⇦ | ![]() | ⇨ |
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:
- Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV.
- A ballon rises from rest with a constant acceleration g/8. A stone is released
- An aicraft executes a horizontal loop of radius 1 km with a speed of 900 kmh⁻¹
- In common base mode of a transistor, the collector current is 5.488 mA
- A charge Q is uniformly distributed over a large plastic plate. The electric field
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
Leave a Reply