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 =______

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boffeemadrid boffeemadrid · Answer 1 · 2020-03-28T07:43:54+01:00

Answer:

[tex]\dfrac{L_A}{L_B}=16[/tex]

Explanation:

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

n = Number of turns

A = Area

I = Current

Self inductance is given by

[tex]L=\mu_0n^2IA[/tex]

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

For A

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

For B

[tex]L_B=\mu_0n_B^2IA[/tex]

Dividing the above two equations we have

[tex]\dfrac{L_A}{L_B}=\dfrac{\mu_0(4n_B)^2IA}{\mu_0n_B^2IA}\\\Rightarrow \dfrac{L_A}{L_B}=16[/tex]

[tex]\therefore \dfrac{L_A}{L_B}=16[/tex]

Vespertilio Vespertilio · Answer 2 · 2020-03-28T10:33:16+01:00

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

[tex]L = \frac{\mu_{0}N^{2}A}{l}[/tex]

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

[tex]\frac{L_{A}}{L_{B}}=\frac{N_{A}^{2}}{N_{B}^{2}}[/tex]

[tex]\frac{L_{A}}{L_{B}}=\frac{16N^{2}}{N^{2}}[/tex]

So, LA : LB = 16 : 1