Q 12-04-007NEETNEET 2024Top questionEasy
A tightly wound $100$ turns coil of radius $10$ cm carries a current of $7$ A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4\pi \times 10^{-7}$ SI units):
Answer: (A) $4.4$ mT
$$B = \frac{\mu_0 NI}{2R} = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1} = 4.4 \times 10^{-3}\ \text{T} = 4.4\ \text{mT}$$
Solution by Sreeraj P, M.Sc Physics