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A coil of 100 turns and area 5 square centimetre is placed in a magnetic field $B=0.2 T$. The normal to the plane of the coil makes an angle of $60^{\circ}$ with the direction of the magnetic field. The magnetic flux linked with the coil is
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The correct answer is:
${ }^{5 \times 10^{-3} \mathrm{~Wb}}$
$\phi=N B A \cos \theta=100 \times 0.2 \times 5 \times 10^{-4} \cos 60^{\circ}$
$=5 \times 10^{-3} \mathrm{~Wb}$
$=5 \times 10^{-3} \mathrm{~Wb}$
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