What is the self inductance of a solenoid of length 50 cm, if area of cross section is $10 cm^2$ and total number of turns is 600?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $9.04 × 10^{-4} H$
The self inductance of an solenoid is -
$L=μ_0\frac{N^2A}{l}$
$∴L=(4π×10^{-7})\frac{(600)^2(10×10^{-6})}{0.5}$
$=9.08×10^{-3}H$