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

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `25^(@)`

Answer

A

Explanation

As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian,

so when it is pointing along the geographic meridian, it will experience a torque due to the horizontal component of earth's magnetic field i.e., `tau=MB_(H) sintheta`
where `theta=` angle between geographical and magnetic meridians called angle of declination
So, `sintheta=(1.2xx10^(-3))/(60xx40xx10^(-6))=1/2 implies theta=30^(@)`