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

Explanation

Here, `M=60Am^2`
`H=40mu Wb//m^2=40xx10^-6 Wb//m^2`,
`tau=1*2xx10^-3 N-m`, `theta=?`
Now, the compass needle is free to rotate in a horizontal plane, and points along the magnetic meridian. When the compass needle is pointing along the geographic meridian, it will experience a torque due to horizontal component of earth's magnetic field (H)
`C=MH sin theta` ...(i)
where `theta` is angle between the geographic and magnetic meridian. This angle is called declination at the place.

From (i) `sin theta=(C)/(MH)=(1*2xx10^-3)/(60xx40xx10^-6)=1/2`
`theta=sin^-1(1//2)=30^@`