A solenoid of length 0.5 m and radius 1 cm has 500 turns carrying a current of 5 A. The magnitude of the field inside it is:

1. Turns per unit length: $n = \dfrac{500}{0.5} = 1000$ m⁻¹. 2. Inside field: $B = \mu_0 n I$. 3. Substitute $\mu_0 = 4\pi\times10^{-7}$, $n = 10^3$, $I = 5$. 4. $B = 4\pi\times10^{-7}\times10^{3}\times5 = 2\pi\times10^{-3}$. 5. $B = 6.28\times10^{-3}$ T. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.153_