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A wire of length $l$ carries a steady current. It is bent first to form a circular plane loop of one turn. The magnetic field at the centre of the loop is $B$. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same is
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Verified Answer
The correct answer is:
$4 B$
The wire of length $l$ is bent to from a circular loop, so $2 \pi r=l$
$$
\Rightarrow r=\frac{l}{2 \pi}
$$
The magnetic field at the centre of the loop is
$$
B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 I \times 2 \pi}{2 l}
$$
Now the same length of the wire is bent to from a double loop
$$
\begin{aligned}
& \therefore 2 \times 2 \pi r^{\prime}=l \\
& \Rightarrow r^{\prime}=\frac{l}{4 \pi}
\end{aligned}
$$
And the magnetic field at the centre
$$
\begin{aligned}
& B^{\prime}=\frac{\mu_0 I \times 2}{2 \times r^{\prime}}=\frac{2 \mu_0 I}{2 \times \frac{l}{4 \pi}}=\frac{2 \times 4 \pi \mu I}{2 l} \\
& \therefore \frac{B^{\prime}}{B}=4 \Rightarrow B^{\prime}=4 B
\end{aligned}
$$
$$
\Rightarrow r=\frac{l}{2 \pi}
$$
The magnetic field at the centre of the loop is
$$
B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 I \times 2 \pi}{2 l}
$$
Now the same length of the wire is bent to from a double loop
$$
\begin{aligned}
& \therefore 2 \times 2 \pi r^{\prime}=l \\
& \Rightarrow r^{\prime}=\frac{l}{4 \pi}
\end{aligned}
$$
And the magnetic field at the centre
$$
\begin{aligned}
& B^{\prime}=\frac{\mu_0 I \times 2}{2 \times r^{\prime}}=\frac{2 \mu_0 I}{2 \times \frac{l}{4 \pi}}=\frac{2 \times 4 \pi \mu I}{2 l} \\
& \therefore \frac{B^{\prime}}{B}=4 \Rightarrow B^{\prime}=4 B
\end{aligned}
$$
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