1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answers (1)
S Sayak

The magnitude of the magnetic field at the centre of a circular coil of radius r carrying current  I is given by,

|B|=\frac{\mu _{0}I}{2r}

For 100 turns, the magnitude of the magnetic field will be,

|B|=100\times \frac{\mu _{0}I}{2r}              

|B|=100\times \frac{4\pi \times 10^{-7}\times 0.4}{2\times 0.08}        (current=0.4A, radius = 0.08m, permeability of free space = 4\pi \times10-7 TmA-1)

=3.14\times10^{-4}T

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