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A wire of length $l$ is bent into a circular coil of one turn of radius $R_1$. Another wire of the same material and same area of cross-section and same lengths is bent into a circular coil of two turns of radius $R_2$. When the same current flows, through the two coils, the ratio of magnetic induction at the centres of the two coils is
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The correct answer is:
$1:4$
Magnetic induction at the centre of a circular coil
$$
\begin{array}{rlrl}
& B =\frac{\mu_0}{2} \cdot \frac{n i}{R} \Rightarrow B \propto \frac{n}{R} \\
\text { Here, } & n =1, n_2=2, \\
\Rightarrow & l =2 \pi R_1=2 \times 2 \pi R_2 \\
& R_2 =\frac{R_1}{2} \\
& \quad \frac{B_1}{B_2} =\frac{n_1}{n_2} \times \frac{R_2}{R_1} \\
& = \frac{1}{2} \times \frac{R_1 / 2}{r}=\frac{1}{4} \\
& \quad B_1: B_2 =1: 4
\end{array}
$$
$$
\begin{array}{rlrl}
& B =\frac{\mu_0}{2} \cdot \frac{n i}{R} \Rightarrow B \propto \frac{n}{R} \\
\text { Here, } & n =1, n_2=2, \\
\Rightarrow & l =2 \pi R_1=2 \times 2 \pi R_2 \\
& R_2 =\frac{R_1}{2} \\
& \quad \frac{B_1}{B_2} =\frac{n_1}{n_2} \times \frac{R_2}{R_1} \\
& = \frac{1}{2} \times \frac{R_1 / 2}{r}=\frac{1}{4} \\
& \quad B_1: B_2 =1: 4
\end{array}
$$
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