Question:

A circular coil with a cross-sectional area of $4\, cm^{2}$ has $10$ turns. It is placed at the centre of a long solenoid that has $15$ turns/cm and a cross-sectional area of $10\, cm^{2}$, as shown in the figure. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?

Updated On: Jul 5, 2022
  • $7.54\, \mu H$
  • $8.54\, \mu H$
  • $9.54\, \mu H$
  • $10.54\, \mu H$
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The Correct Option is A

Solution and Explanation

Let us refer to the coil as circuit $1$ and the solenoid as circuit $2$. The field in the central region of the solenoid is uniform, so the flux through the coil is $\phi_{12}=B_{2}A_{1}=\left(\mu_{0}n_{2}I_{2}\right)A_{1}$ where $n_{2}=N_{2} l =1500$ turns/m The mutual inductance is $M=\frac{N_{1}\phi_{12}}{I_{2}}=\mu_{0}n_{2}N_{1}A_{1}$ $=\left(4\pi\times10^{-7}T\,m\,A^{-1}\right)\left(1500\,m^{-1}\right)\left(10\right)\left(4\times10^{-4}\,m^{2}\right)$ $=7.54\times10^{-6}\,H $ $=7.54\,\mu H$
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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