A straight wire of length $({\pi ^2})$ $metre$ is carrying a current of $2\,A$ and the magnetic field due to it is measured at a point distant $1\, cm$ from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be
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(b) If a wire of length $l$ is bent in the form of a circle of radius $r$ then $2\pi r = l$ $==>$ $r = \frac{l}{{2\pi }}$
$\vec F = i(\vec L \times \vec B)$
Magnetic field due to straight wire ${B_1} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{r} = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2 \times 2}}{{1 \times {{10}^{ - 2}}}}$
$\vec F = i(\vec L \times \vec B)$
Magnetic field due to straight wire ${B_1} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{r} = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2 \times 2}}{{1 \times {{10}^{ - 2}}}}$
also magnetic field due to circular loop ${B_2} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2\pi i}}{r} = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2\pi \times 2}}{{\pi /2}}$
$==>$ $\frac{{{B_2}}}{{{B_1}}} = \frac{1}{{50}}$
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