A steady electric current of 3.2\,\text{A} flows through a conductor for 5.0\,\text{minute… — JAMB UTME

A steady electric current of $3.2\,\text{A}$ flows through a conductor for $5.0\,\text{minutes}$ . Given that the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , how many electrons pass through a cross-section of the conductor during this period?

A
$1.0 \times 10^{20}$
B
$3.6 \times 10^{20}$
$6.0 \times 10^{21}$ Answer
D
$1.536 \times 10^{-16}$

Answer

The number of electrons passing through the cross-section of the conductor is

6.0 \times 10^{21}

Electric current

I

is related to total electric charge

Q

and time

t

Q = I \times t

. Converting time into seconds gives

t = 5.0 \times 60 = 300\,\text{s}

. The total charge passed is

Q = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}

. Using the charge quantization formula

Q = n \cdot e

, the number of electrons

n

is given by

n = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}

Step-by-Step Solution

Convert the time from minutes into seconds.

t = 5.0\,\text{minutes} = 5.0 \times 60\,\text{s} = 300\,\text{s}

SI units require time to be in seconds when calculating electric charge.

Calculate the total electric charge passing through the conductor.

Q = I \times t = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}

Electric current is defined as the rate of flow of charge (

I = Q/t

Determine the number of electrons using charge quantization.

n = \frac{Q}{e} = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}

Total charge is equal to the number of carrier electrons multiplied by the elementary charge (

Q = n \cdot e

Key Concept

Quantization of Electric Charge and Current

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