Question

Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is
8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answer 1

Answer:

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

$L=N\frac\phi I$

 $=N\frac{B.A}{I}$

 $=N\frac{\mu_0NI}{l.I}A$

 $=\frac{\mu_0N^2A}{l}$

 $=\frac{\mu_0N^2}{l}.\pi(\frac d2)^2$

 $=\mu_0\pi\frac{N^2d^2}{4l}$

N= number of turns

$l$ = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

$\phi$ = magnetic flux

$I$= Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

$=\mu_0\pi\frac{N^2D^2}{4L}$

Solenoid B has total number of turns 2N, length  2L and diameter 2D

The inductance of solenoid B is

$=\mu_0\pi\frac{(2N)^2(2D)^2}{4.2L}$

$=\mu_0\pi\frac{16 N^2D^2}{4.2L}$

Therefore,

$\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac{\mu_0\pi\frac{N^2D^2}{4L}}{\mu_0\pi\frac{16 N^2D^2}{4.2L}}$

$\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18$

$\Rightarrow {\textrm{Inductance of A}}=\frac18\times {\textrm{Inductance of B}}$

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.