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A current passing through a circular coil of two turns produces a magnetic field $B$ as its centre. The coil is then rewound so as to have four turns and the same current is passed through it. The magnetic field at its centre now is
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Verified Answer
The correct answer is:
$2 B$
According to questions
$$
\begin{array}{l}
2 \times 2 \pi_{1}=4 \times 2 \pi r_{2} \\
r_{1}=2 r_{2} \\
\text { and } \quad B_{1}=\frac{\mu_{0} 2 \pi I}{4 \pi r_{1}} \\
B_{2}=\frac{\mu_{0} 2 \pi I}{4 \pi r_{2}}=\frac{\mu_{0} 2 \pi I}{4 \pi r_{1} / 2} \\
B_{2}=2 B_{1}
\end{array}
$$
$$
\begin{array}{l}
2 \times 2 \pi_{1}=4 \times 2 \pi r_{2} \\
r_{1}=2 r_{2} \\
\text { and } \quad B_{1}=\frac{\mu_{0} 2 \pi I}{4 \pi r_{1}} \\
B_{2}=\frac{\mu_{0} 2 \pi I}{4 \pi r_{2}}=\frac{\mu_{0} 2 \pi I}{4 \pi r_{1} / 2} \\
B_{2}=2 B_{1}
\end{array}
$$
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