The relation between $\mu$ and $H$ for a specimen of iron is $\mu=\left[\frac{1.4}{H}+12 \times 10^{-4}\right] \mathrm{Hm}^{-1}$. The value of $H$ which produces flux density of 1 T will be ( $\mu=$ magnetic permeability, $H=$ magnetic intensity)

At a place the horizontal component of earth's magnetic field $3 \times 10^{-5} \mathrm{~T}$ and the magnetic declination is $30^{\circ}$. A compass needle of magnetic moment $18 \mathrm{Am}^2$ pointing towards geographic north at this place experiences a torque of

If the vertical component of the earth's magnetic field is 0.45 G at a location and angle of dip is $60^{\circ}$, then magnetic field of earth in that location is

At a place where the magnitude of the earth's magnetic field is $4 \times 10^{-5} \mathrm{~T}$, a short bar magnet is placed with its axis perpendicular to the earth's magnetic field direction. If the resultant magnetic field at a point at a distance of 40 cm from the centre of the magnet on the normal bisector of the magnet is inclined at $45^{\circ}$ with the earth's field, then the magnele moment of the magnet is