Is the following nuclear reaction balanced?

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

vesna_86 [32]

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.

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Explanation:

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