A tightly wound coil of 100 turns and radius 10 cm carries a current of 1 A. The magnitude of the magnetic field at the centre of the coil is:

1. Field at the centre of N turns: $B = \dfrac{\mu_0 N I}{2R}$. 2. Substitute $\mu_0 = 4\pi\times10^{-7}$, $N = 100$, $I = 1$, $R = 0.1$ m. 3. Numerator: $4\pi\times10^{-7}\times100\times1 = 4\pi\times10^{-5}$. 4. Divide by $2R = 0.2$: $B = \dfrac{4\pi\times10^{-5}}{0.2} = 2\pi\times10^{-4}$. 5. $B = 6.28\times10^{-4}$ T. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.147_