Two co axial solenoids having number of turns per unit length 50 and 200, respectively have same length 50 cm. The area of cross-section of the two solenoids are $4\, cm^2$ and $6\, cm^2$, respectively. The mutual inductance of the two solenoids is
Answer & explanation
Correct answer: option 1
The correct answer is Option (3) → $8π × 10^{-7} H$
Given:
Number of turns per unit length: $n_1 = 50 \, \text{turns/cm} = 5000 \, \text{turns/m}$
$n_2 = 200 \, \text{turns/cm} = 20000 \, \text{turns/m}$
Length: $l = 50 \, cm = 0.5 \, m$
Areas: $A_1 = 4 \, cm^2 = 4 \times 10^{-4} \, m^2$, $A_2 = 6 \, cm^2 = 6 \times 10^{-4} \, m^2$
Mutual inductance formula for co-axial solenoids: $M = \mu_0 n_1 n_2 A l$
Use the smaller cross-sectional area as common area: $A = 4 \times 10^{-4} \, m^2$
$M = (4 \pi \times 10^{-7}) (5000) (20000) (4 \times 10^{-4}) (0.5)$
$M = 4 \pi \times 10^{-7} \times 5000 \times 20000 \times 2 \times 10^{-4}$
$M = 4 \pi \times 10^{-7} \times 20 \times 10^3$
$M = 8 \pi \times 10^{-3} \, H$
Answer: The mutual inductance of the solenoids is $8 \pi \times 10^{-3} \, H$.