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A circular coil of radius $40 \mathrm{~mm}$ consists of 250 turns of wire in which the current is $20 \mathrm{~mA}$. The magnetic field in the center of the coil is $\left[\mu=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right]$
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Verified Answer
The correct answer is:
$0.785 \mathrm{G}$
In a circular coil of $n$ turns, magnetic field is
$\mathrm{B}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 250 \times 20 \times 10^{-3}}{2 \times 40 \times 10^{-3}}$
$(\because \mathrm{n}=$ no. of turns, $\mathrm{I}=$ current through coil, $\mathrm{r}=$ radius of coil)
$\begin{aligned}
\Rightarrow B &=\frac{4 \pi \times 250 \times 20 \times 10^{-7-3+3}}{2 \times 40} \\
&=250 \times 3.14 \times 10^{-7} \\
&=785 \times 10^{-7}=0.785 \times 10^{-4} \text { tesla } \\
&=0.785 \text { gauss }
\end{aligned}$
$\mathrm{B}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 250 \times 20 \times 10^{-3}}{2 \times 40 \times 10^{-3}}$
$(\because \mathrm{n}=$ no. of turns, $\mathrm{I}=$ current through coil, $\mathrm{r}=$ radius of coil)
$\begin{aligned}
\Rightarrow B &=\frac{4 \pi \times 250 \times 20 \times 10^{-7-3+3}}{2 \times 40} \\
&=250 \times 3.14 \times 10^{-7} \\
&=785 \times 10^{-7}=0.785 \times 10^{-4} \text { tesla } \\
&=0.785 \text { gauss }
\end{aligned}$
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