A torque of 1.732 times 10^{-5} text{ Nm} is | vedclass

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(A) The torque $	au$ acting on a magnetic dipole in a magnetic field is given by $	au = MB \sin 	heta$. Given,$	au_1 = 1.732 	imes 10^{-5} 	ext{

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$1.5 	imes 10^{-5} 	ext{ Nm}$

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(A) The torque $	au$ acting on a magnetic dipole in a magnetic field is given by $	au = MB \sin 	heta$. Given,$	au_1 = 1.732 	imes 10^{-5} 	ext{ Nm}$ at $	heta_1 = 90^{\circ}$. So,$	au_1 = MB \sin 90^{\circ} = MB(1) = MB$. Therefore,$MB = 1.732 	imes 10^{-5} 	ext{ Nm}$. Now,for $	heta_2 = 60^{\circ}$,the torque $	au_2$ is: $	au_2 = MB \sin 60^{\circ} = (1.732 	imes 10^{-5}) 	imes \frac{\sqrt{3}}{2}$. Since $\sqrt{3} = 1.732$,we have: $	au_2 = 1.732 	imes 10^{-5} 	imes \frac{1.732}{2} = 1.732 	imes 10^{-5} 	imes 0.866 = 1.4999 	imes 10^{-5} \approx 1.5 	imes 10^{-5} 	ext{ Nm}$.

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

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