The magnitude of magnetic field inside a solenoid of length 0.3 m having 800 turns carrying a current of 6 A is

20 mT

The correct answer is Option (3) → 20 mT

Given:

Number of turns, $N = 800$

Length of solenoid, $l = 0.3\ \text{m}$

Current, $I = 6\ \text{A}$

Magnetic field inside a solenoid is given by:

$B = \mu_0 \frac{N}{l} I$

Substitute $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$

$B = 4\pi \times 10^{-7} \times \frac{800}{0.3} \times 6$

$B = 4\pi \times 10^{-7} \times 16000$

$B = 4\pi \times 1.6 \times 10^{-3}$

$B = 20.1 \times 10^{-3}\ \text{T}$

∴ $B = 2.0 \times 10^{-2}\ \text{T}$