Question:

There is a current of $40\, A$ in a wire of $10^{-6} \,m ^{2}$ area of cross-section. If the number of free electrons per cubic metre is $10^{29}$, then the drift velocity is

Updated On: Jun 20, 2022
  • $250 \times 10 ^{-3} \, ms^{-1}$
  • $25.0 \times 10 ^{-3} \, ms^{-1}$
  • $2.50 \times 10 ^{-3} \, ms^{-1}$
  • $1.25 \times 10 ^{-3} \, ms^{-1}$
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The Correct Option is C

Solution and Explanation

Drift velocity
$v_{d}=\frac{i}{n e A} $
$=\frac{40}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}$
$=2.5 \times 10^{-3} \,ms ^{-1}$
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Concepts Used:

Electromagnetic Induction

Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-

  1. When we place the conductor in a changing magnetic field.
  2. When the conductor constantly moves in a stationary field.

Formula:

The electromagnetic induction is mathematically represented as:-

e=N × d∅.dt

Where

  • e = induced voltage
  • N = number of turns in the coil
  • Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
  • t = time

Applications of Electromagnetic Induction

  1. Electromagnetic induction in AC generator
  2. Electrical Transformers
  3. Magnetic Flow Meter