CUET Preparation Today
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? |
$1.44× 10^{-4} H$ $7.20 × 10^{-4} H$ $2.88 × 10^{-4} H$ $9.04 × 10^{-4} H$ |
$9.04 × 10^{-4} H$ |
The correct answer is Option (4) → $9.04 × 10^{-4} H$ The self inductance of an solenoid is - $L=μ_0\frac{N^2A}{l}$ $∴L=(4π×10^{-7})\frac{(600)^2(10×10^{-6})}{0.5}$ $=9.08×10^{-3}H$ |
