A 25 cm long solenoid with 100 turns carries a current of 3 A. The fl…

The correct answer is Option (3) → $1.51 × 10^{-3}\, Wb/m^2$

Given:

Length of solenoid: $l = 25 \, \text{cm} = 0.25 \, \text{m}$

Number of turns: $N = 100$

Current: $I = 3 \, \text{A}$

Magnetic field inside a long solenoid:

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

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

$B = 4 \pi \times 10^{-7} \cdot \frac{100}{0.25} \cdot 3$

$\frac{100}{0.25} = 400$

$B = 4 \pi \times 10^{-7} \cdot 400 \cdot 3 = 4 \pi \times 10^{-7} \cdot 1200 = 4800 \pi \times 10^{-7}$

$B = 4.8 \pi \times 10^{-4} \, \text{T} \approx 1.507 \times 10^{-3} \, \text{T}$

Answer: $B \approx 1.51 \times 10^{-3} \, \text{T}$