Ncert Exercises & Solutions

4.1 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?

NEETprep Answer:

Number of turns on the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.8 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given by the relation,

|\vec{B}| = \frac{μ_{0}}{2 π} \frac{2 πnl}{r}

where, μ_{0} = Permeability of free space = 4 π \times 10^{- 7} T m A^{- 1}

So,

|\vec{B}| = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{2 π \times 100 \times 0.4}{r} = 3.14 \times 10^{- 4} T

Hence, the magnitude of the magnetic field is

3.14 \times 10^{- 4} T

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