The magnitude of magnetic field inside a solenoid of length 0.3 m having 800 turns carrying a current of 6 A is
Answer & explanation
Correct answer: option 3
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 $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$
$B = 4\pi \times 10^{-7} \times \frac{800}{0.3} \times 6$
$B = 4\pi \times 10^{-7} \times 16000$
$B = 4\pi \times 1.6 \times 10^{-3}$
$B = 20.1 \times 10^{-3}\ \text{T}$
∴ $B = 2.0 \times 10^{-2}\ \text{T}$