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Home » 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?
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?
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
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?

The number of turns on the coil (n) is given as 100

The radius of each turn (r) is given as 8 \mathrm{~cm} or 0.08 \mathrm{~m}

The magnitude of the current flowing in the coil (I) is given as 0.4 \mathrm{~A}

The magnitude of the magnetic field at the centre of the coil can be calculated by the following relation:

|\bar{B}|=\frac{\mu_{0} 2 \pi n I}{4 \pi r}

where \mu_{0} is the permeability of free space equal to 4 \pi \times 10^{-7} T m A^{-1}

hence,

|\bar{B}|=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \pi \times 100 \times 0.4}{0.08}

=3.14 \times 10^{-4} T

The magnitude of the magnetic field is 3.14 \times 10^{-4} T.

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