Question
A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of $3.14 \times 10^{- 3}\text{ }T$ at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively:
(Take $\mu_{0} = 4\pi \times 10^{- 7}\text{ }T\text{ }m/A$ )
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Solution
Sol. [Key Concept: Magnetic field at centre of a circular coil] Magnetic field of a circular loop:
$${\begin{matrix} & B_{0} = \frac{\mu_{0}Ni}{2R} \\ & i = \frac{2RB_{0}}{\mu_{0}N} = \frac{2 \times 5 \times 10^{- 2} \times 3.14 \times 10^{- 3}}{4\pi \times 10^{- 7} \times 100} \\ & i = 2.5\text{ }A \end{matrix} } {Magnetic\ moment } \begin{matrix} & M = NiA \\ & \ = 100 \times 2.5 \times 3.14 \times \left( 5 \times 10^{- 2} \right)^{2}\left\lbrack \because A = \pi R^{2} \right\rbrack \\ & \ = 2\text{ }A{\text{ }m}^{2} \end{matrix}$$
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