A compass needle whose magnetic moment is `60Am^(2)` pointing geographical north at a certain place where the horizontal component of earth's magnetic field is `40 xx 10^(-6) Wbm^(-2)` experiences a torque of `1.2 xx 10^(-3)Nm`. The declination of the place is

Explanation

The horizontal component `(B_(B))` of earth.s field points towards geomagnetic north pole (GEN).
While the magnet (say magnetic moment M) is pointing towards geographical north (GN). The angle between `B_(B)` and M is angle of declination `theta`.
Due to `B_(H)`, magnet will experience a torque.`tau = M B_(B) sin theta`
`B_(B) = B cos delta = 80 xx 10^(-6) xx cos 60^(@)`
`1.2xx10^(-3)=60xx40xx10^(-6)sinq`
Solving we get `sin theta= 1//2`
`theta=30^(@)`