Question

The number of turns in an air-core solenoid of length 25 cm and radius 4 cm is 100. Its self inductance will be
(A) $5\times {10}^{-4}H$
(B) $2.5\times {10}^{-4}H$
(C) $5.4\times {10}^{-3}H$
(D) $2.5\times {10}^{-3}H$

Answer 1

Hint
Firstly, we need to derive the expression of the self inductance in terms of the given parameters. By converting the values of the given parameters to SI units and substituting the values, we will obtain the value of the self inductance.
Formula used: In this solution we will be using the following formula,
 $\Rightarrow L={\mu }_0{N}^2{\displaystyle \frac{A}{l}}$
Where $L$ is the self-inductance,
 ${\mu }_0$ is the permittivity,
 $N$ is the number of turns,
 $I$ is the current and $l$ is the length of the solenoid.

Complete step by step answer
First, we need to derive the expression to find the self inductance of the air core solenoid.
The magnetic field of the solenoid is given by the expression as follows.
 $\Rightarrow B={\displaystyle \frac{{\mu }_{0}NI}{l}}$
Now the current due to this magnetic field is given by,
 $\Rightarrow I={\displaystyle \frac{Bl}{{\mu }_{0}N}}$ …… (1)
The self inductance of the solenoid is given by the formula as follows.
 $\Rightarrow N\mathrm{\Phi }=LI$ …… (2)
Where $\mathrm{\Phi }$ is the electric flux.
So the electric flux is given by the formula as,
 $\Rightarrow \mathrm{\Phi }=BA\cos \theta$
Now since the angle between the magnetic field and the area vector is 0, so we can write,
 $\Rightarrow \mathrm{\Phi }=BA\cos 0^{\circ }=BA$ …… (3)
Now combining the equations (1), (2) and (3) we can obtain expression in terms of the self inductance.
So, we have the formula for calculating the self inductance as,
 $\Rightarrow L={\mu }_0{N}^2{\displaystyle \frac{A}{l}}$
The values of the parameters given in the question as,
 $\Rightarrow N=100$
Again, here we will change the units of the parameters to SI units.
 $\Rightarrow l=25cm=0.25m$
 $\Rightarrow r=4cm=0.04m$
We can then calculate the area of the air core solenoid as follows,
 $\Rightarrow A=\pi r^2$
Substituting the radius,
 $\Rightarrow A=\pi (0.04{)}^2$
On calculating we get,
 $\Rightarrow A=1.6\times {10}^{-3}\pi$
The value of the permittivity is constant and given as ${\mu }_0=4\pi \times {10}^{-7}$
Now, we will obtain the value of the self inductance using the formula that we have derived earlier.
So, we have,
 $\Rightarrow L={\mu }_0{N}^2{\displaystyle \frac{A}{l}}$
Substituting the values,
 $\Rightarrow L=4\pi \times {10}^{-7}\times {100}^2\times \left({\displaystyle \frac{1.6\times {10}^{-3}}{0.25}}\right)$
On doing the calculation above, we get
 $\Rightarrow L={\displaystyle \frac{6.3165\times {10}^{-5}}{0.25}}$
Hence the self inductance has a value of,
 $\Rightarrow L=2.52\times {10}^{-4}H$
$\therefore$ The value of the self inductance is $2.52\times {10}^{-4}H$ , thus, option (B) is correct.

Note
The term self inductance means that voltage is induced in a current carrying wire, when the current of the wire is itself changing. It is a form of electromagnetic induction. The current that is induced in the circuit always opposes the change in current in the circuit according to Lenz’s law.