Question:

$A$ $2 \,m$ long solenoid with diameter $2\, cm$ and $2000$ turns has a secondary coil of $1000$ turns wound closely near its midpoint. The mutual inductance between the two coils is

Updated On: Jul 2, 2022
  • $2.4 \times 10^{-4}\,H$
  • $3.9 \times 10^{-4}\,H$
  • $1.28 \times 10^{-3}\,H$
  • $3.14 \times 10^{-3}\,H$
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The Correct Option is B

Solution and Explanation

Here, $l= 2 m$, diameter $= 2\, cm$ $\therefore$ radius, $ r=\frac{2}{2}=1\,cm =1\times10^{-2}\,m$ $N_{1}=2000, N_{2}=1000$ Area $=\pi r^{2}=\pi\times\left(1\times10^{-2}\right)^{2} $ $=3.14\times10^{-4}\,m^{2}$ Mutual inductance, $M =\frac{\mu_{0} N_{1}N_{2}A}{l}$ $=\frac{4\pi\times10^{-7}\times2000\times1000\times3.14\times10^{-4}}{2}$ $=3.9\times10^{-4}\,H$
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Top Questions on Electromagnetic induction

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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