Question

n number of electrons flowing in a copper wire for 1 minute constitute a current of 0.5 A. Twice the number of electrons flowing through the same wire for 20 s will constitute a current of:

• A. 0.25 A
• B. 3 A
• C. 1 A
• D. 1.25 A
• E. 2.25 A

Hint:

To calculate the current, use the formula \( I = \frac{Q}{t} \), where \( Q \) is the total charge and \( t \) is the time. If the number of electrons is doubled, the charge doubles, which increases the current.

Answer 1

The Correct Option is B

We know the formula for current \( I \) is: \[ I = \frac{Q}{t} \] where \( Q \) is the charge and \( t \) is the time. The current \( I_1 = 0.5 \, {A} \) is given for a time of \( t_1 = 1 \, {minute} = 60 \, {seconds} \). 
Now, the current is given by: \[ I_1 = \frac{Q_1}{t_1} \] So, the charge \( Q_1 \) is: \[ Q_1 = I_1 \times t_1 = 0.5 \, {A} \times 60 \, {s} = 30 \, {C} \] If the number of electrons is doubled, the total charge will also double. So the new charge \( Q_2 \) will be: \[ Q_2 = 2 \times Q_1 = 2 \times 30 \, {C} = 60 \, {C} \] The new current \( I_2 \) is given by: \[ I_2 = \frac{Q_2}{t_2} \] where \( t_2 = 20 \, {s} \). Substituting the values: \[ I_2 = \frac{60 \, {C}}{20 \, {s}} = 3 \, {A} \] Thus, the current when twice the number of electrons flows for 20 seconds is \( 3 \, {A} \).
Therefore, the correct answer is option (B), 3 A.