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viva [34]
2 years ago
7

A copper wire has a diameter of 2.097 mm. what magnitude current flows when the drift velocity is 1.54 mm/s? take the density of

copper to be 8.92 103 kg/m3.
Physics
1 answer:
frozen [14]2 years ago
6 0

By using drift velocity of the electron, the current flow is 7.20 ampere.

We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as

v = I / (n . A . q)

where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.

From the question above, we know that

d = 2.097 mm

r = (0.002097 / 2) m

v = 1.54 mm/s = 0.00154 m/s

ρ = 8.92 x 10³ kg/m³

q = e = 1.6 x 10¯¹⁹C

Find the atom density

n = Na x ρ / Mr

where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).

n = 6.022 x 10²³ x 8.92 x 10³ / 0.635

n = 8.46 x 10²⁷ /m³

Find the current flows

v = I / (n . A . q)

0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)

0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)

I = 7.20 ampere

For more on drift velocity at: brainly.com/question/25700682

#SPJ4

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Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  \eta  =  -\epsilon _o E

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮E \cdot dA =  \frac{-Q}{\epsilon _o}

Where E is the electric field

          dA change in unit area

           -Q is the negative charge

          \epsilon _o  is the permittivity of free space

So

          EA  =  \frac{-Q}{\epsilon _o }

           \frac{Q}{A}  =  -\epsilon _o E

          \eta  =  -\epsilon _o E

Where \eta is the charge density

   

8 0
3 years ago
If the potential difference across the bulb in a camping lantern is 9.0 V, what is the
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Answer:

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4 0
3 years ago
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Answer:

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Explanation:

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So the foumula goes like:

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Therefore;

95=(x/200)x100

0.95=x/100

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8 0
3 years ago
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3 0
3 years ago
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vovangra [49]

Answer

Hi,

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Explanation

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Good luck!

4 0
3 years ago
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