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