A galvanometer coil has $500$ turns and each turn has an average area of $3 \times 10^{-4}\, m ^{2}$. If a torque of $1.5\,Nm$ is required to keep this coil parallel to magnetic field when a current of $0.5\, A$ is flowing through it, the strength of the field (in $T )$ is
JEE MAIN 2020, Medium
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$\vec{\tau}=\vec{m} \times \vec{B}$
$\tau=N I \times A \times B$
$105=500 \times 3 \times 10^{-4} \times \frac{1}{2} \times B$
$B=20$
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