A solenoid of 2.5 m length and 2.0 cm diameter possesses 10 turns per cm. A current of 0.5 A is flowing through it. The magnetic induction at axis inside the solenoid is

Magnetic induction of solenoid along the axis is given by,

$B={\mu }_{0}nI$

n is the number of turns and I is the current.

$$\begin{gathered} \frac{{\mu }_0}{4\pi }={10}^{-7}T/A \\ \Rightarrow {\mu }_0=4\pi \times {10}^{-7}T/A \end{gathered}$$

$n=10c{m}^{-1}=1000m^{-1}$

$I=0.5A$

$\therefore B=4\pi \times {10}^{-7}\times 1000\times 0.5$

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

Hence the magnetic induction is $2\pi \times {10}^{-4}T$ .

Option A is the correct answer.