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Question: The self-inductance of an air core solenoid of \(100\) turns is \(1\,mH\). The self-inductance of an...

The self-inductance of an air core solenoid of 100100 turns is 1mH1\,mH. The self-inductance of another solenoid of 5050 turns (with the same length and cross-sectional area) with a core having relative permeability 500500 is:
A) 125mH125\,mH
B) 24mH24\,mH
C) 60mH60\,mH
D) 30mH30\,mH
E) 45mH45\,mH

Explanation

Solution

Here we have first use the formula of the self inductance of a solenoid in air, then the self-inductance of a solenoid without air and then at last divide them with each other to find the self-inductance of the solenoid without air.

Complete step-by-step answer:
Self-inductance is defined as the induction of a voltage in a current-carrying wire as the current in the wire itself varies. In case of self-inductance, the magnetic field produced by the changing current in the circuit itself causes a voltage in the same circuit. The voltage is also self-induced.
Self-inductance is the property of the current-carrying loop that opposes or restricts the difference in current moving through it. This happens essentially because of self-induced emf delivered in the loop itself. In straightforward terms, we can likewise say that self-inductance is where there is the induction of a voltage in a current-conveying wire.
The above property of the coil exists just for the changing current which is the alternating current and not for the immediate or consistent current. Self-inductance is continually restricting the changing current and is estimated in Henry.
Self-inductance of a solenoid LL in presence of air is given as:
L=μN2AIL = \dfrac{{{\mu _ \circ }{N^2}A}}{I}
The self-inductance of a solenoid other than air is given as:
L=μN2AI =μrμN2AI L' = \dfrac{{\mu {N^2}A}}{I} \\\ = \dfrac{{{\mu _r}{\mu _ \circ }{N^2}A}}{I}
Since, the length and area is same
LαμrN2L\alpha {\mu _r}{N^2}
Hence, we can write the equations as:
LL=μ×1002500×μ×502 L=5004×1mH L=125mH  \dfrac{L}{{L'}} = \dfrac{{{\mu _ \circ } \times {{100}^2}}}{{500 \times {\mu _ \circ } \times {{50}^2}}} \\\ L' = \dfrac{{500}}{4} \times 1\,mH \\\ L' = 125\,mH \\\

Hence, option A is correct.

Note: Self-induced emf present in the loop will oppose the ascent of current when the current increases and it will likewise oppose the fall of current if the current reduces. Basically, the course of the induced emf is inverse to the applied voltage if the current is expanding and the heading of the induced emf is a similar direction as the applied voltage if the current is falling.
Here we have to pay attention to the question whether the value of permeability in free space or air is given. And also the value of relative permeability is given or not.