Due to $10\, ampere$ of current flowing in a circular coil of $10\, cm$ radius, the magnetic field produced at its centre is $3.14 \times {10^{ - 3}}\,Weber/{m^2}$. The number of turns in the coil will be
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(c) $B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2\pi Ni}}{r}$
$==>$ $3.14 \times {10^{ - 3}} = \frac{{{{10}^{ - 7}} \times 2 \times 3.14 \times N \times 10}}{{(10 \times {{10}^{ - 2}})}}$ $==>$ $N = 50$
$==>$ $3.14 \times {10^{ - 3}} = \frac{{{{10}^{ - 7}} \times 2 \times 3.14 \times N \times 10}}{{(10 \times {{10}^{ - 2}})}}$ $==>$ $N = 50$
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