Question

A compass needle whose magnetic moment is $60A{m}^2$ pointing geographical north at a certain place where the horizontal component of earth’s magnetic field is $40\times 10^{-6}Wb{m}^{-2}$ experiences a torque of $1.2\times 10^{-3}Nm$. The declination of the place is

• A. $20^{\circ }$
• B. $45^{\circ }$
• C. $60^{\circ }$
• D. $30^{\circ }$

Answer 1

Answer: (D)

In stable equilibrium, a compass needle points along magnetic north and experiences no torque.
When it is turned through declination $\alpha$, it points along geographic north and experiences torque,
$\tau =mBsin\alpha$
$\therefore sin\alpha =\frac{\tau }{mB}$
$=\frac{1.2\times 10^{-3}}{60\times 40\times 10^{-6}}$
$=\frac{1}{2}$ or $\alpha =30^{\circ }$