Question

Two boys A and B are sitting at two points in a field. Both boys are sitting near an assemblage of charged balls each carrying a charge of +3e . A throws 100 balls per second towards B and B throws 50 balls per second towards A. Find the current at the midpoint of A and B.
$\begin{align}
  & \left( A \right)i=3.4\times {{10}^{-17}}A\text{ from A}\text{ to B} \\
 & \left( A \right)i=2.4\times {{10}^{-17}}A\text{ from A}\text{ to B} \\
 & \left( A \right)i=3.4\times {{10}^{-17}}A\text{ from B}\text{ to A} \\
 & \left( D \right)\text{none of the above} \\
\end{align}$

Answer 1

Hint: Since, all the balls are charged, the throwing of ball can be considered as charge flowing from A to B and from B to A. Also, since A is throwing more balls from towards B, the direction of current will be from A to B. To calculate the net current at the midpoint of A and B, we will use its definition, that is, current is the amount of charge transferred per second.

Complete step-by-step solution:
We will calculate the amount of charge transferred from A to B and vice versa and then find the net charge transferred. Since the charged balls are thrown per second, the net charge transferred between A and B will be equal to the current that flows between A and B.
Let the direction from A towards B be taken as positive. Then, the amount of charge transferred from A to B is equal to the product of the number of balls thrown per second and the charge on each ball. This can be calculated as follows:
$\Rightarrow {{q}_{A\to B}}=100\times 3e$
Where, ‘e’ is the charge on one electron which is equal to $1.6\times {{10}^{-19}}C$. Putting this in the above equation, we get:
$\Rightarrow {{q}_{A\to B}}=100\times 3\times 1.6\times {{10}^{-19}}C$
Now, let us calculate the amount of charge transferred from B to A. This is also equal to the product of number of balls thrown per second and the charge on each ball which can be calculated as follows:
$\begin{align}
  & \Rightarrow {{q}_{B\to A}}=-50\times 3e \\
 & \Rightarrow {{q}_{B\to A}}=-50\times 3\times 1.6\times {{10}^{-19}}C \\
\end{align}$
Therefore, the current at the midpoint of A and B will be equal to the net charge transferred per second. This can be calculated as follows:
$\Rightarrow \left[ \dfrac{\left( {{q}_{A\to B}}+{{q}_{B\to A}} \right)C}{1s} \right]=i$
Where, ‘I’ is the current at the midpoint of A and B. Putting the values of all the known terms, we get:
$\begin{align}
  & \Rightarrow i=\left[ \dfrac{100\times 3\times 1.6\times {{10}^{-19}}C+\left( -50\times 3\times 1.6\times {{10}^{-19}}C \right)}{1s} \right] \\
 & \Rightarrow i=\left( 100\times 3\times 1.6\times {{10}^{-19}}-50\times 3\times 1.6\times {{10}^{-19}} \right)A \\
 & \Rightarrow i=50\times 3\times 1.6\times {{10}^{-19}}A \\
 & \therefore i=2.4\times {{10}^{-17}}A \\
\end{align}$
Since, the current is positive, it means the direction of current is from A towards B.
Hence, the current at the midpoint of A and B is, $i=2.4\times {{10}^{-17}}A\text{ from A}\text{ to B}$.
Hence, option (B) is the correct option.

Note: It is very important to write current at the midpoint as it is the only point of equal distance between the two boys A and B. If we were to calculate current at any point other than the midpoint, then the uniformity of balls would have been lost, that means we would need to first calculate the number of balls as a function of distance and then solve it, which would make the problem very complex.