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A torque of $1 \cdot 732 \times 10^{-5} \mathrm{Nm}$ is required to hold a magnet at $90^{\circ}$ with the
horizontal component of earth's magnetic field. The torque required to hold it at
$60^{\circ}$ will be $\left[\sin \frac{\pi}{2}=1, \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right][\sqrt{3}=1 \cdot 732]$
Options:
horizontal component of earth's magnetic field. The torque required to hold it at
$60^{\circ}$ will be $\left[\sin \frac{\pi}{2}=1, \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right][\sqrt{3}=1 \cdot 732]$
Solution:
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Verified Answer
The correct answer is:
$1 \cdot 5 \times 10^{-5} \mathrm{Nm}$
$\begin{array}{ll}\tau_{1}=1.732 \times 10^{-5} \mathrm{Nm} & \theta=90^{\circ} \\ \tau_{2}=? & \theta=60^{\circ}\end{array}$
$\tau_{1}=\mathrm{BM} \sin \theta=\mathrm{BM} \sin 90^{\circ}=\mathrm{BM}$
$\tau_{2}=\mathrm{BM} \sin 60^{\circ}=\mathrm{BM} \frac{\sqrt{3}}{2}=1.732 \times 10^{-5} \times \frac{1.732}{2}$
$=1.4999 \times 10^{-5}=1.5 \times 10^{-5} \mathrm{Nm}$
$\tau_{1}=\mathrm{BM} \sin \theta=\mathrm{BM} \sin 90^{\circ}=\mathrm{BM}$
$\tau_{2}=\mathrm{BM} \sin 60^{\circ}=\mathrm{BM} \frac{\sqrt{3}}{2}=1.732 \times 10^{-5} \times \frac{1.732}{2}$
$=1.4999 \times 10^{-5}=1.5 \times 10^{-5} \mathrm{Nm}$
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