Question

A wire contains a steady current of 2 A. The number of electrons that pass a cross section in2 s is:A. 2B. 4C. 6.3 × 10^18D. 1.3 × 10^19E. 2.5 × 10^19

Answer 1

Answer:

Number of electrons, $n=2.5\times 10^{19}\ electrons$

Explanation:

Given that,

Current flowing in wire, I = 2 A

Time, t = 2 s

We need to find the number of electrons that passes through a cross section. The electric current is defined as the electric charge per unit time. It is given by :

$I=\dfrac{q}{t}$

Since, q = ne

$I=\dfrac{ne}{t}$

n is the number of electrons

$n=\dfrac{It}{e}$

$n=\dfrac{2\times 2}{1.6\times 10^{-19}}$

$n=2.5\times 10^{19}\ electrons$

So, there are $2.5\times 10^{19}\ electrons$ passes through a cross section. Hence, this is the required solution.

Answer 2

The number of electrons that pass a cross section = 2.5 × 10-19 electrons.

Calculation of current flow

Current of the wire (I) = 2A

Time (t) = 2 seconds

But formula for quantity of charge (Q)= It

Q = 2×2

= 4C

Basic formula to calculate electrons is n = Q/e.

Where 'e' is standard electron charge = 1.6X10^-19C

Therefore 'n' which is number of electrons

= 4/1.6X10^-19C

= 2.5 × 10-19 electrons

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