Question

Two coils A and B have 180 and 360 turns respectively. The current of 1A flows through both the coils. Due to current of 1A in coil A, flux per turn of \(0.8\times 10^{-3}\) Wb is linked with coil A. Due to current of 1A in coil B, flux per turn of \(1\times 10^{-3}\) Wb is linked with coil B. The self inductance of coil A is \(L_A\) and the self inductance of coil B is \(L_B\). The ratio \(L_A\) to \(L_B\) is

• A. \(\frac{1}{5}\)
• B. \(\frac{2}{5}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{5}{2}\)

Hint:

Self inductance L = N times flux per turn divided by current.

Answer 1

The Correct Option is B

Step 1: Compare factor by factor:
$\dfrac{L_A}{L_B} = \dfrac{N_A}{N_B}\cdot\dfrac{\phi_A}{\phi_B}$ since the currents are equal.

Step 2: Substitute:
$\dfrac{N_A}{N_B} = \dfrac{180}{360} = \dfrac12$ and $\dfrac{\phi_A}{\phi_B} = \dfrac{0.8}{1} = \dfrac45$.
$\dfrac{L_A}{L_B} = \dfrac12\times\dfrac45 = \dfrac25$.

Step 3: Check:
$L_A = 0.144$ H and $L_B = 0.36$ H, which have the ratio 0.4.

Final Answer:
Option (B). \[ \boxed{\frac{2}{5} \text{ (B)}} \]