Two long coaxial solenoids of the same length \( L \), have inner and outer coils with diameters \( D_1 \) and \( D_2 \) and number of turns \( N_1 \) and \( N_2 \), respectively. What will be the ratio of the self-inductance of the inner solenoid to the mutual inductance of both the solenoids?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( N_1 N_2^{-1} \)
Let \( A_1 \) be the cross-section of the inner solenoid. The self-inductance of the inner solenoid is \( L_{11} = \mu_0 N_1^{2}A_1/L \).
For the mutual inductance, the field of the inner solenoid, \( B_1 = \mu_0 N_1 I_1/L \), exists only inside it, so the flux linked with each turn of the outer solenoid is \( B_1 A_1 \) (the outer solenoid is larger, but the field is zero outside the inner one).
\( \displaystyle M = \frac{N_2\,B_1 A_1}{I_1} = \frac{\mu_0 N_1 N_2 A_1}{L} \)
\( \displaystyle \frac{L_{11}}{M} = \frac{\mu_0 N_1^{2}A_1/L}{\mu_0 N_1 N_2 A_1/L} = \frac{N_1}{N_2} \)