Question

A circular coil of 100 turns and radius 10 cm carries a current of 3.2 A. The magnetic field at the centre of the coil is:

पर्याय

• \[4 \times 10^{-3}\,\text{T}\]

• \[2 \times 10^{-4}\,\text{T}\]

• \[2 \times 10^{-2}\,\text{T}\]

• \[2 \times 10^{-3}\,\text{T}\]

Answer 1

The field at the centre of a circular coil is \[B_{centre} = \frac{\mu_0 NI}{2R} = \frac{4\pi \times 10^{-7} \times 100 \times 3.2}{2 \times 0.1} = 2 \times 10^{-3}\,\text{T}\].