A solenoid of length 30 cm and 3000 turns carries a current of 10 A. The value of the magnetic field ($\vec B$) at the centre of solenoid is:

(Given $μ_0 = 4π × 10^{-7} Wb/A-m$)

$0.13\, Wb/m^2$

The correct answer is Option (1) → $0.13\, Wb/m^2$

Magnetic field inside a solenoid: $B = \mu_0 n I$

Length of solenoid, $l = 30 \, \text{cm} = 0.30 \, \text{m}$

Number of turns, $N = 3000$

Turns per unit length, $n = \frac{N}{l} = \frac{3000}{0.30} = 10000 \, \text{turns/m}$

Current, $I = 10 \, \text{A}$

Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}$

$B = \mu_0 n I = (4\pi \times 10^{-7})(10000)(10)$

$B = 4\pi \times 10^{-7} \times 10^5$

$B = 4\pi \times 10^{-2} \, \text{T}$

$B \approx 0.126 \, \text{T}$

Answer: $B = 0.13 \, \text{T}$