Question

The free electrons of a copper wire of cross sectional area 10^-6m^2 acquires a drift velocity of 10^-4 when a certain potential difference is applied across the wire find the current flowing in the wire if the density of the wire is 8.5×10^28

Answer 1

Explanation:

-6 2

A=10 m

vd=10-4m/s

28 3

ed=8.5^10 e/m

i=neAvd

28 -19 -6 -4

i=8.5^10 ^ 1.6^10 ^10 ^10

-1

i=8.5^1.6^10

.: i=1.36Amp

Answer 2

Answer:

The current flowing in the wire will be $i = 1.36 \ Ampere$

Explanation:

Given:

We have been given that the free electrons of a copper wire of cross-sectional area $10^{-6} \ m^2$ acquires a drift velocity of $10^{-4$ when a certain potential difference is applied across the wire.

To find:

We have to find the current flowing in the wire if the density of the wire is  $8.5 \times 10^{28$

Solution:

We know that,

$A =$ $10^{-6} \ m^{2}$

$v_{d}=10^{-4}$

$e_{d} = 8.5 \times 10^{28} \ e/m^{3}$

Now, applying the following formula and substituting the given values we get,

$i = neAv_{d}$

$i = 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6} \times 10 ^{-4}$

$i = 8.5 \times 1.6 \times 10^{-1}$

$i = 1.36 \ Ampere$

Final Answer:

Hence, the current flowing in the wire will be $i = 1.36 \ Ampere$

Link:

https://brainly.in/question/39420789?referrer=searchResults

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