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A $20\,\mu A$ current flows for 30 seconds in a wire, transfer of charge will be
A. $2 \times {10^{ - 4}}C$
B. $4 \times {10^{ - 4}}C$
C. $6 \times {10^{ - 4}}C$
D. $8 \times {10^{ - 4}}C$

Answer
VerifiedVerified
164.4k+ views
Hint: Start with finding the relation between the given quantities in the question that is here current that flows in the wire, take taken by the current of given amount to flows in the wire given and the transfer of charge due to this flows of the current in the wire. Then put all the values in that relation and get the required answer.

Formula used:
Transfer of charge;
$Q = i \times t$
Where, i is current that is flowing through the given wire and t is the time taken by the current.

Complete step by step solution:
First start with the given information in the question:
Amount of current flowing through the given wire, $i = 20\mu A$
Changing $\mu A$ into $A$, we get;
We know, $1\mu A = {10^{ - 6}}A$
So, $i = 20\mu A = 20 \times {10^{ - 6}}A$
Time taken by the $20\,\mu A$ current to flow through the given wire, $t = 30\,s$.

Now, we need to find the transfer of charge while $20\,\mu A$ current flows through the given wire in time $t = 30\,s$.
Let the transfer of charge be Q. We know that transfer of charge Q is equal to the multiplication of the current flowing through the wire and the time taken. So, charge Q is given by:
$Q = i \times t$
Putting the values of current and time taken from the question, we get;
$Q = 20 \times {10^{ - 6}} \times 30$
By solving;
$\therefore Q = 6 \times {10^{ - 4}}C$

Hence, the correct answer is option C.

Note: Be careful about the units of the given quantities in the question. Change all the units of the given quantity in the SI unit and then only use them in the found relation of the transfer of charge and hence solve them and get the required answer in the same SI unit.