Question

While keeping the area of cross-section of a solenoid the same, the number of turns and length of the solenoid one both doubled. The self-inductance of the coil will be?
A. Halved.
B. Doubled
C. 1/4 times the original value
D. Unaffected

Answer 1

Hint: The self-inductance of the solenoid is due to the effect of current flowing through the other loops in the same solenoid. A changing current in one of them can cause a back emf in the coil. This is the cause of self-inductance. Find the self-inductance of the solenoid. Find the dependency on length and number of terms.

Formula Used:
The magnetic field inside a solenoid is given by,
$B=\dfrac{{{\mu }_{0}}NI}{l}$......................(1)

Where,
$N$ is the number of turns on the coil
$I$ is the current passing through the coil.
$B$ is the magnetic field inside the solenoid
$l$ is the length of the solenoid
${{\mu }_{0}}$ is the magnetic permeability

Complete step by step answer:

Consider a solenoid of length ‘l’, area of cross-section ‘A’, number of turns ‘N’, and the current flowing through the
From the definition of magnetic flux, we can write,
$\phi =BA$........................(2)

Using equation (1), the magnetic field inside a solenoid is given by,
$B=\dfrac{{{\mu }_{0}}NI}{l}$
So, we can write equation (2) as,
$\phi =(\dfrac{{{\mu }_{0}}NI}{l})A$

This is the amount of flux through each turn of the solenoid.

So, the total magnetic flux is given by,
$\phi =(\dfrac{{{\mu }_{0}}NI}{l})A$
Where we can define L as the self-inductance of the solenoid.
$L=N\dfrac{\phi }{A}$

So, we can write,
$L=N(\dfrac{{{\mu }_{0}}NI}{l})A\times \dfrac{1}{A}$
$\Rightarrow L=\dfrac{{{\mu }_{0}}{{N}^{2}}A}{l}$.......................(3)

Now, let's assume the new inductance is,
${{L}_{1}}$

The area is kept the same, but the number of terms and length has been doubled.

So, we can write,
${{l}_{1}}=2l$
${{N}_{1}}=2N$

 So, using equation (3) we can write,
$\Rightarrow {{L}_{1}}=\dfrac{{{\mu }_{0}}{{(2N)}^{2}}A}{2l}=2\dfrac{{{\mu }_{0}}{{N}^{2}}A}{l}=2L$

Hence, the self-inductance is twice the previous value.

The correct option is - (B).

Note: You don’t need to derive equation (3) every time for these problems. Memorize equation (3) to do it faster.
We can also find the magnitude of the induced EMF. The work done to increase the current from zero to a specific value is given by,
$W=\dfrac{1}{2}LI_{0}^{2}$
Where,
$L$ is the self-inductance of the solenoid
${{I}_{0}}$ is the final current.

This energy is stored as the potential energy of the inductor.