Question

If circular coil with $N_1$ turns is changed in to a coil of $N_2$ turns. What will be the ratio of self inductances in both cases.

• A. $\frac{N_1}{N_2}$
• B. $\frac{N_2}{N_1}$
• C. $\frac{N_1^2}{N_2^2}$
• D. $\sqrt{\frac{N_1}{N_2}}$

Answer 1

The correct option is A $\frac{N_1}{N_2}$

If circular coil with N1 turns is changed in to a coil of N2 turns.

$L=\frac{Nd\varphi }{dt}$

$L_1=\frac{N_{1}d\varphi }{dt}$

$L_2=\frac{N_{2}d\varphi }{dt}$

L is in Henries

N is the Number of Turns

Φ is the Magnetic Flux

Ι is in Amperes

$\frac{L_1}{L_2}=\frac{N_1}{N_2}$