=tNe​ or tN​=eI​=1.6×10−194.8​=3×1019 electrons per second

$1 0^{19}$

$2 \times 1 0^{19}$

$3 \times 1 0^{19}$

$7 \times 1 0^{20}$

Let $N$ electrons are flowing through a conductor in time $t$ .
$∴$ Current $, I = \frac{N e}{t}$
or $\frac{N}{t} = \frac{I}{e} = \frac{4.8}{1.6 \times 1 0 ^{- 19}} = 3 \times 1 0^{19}$ electrons per second