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. ${20}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${30}^o$

Answer 1

The correct option is D ${30}^o$

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