​Consider a solenoid of 5.0 cm length and radius 0.40 cm. It consists of 500 turns of wire and carries a current of 3.0 A. The magnitude of magnetic field at the centre of the solenoid is close to [($$\frac{μ_o}{4π})=10^{-7}$$​ Tm/A]:​

Correct option is A

​$$\text{Given:} \\\quad \text{Length of the solenoid, } l = 5.0 \, \text{cm} = 0.05 \, \text{m} \\\quad \text{Radius of the solenoid, } r = 0.40 \, \text{cm} = 4 \times 10^{-3} \, \text{m} \\\quad \text{Number of turns, } N = 500 \\\quad \text{Current, } I = 3.0 \, \text{A} \\\\\\\text{Solution:} \\\quad 1. \, \text{Magnetic field at the center of the solenoid:} \\\quad \text{The magnetic field in a solenoid is given by the formula:} \\B = \mu_0 \cdot n \cdot i \\\quad \text{where:} \\\quad \mu_0 = 4\pi \times 10^{-7} \, \text{T.m/A} \, (\text{permeability of free space}), \\\quad n = \frac{N}{l} \, \text{is the number of turns per unit length}, \\\quad i = \text{the current}.$$

$$\text{2. Calculate } n: \\n = \frac{N}{l} = \frac{500}{0.05} = 10000 \, \text{turns/m} \\\\\text{3. Substitute into the formula for } B: \\B = 4\pi \times 10^{-7} \times 10000 \times 3 \\\\\text{4. Simplify:} \\B = 4 \times \pi \times 10^{-7} \times 10000 \times 3 = 37.68 \, \text{mT}$$​​​