Q: A compass needle of magnetic moment 60 A-m^{2}, pointing towards geographical north at a certain place where the horizontal component of earth’s magnetic field is 40 μwb/m^{2} experience a torque of 1.2 × 10^{-3} Nm. Find the declination at that place.

Sol: If θ is the declination of the place, then the torque acting on the needle is

τ = M B_{H} sin θ

sin θ = τ/MB_{H}

sin θ = (1.2 ×10^{-3})/(60 × 40×10^{-6})

sin θ = 1/2 ∴ θ = 30°