In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

  • A
    In the direction of the magnetic dipole moment ( $\overrightarrow{M}$ ) of the magnet.
  • B
    In the opposite direction of the magnetic dipole moment ( $\overrightarrow{M}$ ) of the magnet.
  • C
    In the perpendicular direction of the magnetic moment ( $\overrightarrow{M}$ ) of the magnet.
  • D
    Its direction depends on the magnitude of the magnetic moment ( $\overrightarrow{M}$ ) of the magnet.

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$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

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