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Two long, single layered solenoids 'a' and 'b' have the same length and the same number of turns. The cross sectional areas of two are'xa' and 'xb', respectively, with xb < xa. They are placed coaxially, with solenoid 'b' placed within the solenoid 'a'. Determine the coefficient of coupling between them. 
1. \(\sqrt {\dfrac{X_b}{X_a}}\)
2. \(\sqrt {\dfrac{X_a}{X_b}}\)
3. 1
4. 0

1 Answer

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Best answer
Correct Answer - Option 2 : \(\sqrt {\dfrac{X_a}{X_b}}\)

Concept:

Coefficient of Coupling (k):
The coefficient of coupling (k) between two coils is defined as the fraction of magnetic flux produced by the current in one coil that links the other.

Two coils have self-inductance L1 and L2, then mutual inductance M between them then Coefficient of Coupling (kis given by 

\(k=\frac{M}{\sqrt {L_1L_2}}\)

Where,

\(M=\frac{N_1N_2\mu_o \mu_rA}{ l}\)

\(L_1=\frac{N_1^2\mu_o \mu_rA}{ l}\)

\(L_2=\frac{N_2^2\mu_o \mu_r A}{l}\)

N1 and N2 is the number of turns in coil 1 and coil 2 respectively
A is the cross-section area
l is the length

Calculation:

Given,

N1 = N2 = N

l1 = l2 = l

A1 = Xa

A2 = Xb

From the above concept,

\(M=\frac{N^2\mu_o \mu_rX_a}{ l}\)

\(L_1=\frac{N^2\mu_o \mu_r X_a}{l}\)

\(L_2=\frac{N^2\mu_o \mu_r X_b}{l}\)

We know that,

\(k=\frac{M}{\sqrt {L_1L_2}}=\frac{X_a}{\sqrt {X_aX_b}}\)

\(k=\frac{X_a}{\sqrt {X_aX_b}}\times\frac{\sqrt X_a}{\sqrt X_a}\)

\(k=\sqrt {\dfrac{X_a}{X_b}}\)

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