Maharashtra State BoardHSC Science (General) 11th
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Solve the following problem. The magnitude of current density in a copper wire is 500 A/cm2. If the number of free electrons per cm3 of copper is 8.47 × 1022 - Physics

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Sum

Solve the following problem.

The magnitude of current density in a copper wire is 500 A/cm2. If the number of free electrons per cm3 of copper is 8.47 × 1022 calculate the drift velocity of the electrons through the copper wire (charge on an e = 1.6 × 10-19 C)

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Solution

Given: J = 500 A/cm2 = 500 × 104 A/m2, n = 8.47 × 1022 electrons/cm3 = 8.47 × 1028 electrons/m3
e = 1.6 × 10-19 C

To find: Drift velocity (vd)

Formula: vd = `"J"/"ne"`

Calculation: From formula.

vd = `(500xx10^4)/(8.47xx10^28xx1.6xx10^-19)`

= `500/(8.47xx1.6)xx10^-5`

= {antilog [log 500 − log 8.47 − log 1.6]} × 10−5

= {antilog [2.6990 − 0.9279 − 0.2041]} × 10−5

= {antilog [1.5670]} × 10−5

= 3.690 × 101 × 10−5

= 3.69 × 10−4 m/s

The drift velocity of electrons is 3.69 × 10−4 m/s.

Concept: Drift Speed
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APPEARS IN

Balbharati Physics 11th Standard Maharashtra State Board
Chapter 11 Electric Current Through Conductors
Exercises | Q 4. (iv) | Page 220
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