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Question
A steady current of 8 mA flows through a wire. The number of electrons passing through a cross-section of the wire in 10 s is ______.
Options
4.0 × 1016
5.0 × 1017
1.6 × 1016
1.0 × 1017
Solution
A steady current of 8 mA flows through a wire. The number of electrons passing through a cross-section of the wire in 10 s is 5.0 × 1017.
Explanation:
Given: I = 8 mA = 8 × 10-3 A
t = 10 s
e = 1.6 × 10-19
We know that,
Q = ne .......(i)
Q = It .......(ii)
Equating (i) and (ii), we get
ne = It
`n xx (1.6 xx 10^-19) = 8 xx 10^-3 xx 10`
n = `(8 xx 10^-2)/(1.6 xx 10^-19)`
n = `80/(16) xx 10^-2/10^-19`
n = `5 xx 10^17`
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