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 $40 \times 10^{-6}\, Wb\, m^{-2}$ experiences a torque of $1.2 \times 10^{-3}\, N\, m$. 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,
$\tau = mB \,sin\alpha$
$\therefore \quad sin\,\alpha = \frac{\tau}{mB} $
$= \frac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}} $
$= \frac{1}{2}$ or $\alpha = 30^{\circ}$