If the dipole moment of a short bar magnet is 1.25 A-m², the magnetic field on its

If the dipole moment of a short bar magnet is 1.25 A-m², the magnetic field on its axis at a distance of 0.5 m from the centre of the magnet is

Options

(a) 1×10⁻⁴ NA⁻¹ m⁻¹
(b) 2×10⁻⁶NA⁻¹ m⁻¹
(c) 4×10⁻²NA⁻¹ m⁻¹
(d) 6.64×10⁻⁸NA⁻¹ m⁻¹

Correct Answer:

2×10⁻⁶NA⁻¹ m⁻¹

Explanation:

The magnetic field at a point on the axis of a bar magnet is, B = 2μ₀M / 4πr³

= 10⁻⁷ × [(2×1.25) / (0.5)³] = 2 x 10⁻⁶ NA⁻¹ m⁻¹

admin:

Related Questions

  1. Two point charge +9e and +e are at 16 cm away from each other.
  2. In Bohr model of hydrogen atom, the force on the electron depends on the principal
  3. The velocity of a particle performing simple harmonic motion, when it passes
  4. A metal conductor of length 1m rotates vertically about one of its ends at angular
  5. The wave function (in SI unit) for a light wave is given as Ψ(x,t)= 10³ sin π