A compass needle whose magnetic moment is `60Am^2` pointing geograhic north at a certain place, where the horizontal component of earth's magnetic field is `40muWbm^-2` experiences a torque `1*2xx10^-3Nm`. What is the declination of the place?

Options

A. `20^(@)`
B. `45^(@)`
C. `60^(@)`
D. `30^(@)`

Answer

D

Explanation

In stable equilibrium, a compass needle points along magnetic north and experience no torque. When it is turned through declination `alpha`, it points along geographic north and experience torque.
`tau=mB sin alpha`
`therefore sin alpha=(tau)/(mB)=(1.2xx10^(-3))/(60xx40xx10^(-6))=(1)/(2)or alpha=30^(@)`