The magnetic field at a point ‘P’ on the axial line of a short bar magnet is $5\mathrm{\mu T}$ ,due South. Now the magnet is turned through ${90}^0$ in anti clockwise direction. Then, the magnetic field at the same point ‘P’ would be

• A

  $2.5\mathrm{\mu T} \mathrm{due} \mathrm{East}$

• B

  $2.5\mathrm{\mu T} \mathrm{due} \mathrm{West}$

• C

  $5\mathrm{\mu T} \mathrm{due} \mathrm{North}$

• D

  $2.5\mathrm{\mu T} \mathrm{due} \mathrm{North}$