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:
Answer & explanation
Correct answer: option 2
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
$∴I=\frac{20×10^{-3}}{(4π×10^{-7})×2000}$
$≃7.99A$