Question

A compass needle whose magnetic moment is 60 A m2 pointing geographical north at a certain place where the horizontal component of earths magnetic field is 40 x 10-6 Wb m-2 experiences a torque of 1.2 x 10-3 N m. The declination of the place is

• A. 20°
• B. 45°
• C. 60°
• D. 30°

Answer 1

Answer: (D)

30°

Answer 2

: 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,

$\tau = \mathrm { mB } \sin \alpha$

$\therefore \sin \alpha = \frac { \tau } { \mathrm { mB } } = \frac { 1.2 \times 10 ^ { - 3 } } { 60 \times 40 \times 10 ^ { - 6 } } = \frac { 1 } { 2 }$

or , $\alpha = 30 ^ { \circ }$