Question

A compass needle whose magnetic moment is $60\, A m ^{2}$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $40\, \mu Wb / m ^{2}$, experiences a torque $1.2 \times 10^{-3} N m$. What is the declination at this place?

• A. $30^{\circ}$
• B. $45^{\circ}$
• C. $60^{\circ}$
• D. $25^{\circ}$

Answer 1

Answer: (A)

As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian,

so when it is pointing along the geographic meridian, it will experience a torque due to the horizontal component of earth's magnetic field,
i.e., $\tau=M B_{H} \sin \theta$ where $\theta=$ angle between geographical and magnetic meridians called angle of declination
So, $\sin \theta=\frac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}}$
$=\frac{1}{2} \Rightarrow \theta=30^{\circ}$