| ⇦ |
| ⇨ |
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: - The ratio of longest wavelength corresponding to Lyman and Blamer series
- When sound is produced in an aeroplane with a velocity of 200 m/s horizontally,
- The interferance pattern is obtained with two coherent light sources of intensity
- When we hear a sound, we can identify its source from
- A spring of spring constant 5 x 10³ N/m is stretched initially by 5 cm
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The ratio of longest wavelength corresponding to Lyman and Blamer series
- When sound is produced in an aeroplane with a velocity of 200 m/s horizontally,
- The interferance pattern is obtained with two coherent light sources of intensity
- When we hear a sound, we can identify its source from
- A spring of spring constant 5 x 10³ N/m is stretched initially by 5 cm
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