A long solenoid is formed by winding insulated copper wire at the rate of 20 turns per cm. The current that is necessary to produce a magnetic field of 20 mT inside the solenoid at its centre would be:

8.0 A

The correct answer is Option (2) → 8.0 A

The magnetic field inside a solenoid,

$B=μ_0nI$

$⇒I=\frac{B}{μ_0n}$

and,

The solenoid has 20 turns per cm,

⇒ n = 20 turns/cm = 20 × 100 turns/m = 2000 turns/m

$∴I=\frac{20×10^{-3}}{(4π×10^{-7})×2000}$

$≃7.99A$