Question

Two solenoids are equal in length and radius, and the cores of both are identical cylinders of iron. However, solenoid A has four times the number of turns per unit length as solenoid B.(a) Which solenoid has the larger self-inductance?A B they are the same(b) What is the ratio of the self-inductance of solenoid A to the self-inductance of solenoid B?LA/LB =______

Answer 1

Answer:

Explanation:

Length of both the solenoids = l

Area of crossection of both the solenoids = A

Current in both the solenoids = i

Let the number of turns in coil A is 4N and the number of turns in coil B is N.

The self inductance due to the long solenoid is given by

$L = \frac{\mu_{0}N^{2}A}{l}$

As the current, area of crossection and the length is same so

$\frac{L_{A}}{L_{B}}=\frac{N_{A}^{2}}{N_{B}^{2}}$

$\frac{L_{A}}{L_{B}}=\frac{16N^{2}}{N^{2}}$

So, LA : LB = 16 : 1

Answer 2

Answer:

$\dfrac{L_A}{L_B}=16$

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

n = Number of turns

A = Area

I = Current

Self inductance is given by

$L=\mu_0n^2IA$

Here, A has more turns so the self-inductance of A will be higher

For A

$L_A=\mu_0n_A^2IA=\mu_0(4n_B)^2IA$     $[\because n_A=4n_B]$

For B

$L_B=\mu_0n_B^2IA$

Dividing the above two equations we have

$\dfrac{L_A}{L_B}=\dfrac{\mu_0(4n_B)^2IA}{\mu_0n_B^2IA}\\\Rightarrow \dfrac{L_A}{L_B}=16$

$\therefore \dfrac{L_A}{L_B}=16$