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Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A=10c{{m}^{2}}$ and length $=20cm$. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is $\left( {{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}} \right)$
A.$4.8\pi \times {{10}^{-4}}H$
B.$4.8\pi \times {{10}^{-5}}H$
C.$2.4\pi \times {{10}^{-4}}H$
D.$2.4\pi \times {{10}^{-5}}H$

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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Hint:We are two solenoids in each other’s vicinity due to which magnetic field will be created in one coil due to current in the other. Therefore, we shall directly apply the formula of mutual inductance between two objects to find mutual inductance between these two solenoids.

Complete answer:
For mutual inductance, we take two coils. One with ${{N}_{1}}$ turns and the other with ${{N}_{2}}$ turns. When we pass current through coil 1, the magnetic field created by the coil increases in magnitude and the magnetic field lines will also go through coil 2 and create an increasing magnetic flux. The increase in magnetic flux in one direction causes the induced current to create a magnetic field going in the opposite direction to reduce the increasing magnetic flux.
Therefore, as we increase current in coil 1, the magnitude of magnetic field goes up and the flux in one direction increases creating an induced EMF across the second coil.
The induced EMF in coil 2, ${{\varepsilon }_{2}}$ is given as:
${{\varepsilon }_{2}}={{M}_{21}}\dfrac{d{{i}_{1}}}{dt}$
Where,
${{M}_{12}}=$ mutual inductance coefficient
$\dfrac{di}{dt}=$ change in current per unit time in coil 1
The unit of mutual inductance is Henry.
This mutual inductance is expressed as:
$M=\dfrac{{{\mu }_{0}}{{N}_{1}}{{N}_{2}}A}{l}$
Where,
${{N}_{1}}=$ number of turns in coil 1
${{N}_{2}}=$ number of turns in coil 2.
$A=$ area of coils
$l=$ length of coil
Here, we have ${{N}_{1}}=300,{{N}_{2}}=400,A=10c{{m}^{2}}=10\times {{10}^{-4}}{{m}^{2}}$ and $l=20cm=0.2m$. Also given that ${{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}$.
Substituting these values in formula of mutual inductance, we get
\[\Rightarrow M=\dfrac{\left( 4\pi \times {{10}^{-7}} \right)300.400\left( {{10}^{-3}} \right)}{0.2}\]
On simplifying, we get
\[\begin{align}
  & \Rightarrow M=4\pi \left( 6 \right){{10}^{-5}} \\
 & \Rightarrow M=24\pi \times {{10}^{-5}} \\
\end{align}\]
$\therefore M=2.4\pi \times {{10}^{-4}}H$
Hence, the mutual inductance is equal to $2.4\pi \times {{10}^{-4}}H$.

Therefore, the correct option is (C) $2.4\pi \times {{10}^{-4}}H$.

Note:
The direction of the induced, induced current and induced magnetic field is such that it opposes the magnetic field they have been originated from and as a result the net value of the magnetic field decreases. This has been well-defined in Lenz's law that the induced magnetic field opposes its source.
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