Comment
Share
Q)

# A compass needle whose magnetic moment is $60Am^2$ pointing geographic north at a certain place, Where the horizontal component of earth's magnetic field is $40 \mu wbm^{-2}$ experiences a torque $1.2 \times 10^{-3}\;Nm$ What is the declination of the place?

$\begin{array}{1 1} 60^{\circ} \\ 30^{\circ} \\ 15^{\circ} \\ 65^{\circ} \end{array}$

Comment
A)
Solution :
$M= 60 m^2$
$H= 40 \mu wbm^{-2}$
$\quad= 40 \times 10^{-6} wb/m^2$
$T= 1.2 \times 10^{-3} Nm$
$T= MH \sin \theta$
$\sin \theta= \large\frac{T}{MH}$
$\qquad= \large\frac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}}$
$\qquad= \large\frac{1}{2}$
$\sin \theta = \large\frac{1}{2}$
$\theta= 30^{\circ}$
Answer : $30^{\circ}$