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

Related Questions:

1. Two cities are 150 km apart. Electric power is sent from one city to another city
2. A bicyclist comes to a skidding stop in 10 m.During this process, the force
3. Light from the constellation Virgo is observed to increase in wavelength by 0.4%
4. A drum of radius R and mass M, rolls down without slipping along an inclined plane
5. The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV

Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends