Question: Q. 2. Define ‘self-inductance’ of a coil. Obtain an expression for self inductance of a long solenoid of cross sectional area $A$, length $L$ having $n$ turns for unit length. Prove that self inductance is the analogue of mass in mechanics. $R$ [SQP II 2017]
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Solution:
Ans. Try yourself, Definition : Similar to Q. 2, Short Answer Type Questions-I
[CBSE Marking Scheme 2017]
Expression :
Consider a long solenoid of length $l$, area of crosssection $A$ with $N$ number of closely wound turns. If $I$ is the amount of current flowing through the solenoid, then magnetic field $B$ inside the solenoid will be :
$$ B=\frac{\mu_{0} N I}{l} $$
Now magnetic flux through each turn of solenoid is :
Since
$$ \phi=B A=\frac{\mu_{0} N^{2} A I}{l} $$
$$ \begin{align*} \phi & =L I \ L I & =\frac{\mu_{0} N^{2} A I}{l} \ L & =\frac{\mu_{0} N^{2} A}{l} \end{align*} $$
Hence, If a coil of wire with few turns around metal core carries a charge is passed through. The current will create a magnetic field that runs through the center of the coil pointing downward. If the current is stopped suddenly, then the magnitude of magnetic field tends to be zero.
It is known that changing magnetic fields will influence the charges in loops of wire. If magnitude of magnetic field in the coil approaches to zero, it induces a voltage in the coil which creates the magnetic field. So as per Lenz’s law, voltage induced by changing magnetic field gives rise to a current which counteract the changes.
Mathematically, voltage across the inductor, the loop of coil, is
$$ \begin{equation*} V_{\text {Ind }}=L \frac{d I}{d t} \tag{1} \end{equation*} $$
The above expression looks similar to expression of force :
$$ \begin{aligned} E_{\text {Ind }} & =\frac{1}{2} L I^{2} \ E_{\text {Kinetic }} & =\frac{1}{2} m v^{2} \end{aligned} $$
Now,
From the above description, there appears an analogy between mechanical motion and flow of electricity. Here, self-inductance $L$ and mass $m$, both provide inertia that will resists in changing the current $I$ or velocity $v$ suddenly.
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AI Q. 3. Define mutual inductance of a coil. On what factors it depends? $\quad$ [SQP II 2017] Two concentric circular coils, one of small radius $r_{1}$ and the other of large radius $r_{2}$, such that $r_{1}«$ $r_{2}$, are placed co-axially with centers coinciding. Obtain the mutual inductance of the arrangement.
Ans. Try yourself, Definition : Similar to Q. 2, Short Answer Type Questions-II.
Try yourself, Factors : Similar to Q. 8 (ii) b, Short Answer Type Questions-II.
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Expression :
If a current $I_{2}$ flows through the outer circular coil of radius $r_{2}$, the magnetic field at the centre of the coil will be,
$$ B_{2}=\frac{\mu_{0} I_{2}}{2 r_{2}} $$
As $r_{1}«r_{2}$, so field $B_{2}$ will be considered to be constant over the entire cross-sectional area of inner coil with radius $r_{1}$. Hence, magnetic flux linked with the smaller coil will be :
$$ \begin{align*} \phi_{1} & =B_{2} A_{1} \ & =\frac{\mu_{0} I_{2}}{2 r_{2}} \times \pi r_{1}^{2} \end{align*} $$
From the definition :
$$ \phi_{1}=M_{12} I_{2} $$
Hence, mutual inductance will be,
$$ M_{12}=\frac{\phi_{1}}{I_{2}}=\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}} $$
But $\quad M_{12}=M_{21}=M$
$\therefore$ Mutual inductance, $M=\oint E \cdot d l=-\frac{d \phi}{d t}$