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A circular coil of wire consisting of 100 turns, each of radius $8.0 \mathrm{~cm}$ carries a cur rent of $0.40 \mathrm{~A}$. What is the magnitude of the magnetic field $B$ at the centre of the coil?
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Verified Answer
Given, $\mathrm{N}=100, \mathrm{r}=8 \mathrm{~cm}=0.08 \mathrm{~m}$
$$
\mathrm{I}=0.4 \mathrm{~A}, \mathrm{~B}=\text { ? }
$$
As we know, $B=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{r}}$
$$
=\frac{4 \pi \times 10^{-7} \times 100 \times 0.4}{2 \times 0.08}=\pi \times 10^{-4} \mathrm{~T}
$$
$$
\mathrm{I}=0.4 \mathrm{~A}, \mathrm{~B}=\text { ? }
$$
As we know, $B=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{r}}$
$$
=\frac{4 \pi \times 10^{-7} \times 100 \times 0.4}{2 \times 0.08}=\pi \times 10^{-4} \mathrm{~T}
$$
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