Solveeit Logo
Login

Question

Physics Question on Magnetism and matter

A compass needle whose magnetic moment is 60 A m260\, A \,m^2 pointing geographical north at a certain place where the horizontal component of earth?s magnetic field is 40×10−6 Wb m−240 \times 10^{-6}\, Wb\, m^{-2} experiences a torque of 1.2×10−3 N m1.2 \times 10^{-3}\, N\, m. The declination of the place is

A

20∘20^{\circ}

B

45∘45^{\circ}

C

60∘60^{\circ}

D

30∘30^{\circ}

Answer

30∘30^{\circ}

Explanation

Solution

In stable equilibrium, a compass needle points along magnetic north and experiences no torque. When it is turned through declination α\alpha, it points along geographic north and experiences torque, τ=mB sinα\tau = mB \,sin\alpha ∴sin α=τmB\therefore\quad sin\,\alpha = \frac{\tau}{mB} =1.2×10−360×40×10−6= \frac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}} =12= \frac{1}{2} or α=30∘\alpha = 30^{\circ}