Q. Two coils $A\ and\ B$ have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of $10^{- 3}$ Wb to link with $A$ and a flux per turn of 0.8$\times 10^{- 3}$ Wb through B. The ratio of mutual inductance of $A\ and\ B$ is

A) 0.625
B) 1.25
C) 1.5
D) 1.625
Explanation

(a) From $\phi = Mi$
$\frac{M_{1}}{M_{2}} = \frac{\phi_{1}}{\phi_{2}} = \frac{10^{- 3} \times 200}{0.8 \times 10^{- 3} \times 400} = \frac{10}{16} = 0.625$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.