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. ${20}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${30}^0$

Answer 1

Answer: (D)

${30}^0$

In stable equilibrium, a compass needle points along the magnetic north and experiences no torque.
When it is turned through declination $\alpha$, it points along geographic north and experiences torque,
$T=mBsin\alpha$
$\therefore sin\alpha =\frac{T}{mB}=\frac{1.2\times {10}^{-3}}{60\times 40\times {10}^{-6}}=\frac{1}{2}$ or $\alpha ={30}^0$