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The number of free electrons per 10 mm of an ordinary copper wire is 2 x 1021. The average drift speed of the electrons is 0.25 mm/s. The current flowing is: A. 0.8 A B. 8 A C. 80 A D. 5 A
Solution: Here, we are given that the number of free electrons = 2 x 10 21 and average drift speed is 0.25mm/s. Thus, the passing strength thr the cross section in one second n = 2 x 10 21 / 10 x 0.25 = 5 x 1019 And I = charge per second = ne = 5 x 1019 x 1.6 x 10–19 The average drift speed of current flowing = 8A Thanks & Regards, Vasantha Sivaraj, askIITians faculty
Solution:
Here, we are given that the number of free electrons = 2 x 10 21 and average drift speed is 0.25mm/s.
Thus, the passing strength thr the cross section in one second n = 2 x 10 21 / 10 x 0.25 = 5 x 1019
And I = charge per second = ne
= 5 x 1019 x 1.6 x 10–19
The average drift speed of current flowing = 8A
Thanks & Regards,
Vasantha Sivaraj,
askIITians faculty
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