
The self-inductance of an air core solenoid of $100$ turns is $1\,mH$. The self-inductance of another solenoid of $50$ turns (with the same length and cross-sectional area) with a core having relative permeability $500$ is:
A) $125\,mH$
B) $24\,mH$
C) $60\,mH$
D) $30\,mH$
E) $45\,mH$
Answer
589.2k+ views
Hint: 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 $L$ in presence of air is given as:
$L = \dfrac{{{\mu _ \circ }{N^2}A}}{I}$
The self-inductance of a solenoid other than air is given as:
$
L' = \dfrac{{\mu {N^2}A}}{I} \\
= \dfrac{{{\mu _r}{\mu _ \circ }{N^2}A}}{I}
$
Since, the length and area is same
$L\alpha {\mu _r}{N^2}$
Hence, we can write the equations as:
$
\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.
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 $L$ in presence of air is given as:
$L = \dfrac{{{\mu _ \circ }{N^2}A}}{I}$
The self-inductance of a solenoid other than air is given as:
$
L' = \dfrac{{\mu {N^2}A}}{I} \\
= \dfrac{{{\mu _r}{\mu _ \circ }{N^2}A}}{I}
$
Since, the length and area is same
$L\alpha {\mu _r}{N^2}$
Hence, we can write the equations as:
$
\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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Give 10 examples of unisexual and bisexual flowers

Coming together federation is practiced in A India class 12 social science CBSE

Write the formula to find the shortest distance between class 12 maths CBSE

