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

If ` theta ` is the declination of the place , then the torque acting on the needle is ` tau = M B_H sin theta `
` implies sin theta = (tau)/(MB_H) = (1.2 xx 10^(-3))/(60 xx 40 xx 10^(-6)) =(1)/(2)`
`implies :. theta = 30^(@)`