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## QuestionPhysicsClass 12

Consider a tightly wound  $$100$$  turn coil of radius $$10 \text{ cm}$$ carrying a current of  $$1\;A$$ . What is the magnitude of the magnetic field at the centre of the coil?

$$6.28\times 10^{-4}\;T$$
4.6
4.6

## Solution

Since the coil is tightly wound. We may take each circular element to have the same radius  $$R=10 \text{ cm} = 0.1\text{ m}$$
The number of turns  $$N=100$$
The magnitude of the magnetic field is.
$$B=\frac{\mu _o\mathit{NI}}{2R}$$
$$=\frac{4\Pi \times 10^{-7}\times 10^2\times 1}{2\times 10^{-1}}$$
$$=6.28\times 10^{-4}\;T$$